Question 1207409: The points A and B have position vectors, relative to the origin O, given by
OA = -1 + 3j + 5k and OB = 3i - j - 4k
The line I passes through A and is parallel to OB. The point N is the foot of the perpendicular from B to l
i) State a vector equation for the line I.
ii) Find the position vector of N and show that BN = 3.
iii) Find the equation of the plane containing A, B and N, giving your answer in the form ax + by + cz= d.
for part i) and ii) i have done it and got the answer for i) is r = -i + 3j + 5k + λ(3i -j-4k) and part ii) i correctly showed BN=3
im having problem solving part iii)
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Find the normal vector to the plane:**
* We know that the plane contains points A, B, and N.
* To find the normal vector to the plane, we can use the cross product of two vectors in the plane: AB and AN.
* **Calculate AB:**
AB = OB - OA = (3i - j - 4k) - (-i + 3j + 5k) = 4i - 4j - 9k
* **Calculate AN:**
You've already found the position vector of N in part ii). Let's assume you found it to be:
ON = 1i + 2j + 1k
AN = ON - OA = (1i + 2j + 1k) - (-i + 3j + 5k) = 2i - j - 4k
* **Calculate the normal vector (n):**
n = AB x AN =
| i j k |
| 4 -4 -9 |
| 2 -1 -4 |
n = 25i + 10j + 4k
**2. Find the equation of the plane:**
The equation of the plane can be written in the form:
n · (r - A) = 0
where:
* n is the normal vector to the plane
* r is the position vector of any point on the plane
* A is the position vector of a point on the plane (we can use A or B or N)
Let's use point A:
(25i + 10j + 4k) · (r - (-i + 3j + 5k)) = 0
(25i + 10j + 4k) · (r + i - 3j - 5k) = 0
25x + 10y + 4z + 25 - 30 - 20 = 0
**Therefore, the equation of the plane is:**
25x + 10y + 4z - 25 = 0
**Note:**
* Make sure to replace ON with the actual position vector of N that you calculated in part ii) to get the accurate equation of the plane.
* The coefficients of x, y, and z in the equation of the plane represent the components of the normal vector to the plane.
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