SOLUTION: I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. (I haven’t learned any of this, but I am trying to become familiar with it.)

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Question 1121153: I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. (I haven’t learned any of this, but I am trying to become familiar with it.) I found a website full of problems of the content we will be learning, but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:
I’m going to do my best to talk through each part and tell you how I think it should be solved!
A line L1 passes through points P(−1,6,−1) and Q(0,4,1).
a. (i) Show that vector PQ=(1,−2,2).
So, I can see that if you take the x, y, and z of Q, and subtract the x, y and z of P, then you get PQ. But, is there some sort of formula for this that you would use to answer this question?
(ii) Hence, write down an equation for L1 in the form r=a+tb.
Would this be: r=(−1,6,−1)+t(0,4,1)? (I just plugged in P for a and Q for b.
b. A second line L2 has equation r=(4,2,−1)+s(3,0,−4).
Find the cosine between vector PQ and L2.
I’m super confused on how you could find the cosine of this when all you have is some points? Or, if I draw this out somehow would I find the sides to do the Law of Cosines?
c. The lines L1 and L2 intersect at the point R. Find the coordinates of R.
Would I use a midpoint formula to discover this? Since R would be in the middle of both lines?
Found 2 solutions by Theo, math_helper:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
A line L1 passes through points P(−1,6,−1) and Q(0,4,1).

a. (i) Show that vector PQ=(1,−2,2).

vector PQ is created from line segment PQ.

a vector has both magnitude and direction.

vector PQ = = <0--1,4-6,1--1> = <1,-2,2>

note that -- means plus.
you are subtracting a negative number, which is the same as adding a positive number.

it has direction because from, any point in space, it can only go 1 unit in the direction of the x-axis, -2 units in the direction of the y-axis, and 2 units in the direction of the z-axis.

the length of vector V is sqrt(1^2 + (-2)^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9).

(ii) Hence, write down an equation for L1 in the form r=a+tb.
Would this be: r=(−1,6,−1)+t(0,4,1)? (I just plugged in P for a and Q for b.

the vector equation of a line is in the form of R = R0 + T * vector V.

R is any point on the line which is the same as any position vector that terminates on the line.

R0 is a specific point on the line which is the same as a a position vector that terminates on a specific point on the line.

V is a direction Vector.

the point P from the line PQ serves as the point R0 which is the same as position vector R0.

note that any point on the graph also serves as a position vector.
a position vector is a vector that has its tail at the origin and its head terminating at the point.

therefore point (-1,6,-1) is also position vector <-1,6,-1).

this means that vector arithmetic applies to it just as with any other vector.

vector V is the same as vector PQ that we just created.

therefore, in the equation R = R0 + T * vector V, we get:

vector R0 = <-1,6,-1>
vector V = <1,-2,2>

vector R = <-1,6,-1> + T * <1,-2,2> which becomes:

vector R = <-1,6,-1> + which becomes:

vector R = <-1+T,6-2T,-1+2T>, where T is any real number.

that's the equation of the line R in vector equation form.

x = -1 + T
y = 6 - 2T
z = -1 + 2T

b. A second line L2 has equation r=(4,2,−1)+s(3,0,−4).

the equation of the first line is R = (-1,6,-1) + T * <1,-2,2>
the equation of the second line is L1 = (4,2,-1) + S * <3,0,-4>

the vector portion of these two lines is <1,-2,2> and <3,0,-4>

we'll call <1,-2,2> vector V and we'll call <3,0,-4> vector W.

these two vectors can be positioned anywhere on the graph, so they can be brought to intersect at a common point, allowing us to find the angle between them.

for example, you can place the tail end of both vectors at the origin.

the dot product of these two vectors can be found by the formula

vector V dot vector W = vx*wx + vy*wy + vz*wz.

that makes vector V dot vector W equal to 1*3 - 2*0 - 2*4 = -5.

there is an alternate formula for the dot product.

it is vector V dot vector W = magnitude of vector V * magnitude of vector W * cosine (theta), where theta is the angle between the two vectors.

you can use this formula to solve for the angle between the two vectors.

you get cosine (theta) = (vector V dot vector W) / (magnitude of vector V * magnitude of vector W).

we already calculated the dot product of vector V and vector W, which is equal to -5.

the magnitude of vector V is equal to sqrt(1^2 + (-2)^2 + 2^2) = sqrt(9)

the magnitude of vector W is equal to sqrt(3^2 + 0^2 + (-4)^2) = sqrt(25) = 5

cosine (theta) is therefore equal to -5 / (5*sqrt(9)) which is equal to -1/3.

in the first quadrant, the angle would be arcosine (1/3) = 70.52877936.

since the cosine is negative, the angle could be in the second quadrant or the third quadrant.

in the second quadrant, the angle would be (180 - 70.52877936) = 109.4712206.

in the third quadrant, the angle would be (180 + 70.52877936) = 250.5287794.

it's difficult to see what the angle is on the 3 dimension graph because the angle between the two vectors is in a different plane.

my guess is that the angle is 109.4712206 because 250 degrees just looks too large.

i used an online calculator and it told me the angle was 109 degrees, so i think my guess is right, assuming the calculator did the job correctly.

that calculator can be found at https://www.emathhelp.net/calculators/linear-algebra/angle-between-two-vectors-calculator/?ux=1&uy=-2&uz=2&vx=3&vy=0&vz=-4&steps=on


c. The lines L1 and L2 intersect at the point R. Find the coordinates of R.
Would I use a midpoint formula to discover this? Since R would be in the middle of both lines?

the intersection of the lines is the common point between the lines.

if you have a 3D graphing calculator, you might have the calculator find the point for you.

if not, then you need to find it algebraically by solving the two equations of the lines simultaneously.

a good reference on how to do that can be found at https://www.youtube.com/watch?v=LxDM5-wRAzI

the vector equations of both lines L1 and L2 would be:

L1 = <-1+T,6-2T,-1+2T>

L2 = <4+3S,2,-1-4S>

this means: 



variable            L1                L2

x                   -1 + T            4 + 3S
y                    6 - 2T           2
z                   -1 + 2T          -1 - 4S


if the lines intersect, then the (x,y,z) coordinate of the intersection point will be the same in both lines.

this means that:

-1 + T = 4 + 3S
6 - 2T = 2
-1 + 2T = -1 - 4S

take any two of these equations and solve for one of the variables.
then take the value of that variable and replace it in any one of the equations.

you will get S = -1 and T = 2

when S = -1 and T = 2, your lines will have a common x,y,z coordinate as shhown below.

for L1, x = -1 + 2 = 1
for L2, x = 4 + 3 * -1 = 4 - 3 = 1

for L1, y = 6 - 4 = 2
for L2, y = 2

for L1, z = -1 + 4 = 3
for L2, z = -1 - 4 * -1 = -1 + 4 = 3

the common intersection occurs when T = 2 and S = -1.

that intersection point is at (1,2,3).

since i really didn't know how to do this when i decided to try to answer this question, i had to rely on many references to embellish my understanding sufficiently to provide an answer.

the other tutor who answered these questions was one of those references, namely math_helper(1240).

he was quite helpful.

the other references were on the web and are shown below.

https://www.emathhelp.net/calculators/linear-algebra/angle-between-two-vectors-calculator/?ux=1&uy=-2&uz=2&vx=3&vy=0&vz=-4&steps=on

https://www.youtube.com/watch?v=98C7iv8OcnI

https://www.youtube.com/watch?v=qsgK1d-_8ik

https://www.youtube.com/watch?v=46y2ec1Nuuk

https://www.youtube.com/watch?v=PyPp4QvQY3Q

https://www.youtube.com/watch?v=h2kjjDK1t2s

https://www.youtube.com/watch?v=_YkIivLaVJs

https://www.youtube.com/watch?v=lulSApFPw1M

https://www.youtube.com/watch?v=DXB1PWq8Dg0

https://www.youtube.com/watch?v=MpN8BIci-Ys

https://www.youtube.com/watch?v=MpN8BIci-Ys

https://www.youtube.com/watch?v=H7wre3njI0Y

https://www.youtube.com/watch?v=LxDM5-wRAzI

there may be some redundancy, but better too much than too little so i included all that i found and reviewed that appeared to be pertinent to the problems at hand.

you may find them helpful as well.




Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
a.(i) You are correct. The formula, given points P(x1,y1,z1) and Q (x2,y2,z2) is v = < x2-x1, y2-y1, z2-z1 >. for a vector that points in the direction from P toward Q.

a. (ii) They want a parametric representation of the line. What you wrote is essentially correct, however, be aware that you are specifying vectors, not just coordinates. Sure, once you evaluate a vector equation to a resultant < a,b,c > you can say the tip of the resultant vector is at the point (a,b,c), but be sure to keep vectors straight. Here you can take one point as the start (say P) and then use the vector you found in (a (i)) multiplied by a parameter t: r = < -1, 6, -1 > + t* < 1,-2,2 >, so ( here, -1+t is the x component, 6-2t is the y component, and -1+2t is the z component ). t varies from to

b. The 2nd line has form: = <4+3s, 2, -1-4s>
The "dot" product of two vectors is A ⋅ B = where is the angle between the two vectors, and , gives a SCALAR measure of how much vectors A and B point in the same direction. So you have all the numbers needed. Let A be the vector PQ you found, and let B be the vector <3,0,-4> (the part that 's' scales). This is so because s<3,0,-4> and t<1,-2,2> have the same angle between them as the two lines PQ and L2.
( If this last fact is not clear, watch this video https://www.youtube.com/watch?v=PyPp4QvQY3Q )
You will compute the numbers and solve for

c. At the point of intersection, , and (where is supposed to be r sub 2, sub x, etc.). Each of the lines (a(ii)) and (b) can be thought of as the tip of a vector traced out as t (s) varies.

Setting x components equal: -1+t = 4+3s
Setting y components equal: 6-2t = 2
Setting z components equal: -1+2t = -1-4s

The y component equation gives t=2, substituting into the x component equation
gives s=-1 (and these values work in the z component equation, so it is a good solution).
To find the coordinates, substitute t=2 in part a(ii) and it should match the coordinates when
you set s=-1 in part b. You should find they intersect at (1,2,3).

Hopefully Theo will agree with these solutions. Maybe he can explain it better or make any needed corrections. I was rusty on it as well.


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