Tutors Answer Your Questions about Vectors (FREE)
Question 1159986: A plane is steering at S65°W at an air speed of 625 km/h. The wind is from the NW at 130 km/h. Find the ground speed and track of the plane. Include a vector diagram in your solution.
Answer by ikleyn(52777)   (Show Source):
Question 1160091: 1. A plane is steering at S65°W at an air speed of 625 km/h. The wind is from the NW at 130 km/h. Find the ground speed and track of the plane. Include a vector diagram in your solution.
Answer by ikleyn(52777) (Show Source):
Question 1159991: A plane is steering at S65°W at an air speed of 625 km/h. The wind is from the NW at 130 km/h. Find the ground speed and track of the plane. Include a vector diagram in your solution.
Answer by ikleyn(52777) (Show Source):
Question 1165416: Find vector equation for the line with the following cartesian equation
(a)
X+3y=7
(b)
2x-5y=3
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
(a) vector equation is (x,y) = (7-3t, t), where 't' is the parameter, which may have any real value.
We take 'y' as the parameter 't', and express x via y from the equation,
replacing there 'y' with 't'.
(b) vector equation is (x,y) = ( ,  ), where 't' is the parameter, which may have any real value.
We take 'x' as the parameter 't', and express y via x from the equation,
replacing there 'x' with 't'.
Solved, with explanations.
Question 1167891: A swimmer is capable of swimming at 1.4m/s in still water
(A) how far down the stream will he land y he swims directly across a 180m wide river
(B) how long will it take him to reach the other side
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
A swimmer is capable of swimming at 1.4m/s in still water
(A) how far down the stream will he land y he swims directly across a 180m wide river
(B) how long will it take him to reach the other side
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Some key input information is missed/lost in this post.
Question 1199547: A point travels as described by the following parametric equations x=10t+10cos3.14t, y=20t+10sin3.14t and z=30t, where x,y,z, are in meters, t in seconds, all angles in radians. The vector locating the body at any time is r=ix+jy+kz. Determine the magnitude of the velocity of the body in meters per second at time t = 0.75 second.
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(52777) (Show Source):
Answer by CPhill(1959)  (Show Source):
You can put this solution on YOUR website! To determine the magnitude of the velocity of the body at \( t = 0.75 \) seconds, we follow these steps:
---
### **Step 1: Velocity vector**
The velocity vector \( \mathbf{v} \) is the time derivative of the position vector \( \mathbf{r} \):
\[
\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}
\]
\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{dx(t)}{dt} \mathbf{i} + \frac{dy(t)}{dt} \mathbf{j} + \frac{dz(t)}{dt} \mathbf{k}
\]
---
### **Step 2: Compute derivatives**
The given parametric equations are:
\[
x = 10t + 10\cos(\pi t), \quad y = 20t + 10\sin(\pi t), \quad z = 30t
\]
1. Derivative of \( x \):
\[
\frac{dx}{dt} = 10 - 10\pi \sin(\pi t)
\]
2. Derivative of \( y \):
\[
\frac{dy}{dt} = 20 + 10\pi \cos(\pi t)
\]
3. Derivative of \( z \):
\[
\frac{dz}{dt} = 30
\]
Thus:
\[
\mathbf{v}(t) = \left( 10 - 10\pi \sin(\pi t) \right) \mathbf{i} + \left( 20 + 10\pi \cos(\pi t) \right) \mathbf{j} + 30 \mathbf{k}
\]
---
### **Step 3: Compute magnitude of velocity**
The magnitude of \( \mathbf{v} \) is:
\[
|\mathbf{v}| = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2}
\]
At \( t = 0.75 \):
1. Compute \( \sin(\pi t) \) and \( \cos(\pi t) \):
\[
t = 0.75 \quad \Rightarrow \quad \pi t = 0.75\pi
\]
\[
\sin(0.75\pi) = \cos(0.25\pi) = \frac{\sqrt{2}}{2}, \quad \cos(0.75\pi) = -\sin(0.25\pi) = -\frac{\sqrt{2}}{2}
\]
2. Substitute into derivatives:
\[
\frac{dx}{dt} = 10 - 10\pi \cdot \frac{\sqrt{2}}{2} = 10 - 5\pi\sqrt{2}
\]
\[
\frac{dy}{dt} = 20 + 10\pi \cdot \left(-\frac{\sqrt{2}}{2}\right) = 20 - 5\pi\sqrt{2}
\]
\[
\frac{dz}{dt} = 30
\]
3. Compute magnitude:
\[
|\mathbf{v}| = \sqrt{\left( 10 - 5\pi\sqrt{2} \right)^2 + \left( 20 - 5\pi\sqrt{2} \right)^2 + 30^2}
\]
Expand each term:
\[
\left( 10 - 5\pi\sqrt{2} \right)^2 = 100 - 100\pi\sqrt{2} + 50\pi^2
\]
\[
\left( 20 - 5\pi\sqrt{2} \right)^2 = 400 - 200\pi\sqrt{2} + 50\pi^2
\]
\[
30^2 = 900
\]
Add them together:
\[
|\mathbf{v}|^2 = \left( 100 + 400 + 900 \right) - \left( 100\pi\sqrt{2} + 200\pi\sqrt{2} \right) + \left( 50\pi^2 + 50\pi^2 \right)
\]
\[
|\mathbf{v}|^2 = 1400 - 300\pi\sqrt{2} + 100\pi^2
\]
Finally:
\[
|\mathbf{v}| = \sqrt{1400 - 300\pi\sqrt{2} + 100\pi^2}
\]
---
### **Step 4: Numerical calculation**
Using \( \pi \approx 3.1416 \):
\[
|\mathbf{v}| \approx \sqrt{1400 - 300(3.1416)\sqrt{2} + 100(3.1416)^2}
\]
\[
|\mathbf{v}| \approx \sqrt{1400 - 1331.29 + 986.96} \approx \sqrt{1055.67} \approx 32.50 \, \text{m/s}
\]
---
### **Final Answer**:
The magnitude of the velocity at \( t = 0.75 \) seconds is approximately:
\[
\boxed{32.50 \, \text{m/s}}
\]
Question 1170575: Find the vector equation for the line of intersection of the planes 4x-y+2z=0 and 4x+z=3
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let the given planes be
$$P_1: 4x - y + 2z = 0$$
$$P_2: 4x + z = 3$$
We want to find the vector equation of the line of intersection of these two planes.
**1. Find a point on the line:**
Let's set $z=0$ in the equation for plane $P_2$:
$$4x + 0 = 3 \implies x = \frac{3}{4}$$
Now, substitute $x = \frac{3}{4}$ and $z=0$ into the equation for plane $P_1$:
$$4\left(\frac{3}{4}\right) - y + 2(0) = 0 \implies 3 - y = 0 \implies y = 3$$
So, the point $\left(\frac{3}{4}, 3, 0\right)$ lies on the line of intersection.
**2. Find the direction vector of the line:**
The direction vector of the line is perpendicular to the normal vectors of both planes.
The normal vector of $P_1$ is $\vec{n_1} = \langle 4, -1, 2 \rangle$.
The normal vector of $P_2$ is $\vec{n_2} = \langle 4, 0, 1 \rangle$.
The direction vector $\vec{d}$ is the cross product of $\vec{n_1}$ and $\vec{n_2}$:
$$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 2 \\ 4 & 0 & 1 \end{vmatrix}$$
$$\vec{d} = \mathbf{i}(-1(1) - 2(0)) - \mathbf{j}(4(1) - 2(4)) + \mathbf{k}(4(0) - (-1)(4))$$
$$\vec{d} = \mathbf{i}(-1) - \mathbf{j}(4 - 8) + \mathbf{k}(4)$$
$$\vec{d} = \langle -1, 4, 4 \rangle$$
**3. Write the vector equation of the line:**
The vector equation of a line is given by $\vec{r} = \vec{a} + t\vec{d}$, where $\vec{a}$ is a point on the line and $\vec{d}$ is the direction vector.
Using the point $\left(\frac{3}{4}, 3, 0\right)$ as $\vec{a}$ and $\langle -1, 4, 4 \rangle$ as $\vec{d}$, we have:
$$\vec{r} = \left\langle \frac{3}{4}, 3, 0 \right\rangle + t\langle -1, 4, 4 \rangle$$
In component form:
$$\vec{r} = \left\langle \frac{3}{4} - t, 3 + 4t, 4t \right\rangle$$
**Therefore, the vector equation of the line of intersection is:**
$$\vec{r} = \left\langle \frac{3}{4}, 3, 0 \right\rangle + t\langle -1, 4, 4 \rangle$$
Question 1178892: What is the correct way to write the following function:
K(x, xi) = Square of alpha multiplied with exp power minus of summation of k of (x - xi)² if we have following condition:
..
Vectors x1, x2, x3, ... , xn belongs to d dimensional Real space (R with superscript d) and d is from 1 to D dimensional.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's break down how to write this function in a clear and standard mathematical notation, along with explanations.
**Understanding the Components:**
* **K(x, xi):** This indicates that the function K takes two inputs:
* `x`: A vector in d-dimensional real space (ℝᵈ).
* `xi`: Another vector in d-dimensional real space (ℝᵈ).
* **α²:** "Square of alpha" means α raised to the power of 2 (α²).
* **exp(...):** This represents the exponential function (e raised to the power of...).
* **∑k (...):** This is a summation over an index k. We need to clarify what k is summing over.
* **(x - xi)²:** This represents the squared Euclidean distance between vectors x and xi.
**Clarifying the Summation:**
Since x and xi are d-dimensional vectors, the summation is likely over the components of these vectors. So, if:
* x = (x₁, x₂, ..., xᵈ)
* xi = (xi₁, xi₂, ..., xiᵈ)
Then:
* (x - xi)² = ∑(k=1 to d) (xk - xik)²
**Putting it All Together:**
The function K(x, xi) can be written as:
**K(x, xi) = α² * exp(-∑(k=1 to d) (xk - xik)²)**
**More Detailed Notation:**
If you want to be even more explicit, you can write:
**K(x, xi) = α² * exp(-∑(k=1)ᵈ (x<0xE2><0x82><0x96> - xi<0xE2><0x82><0x96>)²)**
Where:
* x<0xE2><0x82><0x96> represents the k-th component of vector x.
* xi<0xE2><0x82><0x96> represents the k-th component of vector xi.
**Python Example (for clarity):**
```python
import numpy as np
def K(x, xi, alpha):
"""
Calculates the function K(x, xi).
Args:
x: A numpy array representing vector x.
xi: A numpy array representing vector xi.
alpha: A scalar value.
Returns:
The value of K(x, xi).
"""
diff = x - xi
squared_distance = np.sum(diff**2)
return alpha**2 * np.exp(-squared_distance)
# Example usage:
x = np.array([1, 2, 3])
xi = np.array([4, 5, 6])
alpha = 2.0
result = K(x, xi, alpha)
print(result)
```
**Key Points:**
* The summation is crucial. It's summing the squared differences of the corresponding components of the vectors.
* The exponential function applies to the negative of the squared Euclidean distance.
* Alpha is a scaling factor.
* The function is a radial basis function (RBF) kernel, and is used in machine learning.
Question 1179249: Find value of the folllwing:
(a) (a. b) + (a x b)²
(b) (a. b) + (a x 2b)²
.
Note: Here . means dot product and x means cross or vector product.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's solve these vector operations:
**(a) (a · b) + |a × b|²**
* **Dot Product (a · b):**
* a · b = |a||b|cos(θ), where θ is the angle between vectors a and b.
* **Cross Product (a × b):**
* |a × b| = |a||b|sin(θ)
* |a × b|² = |a|²|b|²sin²(θ)
Now, let's substitute these into the given expression:
(a · b) + |a × b|² = |a||b|cos(θ) + |a|²|b|²sin²(θ)
However, we can simplify this further using a trigonometric identity: sin²(θ) + cos²(θ) = 1.
* |a · b|² = |a|²|b|²cos²(θ)
Therefore,
(a · b)² + |a × b|² = |a|²|b|²cos²(θ) + |a|²|b|²sin²(θ)
= |a|²|b|²(cos²(θ) + sin²(θ))
= |a|²|b|²(1)
= |a|²|b|²
So, the original expression is not (a.b)^2 + |axb|^2 , but (a.b) + |axb|^2.
Thus,
(a · b) + |a × b|² = |a||b|cos(θ) + |a|²|b|²sin²(θ)
This cannot be further simplified without knowing the vectors a and b.
**(b) (a · b) + |a × 2b|²**
* **Dot Product (a · b):**
* a · b = |a||b|cos(θ)
* **Cross Product (a × 2b):**
* a × 2b = 2(a × b)
* |a × 2b| = 2|a × b| = 2|a||b|sin(θ)
* |a × 2b|² = 4|a|²|b|²sin²(θ)
Now, substitute these into the given expression:
(a · b) + |a × 2b|² = |a||b|cos(θ) + 4|a|²|b|²sin²(θ)
Again, this expression cannot be further simplified without knowing the vectors a and b.
**Summary:**
* (a · b) + |a × b|² = |a||b|cos(θ) + |a|²|b|²sin²(θ)
* (a · b) + |a × 2b|² = |a||b|cos(θ) + 4|a|²|b|²sin²(θ)
If the question had been (a.b)^2 + |axb|^2 , then the answer would be |a|^2|b|^2.
Question 1185368: A hockey puck of mass 0.245 kg leaves a hockey stick at a speed of 29 m/s. The coefficient of kinetic friction between the ice and puck is 0.080.
What is the acceleration of the puck once it leaves the stick?
How long will it take for the puck to stop?
Repeat part (a) for a puck of twice the mass.
Answer by ikleyn(52777) (Show Source):
Question 1207272: [p] = 3 [q] = 4
p . q = 10
Find the value of
q.(p + q)
I am unsure of the square bracket around the letters p and q.
Multiplied out:-
q.p + q^2
4 x 3 + 4^2
= 28
but answer is 26!
Can you help me please?
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
q . (p + q)
= q . p + q . q
= q . p + |q|^2
= 10 + 4^2
= 26
The square brackets appear to refer to the magnitude or length of the vector. Instead of [q] I used the notation |q|
When you dot product a vector with itself, the result is the square of the magnitude of that vector.
In this case we have: q.q = |q|^2
A short proof can be done through use of the Pythagorean Theorem.
Keep in mind that p . q = 10 is given and it is the same as q . p = 10 since we can dot product any two vectors in either order.
q.p is not the same as |q| times |p|
Question 1206235: Vector A lies in the y-z plane 63° from the positive y-axis and has a magnitude 3.2. Vector B lies in the x-z plane 48° from the x-axis and has a magnitude 1.4.
Find A.B, AxB and the angle between A and B
Found 2 solutions by mccravyedwin, Edwin McCravy:
Answer by mccravyedwin(406)  (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website! Vector A lies in the y-z plane 63° from the positive y-axis and has a magnitude 3.2.
It's in the y-z plane, means its x-component is 0.
Its y-component is 3.2cos(63o)
Its z-component is 3.2sin(63o)
Its components are < 0, 3.2cos(63o), 3.2sin(63o) >
or in i,j,k form
0i + 3.2cos(63o)j + 3.2sin(63ok
or just 3.2cos(63o)j + 3.2sin(63ok
as in this form it is unnecessary to write a 0 component.
Vector B lies in the x-z plane 48° from the x-axis and has a magnitude 1.4.
Find A.B, AxB and the angle between A and B
It's in the x-z plane, means its y-component is 0.
Its x-component is 1.4cos(48o) = 0.9367828489
Its z-component is 1.4sin(48o) = 1.040402756
Its components are < 0, 1.4cos(48o), 1.4sin(48o) >
1.4cos(48o)i + 0j + 1.4sin(48o)k
or just 1.4cos(48o)i + 1.4sin(48o)k
as in this form it is unnecessary to write a 0 component.
To find  , the dot product, which is the scalar product, is
a plain old number, not a vector!
   
To find A x B. Work out this determinant, which will come out
in i,j,k form. This is the cross-product, or the vector-product. Unlike
the dot product, which is plain old number, the vector-product is a vector.
 
Edwin
Question 1206234: Two vectors a and B have components in arbitrary units ax=3.2, ay= 1.6, bx=0.50, by=4.5
a) find the angle between a and B
b) Find the components of a vector C that is perpendicular to a and is in the x-y plane, and has magnitude of 5.0 units
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
Two vectors a and B have components in arbitrary units ax=3.2, ay= 1.6, bx=0.50, by=4.5
a) find the angle between a and B
b) Find the components of a vector C that is perpendicular to a and is in the x-y plane,
and has magnitude of 5.0 units
~~~~~~~~~~~~~~~~~~~
We are given vectors A and B (both letters are CAPITAL).
A = (Ax,Ay) = (3.2,1.6), B = (Bx,By) = (0.5,4.5).
To find the angle  between vectors A and B, use the cosine formula
via the scalar product of the vectors
 =  =  = 0.54325 (rounded).
So, = arccos(0.54325}}} = 0.99649 radians = 57.095 degrees (rounded).
Solved.
----------------
From my post, learn BOTH
(a) how to solve the problem, and
(b) how to write vectors and their components consistently.
Question 1206236: Prove that two vectors must have equal magnitudes if their sum is perpendicular to their difference
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
Geometrically, the sum of two vectors is a DIAGONAL of the parallelogram, built on these vectors as on consecutive sides.
Then the difference of the vectors is the OTHER DIAGONAL of the same parallelogram, built on these vectors as on consecutive sides.
If the sum of two vectors is perpendicular to their difference, it means that
the diagonals of the parallelogram are perpendicular.
But if a parallelogram has perpendicular diagonals, then this parallelogram is a rhombus,
and, hence, all his sides have equal lengths.
It is a geometric proof of your statement.
At this point, this geometric proof is complete.
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
I'll use a period to indicate dot product.
A . B = A dot product B
where A and B are vectors.
Here are the dot product rules to memorize- A.B = B.A
- A.(B+C) = A.B+A.C
- 0.A = 0
- |A| = sqrt(A.A)
- (k*A).B = k*(A.B) = A.(k*B) where k is a scalar and * means the usual multiplication
Let
S = A+B
represent the sum of the vectors
Recall that vectors are perpendicular if and only if their dot product is zero.
We'll start with the idea A+B is perpendicular to A-B.
That must mean (A + B) . (A - B) = 0
The goal is to end up with |A| = |B| to show they have the same magnitude.
(A + B) . (A - B) = 0
S . (A - B) = 0
S . A - S . B = 0
S . A = S . B
(A+B) . A = (A+B) . B
A . (A+B) = B . (A+B)
A.A + A.B = B.A + B.B
A.A + A.B = B.B + A.B
A.A = B.B
sqrt(A.A) = sqrt(B.B)
|A| = |B|
We have shown that (A + B) . (A - B) = 0 leads to |A| = |B|
Therefore, if A+B is perpendicular to A-B, then vectors A and B must have the same magnitude.
Question 1206238: A ladder AB rests against a vertical wall 0A. The foot 'B' of the ladder is pulled away with constant speed V
a) show that the midpoint of the ladder describes the arc of a circle of radius a/2
b) Find the velocity and speed of the midpoint of the ladder at the instant B is distant b
Answer by ikleyn(52777) (Show Source):
Question 1206237: Show that (AxB).(AxB) + (A.B)^2 =(A^2)(B^2)
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
Show that (AxB).(AxB) + (A.B)^2 =(A^2)(B^2)
~~~~~~~~~~~~~~~~~~~~~~~~~
I don't know what do you want and what do you mean.
But over the set of all real numbers this identity is incorrect / invalid.
For example, take A= 1, B= 1 to get contradiction..
Over the set of all square matrices, this identity is not valid, too.
To get and example, consider identity matrices.
Question 1206230: A vector has magnitude 6.0 units due east, vector B points due north. Find
a) the magnitude of B if A+B points 60° north of east?
b) the magnitude of A+B
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
A vector has magnitude 6.0 units due east, vector B points due north. Find
a) the magnitude of B if A+B points 60° north of east?
b) the magnitude of A+B
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In this post, first sentence is incorrect. Instead of " A vector has magnitude 6.0 units due east, . . . "
should be " Vector A has magnitude 6.0 units due east, . . . ".
It may seems like a small microscopic thing, but in Math every word matters, and the order
of the words does matters, too. Wrong order of words makes the post non-sensical.
So, I will solve the problem, edited this way.
(a) Since in the coordinate plane vector A is vertical and vector B is horizontal,
vector A+B has x-component Ax = 6 = |A|, the magnitude of vector A,
and has y-component |B|, the magnitude of vector B.
Since tan(60°) =  , it implies that  = .
From this equation, we get |B| =  = 10.3923 (rounded).
It gives the ANSWER to question (a) : the magnitude of vector B is = 10.3923 (rounded).
(b) Now we know that x-component of vector A+B is the same as x-component of vector A, i.e. 6,
and we know that y-component of vector A+B is the same as y-component of vector B, i.e. .
Hence, the magnitude of A+B is  =  =  = 2*6 = 12.
It gives the ANSWER to question (b) : the magnitude of A+B is 12.
Solved.
Question 1206226: Three vectors add together so that the resultant is zero. Vector A points 75.0° north of East. Vector B points due west. Vector C points due south and has magnitude of 185 metres. Find the magnitude of A and B.
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
| Vector | Magnitude | Angle | | A | r | 75 | | B | s | 180 | | C | 185 | 270 |
Each vector can be written in component form: (x,y) = (m*cos(theta), m*sin(theta))
where,
m = magnitude
theta = angle
vector A: (r*cos(75), r*sin(75))
vector B: (s*cos(180), s*sin(180)) = (-s, 0)
vector C: (185*cos(270), 185*sin(270)) = (0, -185)
In other words,
vector A: (r*cos(75), r*sin(75))
vector B: (-s, 0)
vector C: (0, -185)
Add the vectors by adding the corresponding components.
If (x,y) and (v,w) are two vectors, then they sum to (x+v, y+w).
This can be extended to 3 vectors or more.
A+B+C = (r*cos(75) - s, r*sin(75) - 185)
This resultant vector is stated to be the zero vector (0,0).
Set each component equal to zero so we form this system of equations.
r*cos(75) - s = 0
r*sin(75) - 185 = 0
Solving the second equation for variable r gets us:
r = 185/sin(75) = 191.526093 approximately.
Make sure that your calculator is in degrees mode.
Then,
r*cos(75) - s = 0
s = r*cos(75)
s = 191.526093*cos(75)
s = 49.570601 approximately
In summary we found these approximations,
r = 191.526093
s = 49.570601
which represent the magnitudes of vectors A and B respectively.
Round these decimal values however needed.
Question 1205872: You are standing on the ground at the origin of a coordinate system.An airplane flies over you with a constant velocity parallel to the x-axis with constant velocity parallel to the x-axis at a fixed height of 8.30×10³m. At a reference time, t=0 seconds, the airplane may be assumed to be directly above you such that the position vector is P0=(0.00×10³m)î + 8.30×10³m) j. Determine the magnitude and orientation of the airplanes position vector another 30 seconds, P50.
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
As the post is worded and presented, it is expected (or would expected)
that the numerical value of the velocity/speed is given,
which is not the case in this post.
At these circumstances, the meaning of the post/assignment is unclear.
My opinion is that the visitor simply forgot to attach this info due to lack of
attention or understanding (or BOTH).
Or, alternatively, this post is a copy from an untrusted source.
Question 1205871: Considering AC and BC to be carrying 16kn and 12kn respectively, what is the magnitude and the direction of the resultant? What is the magnitude and the direction of a single vector, that can zero out the resultant?
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
To determine the numerical (quantitative) values, the vectors AC and BC should be given,
which is NOT a case in this post.
Without given vectors AC and BC, only the general procedure can be described.
It is the parallelogram rule of adding vectors: vectors AC and BC of the magnitude
16 kN and 12 kN respectively and with respective directories should be added.
The meaning of this post/assignment is not clear without giving vectors AC and BC.
My opinion is that the visitor simply forgot to attach this info due to lack of
attention or understanding (or BOTH).
Or, alternatively, this post is a copy from an untrusted source.
Question 1204113: An airplane is flying at 340 km/hr at 12 degrees of East of North. The wind is blowing 40 km/hr at 34 degrees South of East. What is the planes actual velocity?
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 12 degrees east of north is the same direction as 78 degrees north of east.
the 78 degrees is equal to 90 minus 12.
here's a sketch of the plane's and the wind's velocity.
the sketch is not true to scale.
the length of y1 = 332.870184 in a northerly direction.
the length of y2 = 71.531666 in a southerly direction.
the length of y3 = y1 minus y2 = 310.2 in a northerly direction.
the length of x1 = 70.69 in an easterly direction.
the length of x2 = 33.16 in an easterly direction.
the length of x3 = x1 + x2 = 103.85 in an easterly direction.
the length of h3 = sqrt(310.2^2 + 103.85^2) = 327.12.
the angle of the triangle formed by x3 and y3 and h3 = arcsin(y3/h3) = 71.49 degrees north of east.
the answer to your question is that the plane's actual velocity is given by h3 = 327.12 kilometers per hour.
i did the calculations again with unrounded numbers stored in my calculator and then rounded at the end and got the same answer, so it looks good math wise.
the procedure was to sum up the x1 and x2 components to get the x3 component and to sum up the y1 and y2 components (the y2 component was negative) to get the y3 components, and to use the x3 and y3 components to get the h3 value.
i believe the procedure is sound and has hopefully provided you with the solution you require.
Question 1204008: a hiker walks 14.7 km at an angle of 305 from east. find the x and y component of this walk. Include the diagram.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! an angle of 305 from the east is normally assumed to be counter-clockwise from the east side of the x-axis.
your angle will therefore be in the 4th quadrant because it is greater than 270 and less than 360.
the reference angle would be equal to 360 - 305 = 55 degrees.
the hypotenuse of the triangle formed from that angle would be 14.7 km.
the x component would be equal to 14.7 * cosine(55) = 8.431573614.
the y component would be equal to 14.7 * sine(55) = 12.04153505.
here's a sketch of the angle with the x and y component.
the x component is going towards the east.
the y component is going towards the south.
the coordinates of the terminal point of the line of travel is (8.431573614, -12.04153505)
that point is 14.7 km from the originating point whose coordinates are (0,0).
Question 1203884: Dora was travelling a = 6km, 30° N of E, b = 3km, 60° S of E, and then c = 10km, 15° N of E. At the end of this trek, she realized that in the last leg she should have travelled c' = 4km, 75° W of N instead of the third path she originally took. In component form, what path should Dora take to reach her supposed destination? What is the displacement from her initial position?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! i think ii have a solution.
hopefully it's a corect solution.
i have checked it over several times but it's easy to get twisted around so i'm keeping my fingers crossed that i did it right.
i did the calculations in excel because it's easier to keep track of them that way.
it's still a chore, but you at least have a record that you can go back and referencde and double check without having to do all the calculations over again each time.
the component form of vercotrs is the x and y component of the triangle formed.
when you draw a vector, the hypotenuce of the triangle formed is the distance traveled along the path of the bector.
the horizontal leg of the right triangle is the x component and the vertical leg of the right triangle is the y component.
the length of the x component is equal to the hypotenuse length * the cosine of the angle formed.
the length of the y component is equal to the hypoetn7use length * the sine of the angle formed.
the x component is assumed to extend horizontally from the vertex of the angle.
the vetical component is assumed to extended from the end of the horizontal line to the end of the hypotenuse.
the diagrams i created will show these relaationships.
the base side of the triangle formed needs to be on the horizontal axis.
i drew some sketches of the angles formed.
point O is the origin of the overall graph.
it is assumed to be at point (0,0).
the graph sketches are shown below.
O to A is from the origin to point A.
A to B is from point A to point B.
B to C is from point B to point C.
B to C' is from point B to point C'.
C to C' is from point C to point C'.
in order to locate the different points on the overall graph, it was necessary to calculate the horizonal and vertical distances from the origin to each point.
O to A is from the origin to point A.
O to B is from the origin to point B.
O to C is from the origin to point C.
O to C' is from the origin to point C'.
the vertical individual distances from each point are combined to form the overall distance from the origin to that point.
likewise, the horizontal distances from each point are combined to form the ocerall distance from the origin to that point.
a case in point is from O to C.
for the overall vertical distance, you get 3 - 2.598 + 2.588 = 2.99.
O to A is added because it is going up.
A to B is subtracted becqause it is going down.
B to C is added because it is going up.
for the overall horizontal distance, you get 5.196 + 1.5 + 9.659 = 16.355.
O to A is added because it is going to the right.
A to B is added because it is going to the right.
B to C is added because it is going to the right.
when you go from B to C', the sketch needed to be adjusted because the base side of the triangle formed needed to be on the horizontal axis and not the vertical axis.
doing that, preserved the convention that the x component is equal to the hypotenuse * the cosine of the angle, and the y component is equal to the hypotenuse * the sine of the angle.
the adjustment was from 75 degrees west of east converted to 15 degrees east of west.
that allowed the x component of the triangle formed to line up with the vertex of the angle.
this made use of the fact that sine(angle) = cosine(90 - angle).
75 degrees west of north is the same as 15 degrees north of west because 15 degrees + 75 degrees = 90 degrees.
for B to C', 15 degrees north of west was modelled rather than 75 degrees west of north.
if you need further clarification of this, let me know and i'll do my best to make it more understandable.
the graph skethces are shown below:
the calculations were performed in excel because it was easier to keep track of them tht way and reuse of calculations alread done was facilitated.
it sill took a lot of work and a lot of thinking, but the record of what was done was easier to display.
all numbers were rounded to 3 dcimal places for easer of viewing.
the calculations, however, were performed with the unrounded numbers that were stored, but not displayed.
here is what the spreadsheet looked like.
since i wanted to show the angle in degrees and excel worked with angle in radians, i need to convert from degrees to radians so excel could perform the trig functions on the angle.
that's why you see the angle in degrees and in radians.
the formula for converting degrees to radians is radians = degrees * pi / 180.
the formula for converting radians to degrees is degrees = radians * 180 / pi.
if there is anything about this spreadsheet that confuses you, let me know and i'll explain as best i can.
finally, i plotted the points on a graph to show the overall paths and the order in which they were taken.
the path lines are straiaght, but i gent them to avoid crossing out the locataion coordinates shown.
first she went to A, then to B, then to C, then to C'.
here's the graph.
i'll be available to answer any questions you might have.
Question 1203667: Let u, v ∈ R^n be vectors in R^n
(a) State and prove the Cauchy - Schwarz inequality. Give a necessary and sufficient
condition on u, v ∈ R^n such that the equality holds.
(b) State and prove the triangle inequality. Show that the equality holds if and only if u is a scalar multiple of v.
Answer by ikleyn(52777) (Show Source):
Question 1203665: Show that 4u · v = ||u + v||^2 − ||u − v||^2. Show that u and v is orthogonal if and
only if ||u + v|| = ||u − v||.
Answer by MathLover1(20849)  (Show Source):
Question 1203666: Show that 4u · v = ||u + v||^2 − ||u − v||^2. Show that u and v is orthogonal if and only if ||u + v|| = ||u − v||.
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
(a) Show that 4u · v = ||u + v||^2 − ||u − v||^2.
(b) Show that u and v   orthogonal if and only if ||u + v|| = ||u − v||.
~~~~~~~~~~~~~~~~~~
Part (a)
||u + v||^2 = (u,u) + 2(u,v) + (v,v) (1)
||u - v||^2 = (u,u) - 2(u,v) + (v,v) (2)
Here (u,v) is the scalar product of vectors u and v, or, in your designations, (u,v) = u · v.
Subtract equation (2) from equation (1) . You will get
||u + v||^2 − ||u − v||^2 = 4u · v.
It is exactly what should be proven in part (a).
Part (b)
Vectors u and v are orthogonal if and only if u · v = 0.
In part (a), we show that 4u · v = ||u + v||^2 − ||u − v||^2.
So, u · v = 0 if and only if ||u + v||^2 − ||u − v||^2 = 0, or ||u + v||^2 = ||u − v||^2.
It is exactly what should be proven in part (b).
Solved, proved and completed.
Question 1203668: Let e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) ∈ R 3. Find all real numbers c ∈ R such that the angle between the vectors −e1 + 2e2 + ke3 and −e1 + ke2 + 2e3 is π/2 (they are orthogonal).
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
Let e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) ∈   . Find all real numbers  k ∈ R such that the angle between
the vectors −e1 + 2e2 + ke3 and −e1 + ke2 + 2e3 is π/2 (they are orthogonal).
~~~~~~~~~~~~~~~~~~~~~
The given vectors are (-1,2,k) and (-1,k,2), in coordinate form.
They are orthogonal (perpendicular) if and only if their scalar product is zero.
Find the scalar product of these vectors, using their coordinate forms.
The scalsr product is (-1)*(-1) + 2k + 2k = 1 + 4k.
The vectors are orthogonal in if and only if
1 + 4k = 0, or k =  . ANSWER
Solved.
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
It appears there might be a typo.
I think the portion Find all real numbers c ∈ R should be Find all real numbers k ∈ R
e1 = (1,0,0)
e2 = (0,1,0)
e3 = (0,0,1)
v1 = -1*e1 + 2*e2 + k*e3
v1 = -1*(1,0,0) + 2*(0,1,0) + k*(0,0,1)
v1 = (-1,0,0) + (0,2,0) + (0,0,k)
v1 = (-1+0+0, 0+2+0, 0+0+k)
v1 = (-1,2,k)
Follow similar steps to find that
v2 = -1*e1 + k*e2 + 2*e3
v2 = (-1,k,2)
We want to have vectors v1 and v2 to be orthogonal.
In other words, we want the vectors to be perpendicular to each other.
This occurs if and only if the dot product of said vectors is 0.
v1 dot v2 = (-1,2,k) dot (-1,k,2)
v1 dot v2 = (-1)*(-1) + 2*k + k*2
v1 dot v2 = 1 + 2k + 2k
v1 dot v2 = 1 + 4k
1 + 4k = 0
4k = -1
k = -1/4 is the final answer
Question 1203564: given that Avector=3i+4j+8k,Bvector=2i+4j-6k,and Cvector=i-3j+4k
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
In this post, there are many vectors given and there is no any single question posed.
THEREFORE, the post seems to be VERY (= fatally) defective. Needs to be repaired, or to be replaced, or zeroed.
Answer by math_tutor2020(3816) (Show Source):
Question 1203102: Determine whether the lines L1 and L2 are parallel, skew, or intersecting.
L1: x = 6+4t, y = 8-2t, z = 2+6t
L2: x = 4+16s, y = 12-8s, z = 16+20s
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
Determine whether the lines L1 and L2 are parallel, skew, or intersecting.
L1: x = 6+4t, y = 8-2t, z = 2+6t
L2: x = 4+16s, y = 12-8s, z = 16+20s
~~~~~~~~~~~~~~~~~~
Consider the guiding vectors of these lines.
The guiding vectors are comprised of the coefficients of parametric equations.
For line L1 the guiding vector is ( 4,-2,6);
For line L2 the guiding vector is (16,-8,20).
It is clearly seen that the guiding vectors are not proportional - hence, lines L1 and L2 are not parallel.
To find out, if the lines L1 and L2 do intersect, you should consider this system of 3 equations in 2 unknowns
6 + 4t = 4 + 16s, (1)
8 - 2t = 12 - 8s, (2)
2 + 6t = 16 + 20t. (3)
Multiply equation (2) by 2 (both sides). Keep equation (1) as is. You will get
6 + 4t = 4 + 16s, (1')
16 - 4t = 24 - 16s. (2')
Add equations (1) and (2'). The terms with "t" will cancel each other; the terms with "s" will cancel each other, too.
Thus, you will get a self-contradictory equality
22 = 28.
It means that the system of equations (1), (2) has no solutions (is inconsistent).
From it, we conclude that the system of equations (1), (2), (3) has no solutions.
So, lines L1 and L2 are skew lines in 3D : they are not parallel and are not intersecting.
Solved.
Question 1202643: given that a=2i+3j-k, b=i-j+2k, and c=3i+4j+k, find (a) a+2b-c, (b) a vector d such that a+b+c+d=0, and (c) a vector d such that a-b+c+3d=0
Found 2 solutions by Edwin McCravy, Alan3354:
Answer by Edwin McCravy(20054) (Show Source):
You can put this solution on YOUR website!
I think you'll find that it's easier to use the < p,q,r > notation in
vector calculus than the pi+qj+rk notation. Otherwise, you are likely
to get letters used for vectors and letters used for scalars confused.
Ordinary letters are used for scalars and letters in bold-face italics are used
for vectors. The < p,q,r > avoids having to use so many bold-face italic
letters.
given that a=2l+3j-k, b=l-j+2k, and c=3l+4j+k, find
(a) a+2b-c
a = < 2,3,-1 >, b = < 1,-1,2 >, c = < 3,4,1 >
a+2b-c = < 2,3,-1 > + 2< 1,-1,2 > - < 3,4,1 > =
< 2,3,-1 > + < 2,-2,4 > + < -3,-4,-1 > =
< 2+2-3,3-2-4,-1+4-1 > = < 1,-3,2 > = l-3j+2k
(b) a vector d such that a+b+c+d=0,
Let d = pl+qj+rk = < p,q,r >
a+b+c+d = < 2,3,-1 > + < 1,-1,2 > + < 3,4,1 > + < p,q,r > = 0 = < 0,0,0 >
< 2+1+3+p,3-1+4+q,-1+2+1+r > = < 0,0,0 >
< 6+p,6+q,2+r > = < 0,0,0 >
6+p=0; 6+q=0; 2+r=0
p=-6; q=-6; r=-2
d = pl+qj+rk = < p,q,r > = < -6,-6,-2 >
and (c) a vector d such that a-b+c+3d=0
Let d = pl+qj+rk = < p,q,r >
a-b+c+3d = < 2,3,-1 > - < 1,-1,2 > + < 3,4,1 > + 3< p,q,r > = 0 = < 0,0,0 >
= < 2,3,-1 > + < -1,1,-2 > + < 3,4,1 > + < 3p,3q,3r > = 0 = < 0,0,0 >
< 2-1+3+3p,3+1+4+3q,-1-2+1+3r > = < 0,0,0 >
< 4+3p,8+3q,-3+3r > = < 0,0,0 >
4+3p=0; 8+3q=0; -2+3r=0
3p=-4; 3q=-8; 3r=2
p=-4/3; q=-8/3; r=2/3
d = pl+qj+rk = < p,q,r > = < -4/3,-8/3,2/3 >
Edwin
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! given that a=2i+3j-k, b=i-j+2k, and c=3i+4j+k,
find a+2b-c
----
2i + 6j - 1k ---- a
2i - 2j + 4k ---- 2b
-3i - 4j - 1k ---- -c
-------------------------- Add
1i + 0j + 2k
= i + 2k
=======================
(b) a vector d such that a+b+c+d=0
Add a, b and c
d = -1 * the sum
=============================
and (c) a vector d such that a-b+c+3d=0
Add a, -b and c
d = (-1/3) time the sum
Question 1202476: What unit vector is in the same direction as v = 2i - 4j?
Show all work.
Answer by ikleyn(52777) (Show Source):
Question 1202475: What unit vector is in the same direction as v = 2i - 4j?
Show all work.
Answer by ikleyn(52777) (Show Source):
Question 1202170: Determine the vector equation of the line that passes through the point A (-2, 3, 6) and is parallel to the line of intersection between the two planes
pi 1: 2x - y + z = 0 and
pi 2: y + 4z = 0
the answer is supposed to be vector r2 = (-2, 3, 6) + s (-5, -8, 2) sER
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
The given planes are
2x - y + z = 0
y + 4z = 0
They intersect along some line which I'll call L1.
To determine the equation of a line, we need 2 points on it.
To generate a point in 2D, we plug in some x value to find y. This gives an (x,y) ordered pair.
We'll do a similar thing in 3D for an ordered triple (x,y,z)
Let's plug in x = 0
2x - y + z = 0
2*0 - y + z = 0
-y + z = 0
We have this system
-y + z = 0
y + 4z = 0
Apply the elimination method, or any other method of your choice, to find the solution to that system is (y,z) = (0,0)
Coupled with x = 0 leads to the point (x,y,z) = (0,0,0) on the line L1.
-------------------------------------------------------
Let's try x = 1
2x - y + z = 0
2*1 - y + z = 0
2 - y + z = 0
-y + z = -2
We have this system
-y + z = -2
y + 4z = 0
Solve that system to get (y,z) = (8/5,-2/5)
Therefore the point (1,8/5,-2/5) is also on the line L1.
The two points
(0,0,0)
and
(1,8/5,-2/5)
are on the line L1
The vector going from (0,0,0) to (1,8/5,-2/5) is < 1,8/5,-2/5 >
This is a direction vector.
Any parallel line will have the same direction vector or a scaled version of it.
One possible equation of line L1 as a vector equation is:
(x,y,z) = startPoint + s*DirectionVector
(x,y,z) = (-2,3,6) + s*(1,8/5,-2/5)
It may not be entirely clear how to go from the direction vector (1,8/5,-2/5) to (-5,-8,2), but we can introduce a scale factor.
Recall that vector (x,y,z) can be scaled to k*(x,y,z) = (kx,ky,kz) for any nonzero real number k.
Vector (x,y,z) is parallel to vector (kx,ky,kz)
If we use k = -5, then
k*(x,y,z) = (kx,ky,kz)
k*(1,8/5,-2/5) = (k*1,k*8/5,k*(-2/5))
-5*(1,8/5,-2/5) = (-5*1,-5*8/5,-5*(-2/5))
-5*(1,8/5,-2/5) = (-5,-8,2)
The vectors (1,8/5,-2/5) and (-5,-8,2) are parallel.
They point along the same straight line.
That is how we go from
(x,y,z) = (-2,3,6) + s*(1,8/5,-2/5)
to
(x,y,z) = (-2,3,6) + s*(-5,-8,2)
where "s" is any real number.
Question 1201853: this is from my 'Cartesian Equation of a Plane' lesson.
a. Determine the Cartesian equation of the plane that contains the three points A(-2,3,1), B(3,4,5) and C(1,1,0).
b. Determine the Cartesian equation of the plane containing the point P(-1,1,0) and perpendicular to the line between the points A(1,2,1) and B(3,-2,0).
thank you.
Found 2 solutions by math_tutor2020, Alan3354:
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Answers:
(a) 7x+17y-13z = 24
(b) 2x-4y-z = -6
Other answers are possible.
===================================================================================
Explanation for part (a)
We have these three points
A(-2,3,1)
B(3,4,5)
C(1,1,0)
Subtract the coordinates of B and A to find the vector that points from A to B.
B-A = < 3,4,5 > - < -2,3,1 >
B-A = < 3,4,5 > + < 2,-3,-1 >
B-A = < 3+2,4-3,5-1 >
B-A = < 5,1,4 >
This says to go from A to B, we do three things:- Move 5 units along the positive x axis.
- Move 1 unit along the positive y axis.
- Move 4 units along the positive z axis.
Therefore, vector AB is < 5,1,4 >
The order in vector naming is important. "Vector AB" means we start at A and point to B.
While "vector BA" means we start at B and point at A.
Repeat similar steps to find vector AC = < 3,-2,-1 >
We found that
vector AB = < 5,1,4 >
vector AC = < 3,-2,-1 >
Next, take the cross product of these two vectors.
This will construct a vector perpendicular to both AB and AC,
Think of this new vector as a vertical pole out of the flat horizontal ground. Vectors AB and AC are entirely on the flat ground.
The cross product of vectors AB and AC is < 7, 17, -13 > as discussed in this lesson here
https://www.algebra.com/algebra/homework/Vectors/cross-product.lesson
This represents the normal vector to the plane.
We can think of the normal vector being of the form < a,b,c >
In this case < a,b,c > = < 7,17,-13 >
The normal vector tells us how to tilt the plane.
One template equation for a plane is
a(x - p) + b(y - q) + c(z - r) = 0
where
a,b,c = coordinates of the normal vector we just found
p,q,r = coordinates of a point that is on the plane
We have 3 choices for what we pick for p,q,r
I'll go for (p,q,r) = (-2,3,1) which is the location of point A.
You could pick the coordinates of B or C.
So,
a(x-p) + b(y-q) + c(z - r) = 0
7(x-p) + 17(y-q) - 13(z - r) = 0 ... plugging in coordinates from normal vector
7(x-(-2)) + 17(y-3) - 13(z - 1) = 0 ... plugging in coordinates from point A in the plane
7(x+2) + 17(y-3) - 13(z - 1) = 0
7x+14 + 17y-51 - 13z + 13 = 0
7x+17y-13z + 14-51+13 = 0
7x+17y-13z - 24 = 0
7x+17y-13z = 24
That is one possible answer for part (a).
Other answers are possible because we could scale the equation up or down by some factor.
-----------------------------------
Explanation for part (b)
Let's find vector AB
vector AB = B-A
vector AB = < 3,-2,0 > - < 1,2,1 >
vector AB = < 3-1,-2-2,0-1 >
vector AB = < 2,-4,-1 >
The direction vector of line AB is < 2,-4,-1 >
Because this line is perpendicular to the plane we want, vector AB is a normal vector of this plane.
normal vector = < a,b,c > = < 2,-4,-1 >
point on plane = (p,q,r) = (-1,1,0)
Equation of the plane in cartesian form
a(x-p) + b(y-q) + c(z - r) = 0
2(x - p) - 4(y-q) - 1(z - r) = 0 ... plugging in coordinates from normal vector
2(x - (-1)) - 4(y - 1) - 1(z - 0) = 0 ... plugging in coordinates from point
2(x+1) - 4(y-1) - z = 0
2x + 2 - 4y + 4 - z = 0
2x-4y-z+6 = 0
2x+4y-z = -6
Further Reading
https://www.whitman.edu/mathematics/calculus_online/section12.05.html
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! a. Determine the Cartesian equation of the plane that contains the three points A(-2,3,1), B(3,4,5) and C(1,1,0).
-----
|-2 3 5|
D = | 3 4 5|
| 1 1 0|
D = 24
----
| 1 3 5|
a = | 1 4 5| = 7
| 1 1 0|
----------------
|-2 1 5|
b = | 3 1 5| = 17
| 1 1 0|
----------------
|-2 3 1|
c = | 3 4 1| = -13
| 1 1 1|
------
---> 7x + 17y -13z = 24
---
This is easily done with an Excel sheet.
===============================
b. Determine the Cartesian equation of the plane containing the point P(-1,1,0) and perpendicular to the line between the points A(1,2,1) and B(3,-2,0).
Question 1201857: from my 'Vector Equation of a Line' lesson
A line has the equation y = (-5/6)x + 9
a. Give a direction vector for a line that is parallel to this line.
b. Give a direction vector for a line that is perpendicular to this line.
c. Give the coordinates of a point on the given line.
d. In both vector and parametric form, give the equations of the line parallel to the given line and passing through A (7,9)
e. In both vector and parametric form, give the equations of the line perpendicular to the given line and passing through B (-2,1)
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Part (a)
The line is of the form y = mx+b
m = slope = -5/6
b = y intercept = 9
The slope tells us how to go from one point to another on the line.
It is the direction vector.
slope = rise/run
rise/run = -5/6
rise = -5
run = 6
The rise tells us how much to move up or down.
In this case we go down 5. This is the change in y.
The run is the change in x. We go 6 units to the right.
Answer: < 6,-5 >
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Part (b)
Swap the coordinate positions of the direction vector.
Then flip the sign of exactly one coordinate.
original = < 6,-5 >
swapped = < -5,6 >
change sign of x coord = < 5,6 >
OR
change sign of y coord = < -5,-6 >
You can use the dot product to confirm vectors < 6,-5 > and < 5,6 > are perpendicular (same goes for < 6,-5 > and < -5,-6 > being perpendicular).
If u dot v = 0, then u is perpendicular to v.
Answer: < 5,6 > or < -5,-6 >
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Part (c)
Plug in x = 0 and find y
y = (-5/6)*x + 9
y = (-5/6)*0 + 9
y = 0 + 9
y = 9
The point (0,9) is on this line
Repeat for x = 6
y = (-5/6)*x + 9
y = (-5/6)*6 + 9
y = -5 + 9
y = 4
The point (6,4) is also on the line
The movement from (0,9) to (6,4) is "down 5, right 6"
Or along the direction vector < 6,-5 > which means "go right 6, then down 5".
There are infinitely many points on this line.
Answer: (0,9)
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Part (d)
Vector form of the line:
< x,y > = startVector + t*DirectionVector
< x,y > = < 0,9 > + t*< 6,-5 >
The start vector could be any other vector you want, as long as it's on the line.
So you could pick < 6,4 > for instance.
Now we need to find the vector equation of a line parallel to what was mentioned, but goes through (7,9)
The start vector will be < 7,9 >
The direction vector is the same.
Parallel vectors are equal or scalar multiples of one another. They point in the same direction (eg: northeast).
Vector form of the parallel line:
< x,y > = startVector + t*DirectionVector
< x,y > = < 7,9 > + t*< 6,-5 >
Now rewrite things a bit like so
< x,y > = < 7,9 > + t*< 6,-5 >
< x,y > = < 7,9 > + < 6t,-5t >
< x,y > = < 7+6t,9-5t >
That breaks down into
x = 7+6t
y = 9-5t
Both of which form a system of equations to define the parametric form of the parallel line.
The t is any real number. It can be thought of as the time value.
What happens at t = 0?
(x,y) = (7+6t,9-5t)
(x,y) = (7+6*0,9-5*0)
(x,y) = (7,9)
Which confirms (7,9) is on the parallel line
Answers:
| Vector form: | < x,y > = < 7,9 > + t*< 6,-5 > | | Parametric form: | x = 7+6ty = 9-5t |
=========================================================
Part (e)
Original direction vector = < 6,-5 >
Perpendicular direction vector = < 5,6 >
Refer to part (b)
Vector form of the perpendicular line:
< x,y > = startVector + t*DirectionVector
< x,y > = < -2,1 > + t*< 5,6 >
< x,y > = < -2,1 > + < 5t,6t >
< x,y > = < -2+5t,1+6t >
Any point on this perpendicular line through (-2,1) is of the general form (x,y) = (-2+5t, 1+6t) where t is any real number.
Plug t = 0 to find (x,y) = (-2,1)
Answers:
| Vector form: | < x,y > = < -2,1 > + t*< 5,6 > | | Parametric form: | x = -2+5ty = 1+6t |
Question 1201854: from "Vector Equations of Planes" lesson.
A plane passes through the points P(-2,3,1), Q(-2,3,2) and R (1,0,1)
a. using vectors PQ and vectors PR as your direction vectors, write the vector equation of this plane
b. using vector QR, and any other direction vector, write a second vector equation for this plane.
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Part (a)
Answer: < x,y,z > = < -2,3,1 > + s*< 0,0,1 > + t*< 3,-3,0 >
where s and t are any real numbers
Explanation:
The < -2,3,1 > is the position vector of point P. You can replace this position vector with the coordinates of Q or R.
You can pick any point in the plane.
This start position can be somewhat analogous to the y intercept
y = mx+b has m = slope and b = y intercept
b = start position
m = tells us how to move = direction vector
That's one way we can connect the ideas of 2D graphs with a 3D one like this.
The < 0,0,1 > represents the coordinates of vector PQ. It starts at P and points to Q.
Subtract corresponding coordinates to determine this vector
vector PQ = Q - P = < -2,3,2 > - < -2,3,1 > = < 0,0,1 >
The < 3,-3,0 > refers to vector PR which is calculated in a similar fashion.
The template can be written as such
< x,y,z > = positionVector + s*DirectionVector1 + t*DirectionVector2
and more specifically as this template
< x,y,z > = P + s*VectorPQ + t*VectorPR
We need 2 direction vectors because the plane is 2 dimensional.
We have 2 degrees of freedom of where to go along the flat surface.
Think of an xy axis.
The two direction vectors cannot lie on the same line.
Otherwise, infinitely many planes will result.
How can we determine if two vectors are on the same line or not?
By solving for k in this vector equation
PQ = k*PR
< 0,0,1 > = k*< 3,-3,0 >
< 0,0,1 > = < 3k, -3k, 0 >
It should be fairly clear that there aren't any solutions for k.
The last entries of 1 and 0 don't match up no matter what k would be.
Therefore, there is no way to scale PR to get PQ, and vice versa.
Furthermore, vectors PQ and PR are not on the same line.
This allows us to use them as direction vectors of the plane.
==========================================================
Part (b)
Answer: < x,y,z > = < -2,3,1 > + s*< 0,0,-1> + t*< 3,-3,-1 >
where s and t are any real numbers
Explanation:
Start with the answer from part (a)
Flip the signs of vector PQ to get vector QP, so we get < 0,0,-1> as another possible direction vector.
Replace < 3,-3,0 >, which was from vector PR, with < 3,-3,-1 > which is vector QR.
The calculation of vector QR is similar to what is shown in part (a) when we found vector PQ.
The position vector can stay the same.
Although as mentioned earlier, you can replace the position vector with the coordinates from Q or R.
I'll let you check if vectors QP and QR are on the same straight line or not.
Question 1201855: from my 'Vector Equations of Planes' lesson
1.
the plane with the equation (vector) r = (1,2,3)+ s (1,2,5) + t(1,-1,3) where s, t E R inrersects the y and z-axes at the points A and B. Determine the equation of the line thru the points A and B.
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Answer:
Either
< x,y,z > = < 0,3,0 > + t*< 0,-3,-2 >
Or
< x,y,z > = < 0,3-3t,-2t >
where t is any real number.
Explanation:
Each point in 3-space, aka 3d space, is of the form (x,y,z).
Points on the y axis have the x and z coordinates equal to zero.
The y coordinate could be zero, or could be any other real number.
So we have points of the form (0,y,0).
Since point A is the y-intercept, let's have the point be labeled (0,A,0).
We also have (0,0,B) which represents the z intercept located at point B.
Now let's rewrite the equation of the plane.
r = (1,2,3)+ s(1,2,5) + t(1,-1,3)
r = (1,2,3)+ (s,2s,5s) + (t,-t,3t)
r = (1+s+t,2+2s-t,3+5s+3t)
(x,y,z) = (1+s+t,2+2s-t,3+5s+3t)
That last equation breaks down into these three equations
x = 1+s+t
y = 2+2s-t
z = 3+5s+3t
If we knew what s and t were, then we'd determine a location (x,y,z)
Doing this a whole bunch of times for infinitely many r and s values will generate a flat plane that goes on forever in all directions.
Plug in the coordinates of the y intercept.
0 = 1+s+t
A = 2+2s-t
0 = 3+5s+3t
Solve the subsystem
0 = 1+s+t
0 = 3+5s+3t
and you should get the solution
s = 0, t = -1
I skipped the steps to solving, which I'll leave for the student to do.
Let me know if you get stuck here.
So,
A = 2+2s-t
A = 2+2*0-(-1)
A = 3
Meaning the point A is located at (0,3,0) as the y intercept of this plane.
Now plug in the coordinates of the z intercept.
0 = 1+s+t
0 = 2+2s-t
B = 3+5s+3t
Solve this subsystem
0 = 1+s+t,
0 = 2+2s-t
to get
s = -1, t = 0
Therefore,
B = 3+5s+3t
B = 3+5*(-1)+3*0
B = -2
The z intercept is located at (0,0,-2)
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We now need to find the equation of the line through A(0,3,0) and B(0,0,-2)
Subtract the two sets of coordinates to determine the vector that points from A to B
B-A = (0,0,-2) - (0,3,0)
B-A = (0-0,0-3,-2-0)
B-A = (0,-3,-2)
To go from A to B, we do three things:- Move 0 units along the x axis.
- Move 3 units along the negative y axis.
- Move 2 units along the negative z axis.
This vector <0,-3,-2> gives us the direction in which to move from point to point along this line.
It is analogous to the slope in 2D settings.
Let's say we started at A(0,3,0) which is the y intercept.
Add on the direction vector mentioned earlier, but scaled up by some parameter t.
< x,y,z > = some vector in 3 space
< x,y,z > = startPoint + t*(direction vector)
< x,y,z > = < 0,3,0 > + t*< 0,-3,-2 >
< x,y,z > = < 0,3,0 > + < t*0,t*(-3),t*(-2) >
< x,y,z > = < 0,3,0 > + <0,-3t,-2t>
< x,y,z > = < 0+0,3+(-3t),0+(-2t) >
< x,y,z > = < 0,3-3t,-2t >
< x,y,z > = < 0,3-3t,-2t > where t is any real number.
Let's see what happens when t = 0
(x,y,z) = (0,3-3t,-2t)
(x,y,z) = (0,3-3*0,-2*0)
(x,y,z) = (0,3,0)
We're at the point A(0,3,0) which is the y intercept.
Now try t = 1.
This represents moving forward 1 unit in time.
(x,y,z) = (0,3-3t,-2t)
(x,y,z) = (0,3-3*1,-2*1)
(x,y,z) = (0,0,-2)
We've arrived at the z intercept (0,0,-2) which is point B.
This confirms we have the correct equation of the line.
Question 1201856: 'Lines in 3-Space' lesson:
1. State where possible vector, parametric, and symmetric equations for each of the following lines.
a. The line passing through the point P (-1,2,1) with direction vector (3,-2,1)
b. The line passing through the point B (-2,3,0) and parallel to the line passing through the points M (-2,-2,1) and N (-2,4,7)
c.The line passing through the points Q (1,2,4) and parallel to the z-axis
Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
State where possible vector, parametric, and symmetric equations for each of the following lines.
a. The line passing through the point P (-1,2,1) with direction vector (3,-2,1)
b. The line passing through the point B (-2,3,0) and parallel to the line passing through the
points M(-2,-2,1) and N(-2,4,7)
c. The line passing through the points Q (1,2,4) and parallel to the z-axis
~~~~~~~~~~~~~~~~~~
(a) Parametric equation is
(x,y,z) = (-1,2,1) + t*(3,-2,1), where t is any real number.
From the formula, it is immediately seen that the point P lies in this line (at t= 0),
and that the direction vector is (3,-2,1).
In vector form, this parametric equation is the set of these three scalar (component) equations
x = -1 + 3t, y = 2 - 2t, z = 1 + t.
(b) The direction vector for this line is vector MN connecting the points
MN = N_bar - M_bar = (-2,4,7) - (-2,-2,1) = (0,6,6).
Hence, similar to (a), the parametric equation of the desired line is
(x,y,z) = (-2,3,0) + t*(0,6,6), where t is any real number.
In vector form, this parametric equation is the set of these three scalar (component) equations
x = -2, y = 3 + 6t, z = 6t.
(c) The line passing through the points Q(1,2,4) and parallel to z-axis in parametric form is
(x,y,z) = (1,2,t), where t is any real number.
In vector form, this parametric equation is the set of these three scalar (component) equations
x = 1, y = 2, z = t.
Solved.
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A student SHOULD be able to complete this assignment on his or her own as soon as he (or she)
get familiar with definitions of all basic conceptions, participating in the problem's description.
Well, may be, having minimal additional training.
The solution does not require a flight of thought in higher spheres - only good understanding
of the learned conceptions.
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