A level question on maths chap vector!?
2 planes, m and n, have eqn x + 2y - 2z = 1 and 2x - 2y + z = 7 respectively.
The line L has eqn
r = i + J - k + t(2i + j + 2k)
1. Show that L is parallel to m
2. Find the position vector of the pt of intersection of L and n.
The point P lying on L is such that its perpendicular distance from m and n are equal.
3. Find the position vectors of 2 possible positions for P and calculate the distance between them.
The perpendicular distance of a pt with position vector xi + yj + zk from the plane ax + by + cz = d is
| ax + by + cz - d |/ sq root(a^2 + b^2 + c^2)
| = modulus
Pls, expl how u arrive at the answer with step by step pls!
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n = {1,2,-2} ..... is a vector normal to m
Answer (1): {2,1,2} ∙ {1,2,-2} = 2+2-4 = 0 .... ⇒ the line is parallel to the plane
2 (1+2t) - 2(1+t) + (-1 + 2t) = 7 ..... plug the line into the plane
t = 2 ... ⇒ the intersection = (5,3,3)
Answer (2): (5,3,3)
x + 2y - 2z - 1 = 2x - 2y + z - 7 .... equate the distances
x = z + 2 ; x = 2y - 1 ..... must satisfy the line
⇒ x = 7 , y = 4 , z = 5 ... by solving the three equations
Answer (3): (7,4,5)