Finding the equation of a line cutting a line at right angles and the equation of a plane...

Find the equation of the line through P(-1,0,1) that cuts the line r=(3,2,1)+t(1,2,2) at right-angles at Q. Also find the length PQ and the equation of the plane containing the two lines.

First we find Q:

P = (-1, 0, 1)
Q = (3+t, 2+2t, 1+2t)

Line PQ has direction vector <3+t+1, 2+2t-0, 1+2t-1> = < 4+t, 2+2t, 2t >
Line r = (3,2,1) + t(1,2,2) has direction vector < 1, 2, 2 >

Since lines are perpendicular, their dot product = 0

< 4+t, 2+2t, 2t > · < 1, 2, 2 > = 0
(4+t) + 2(2+2t) + 2(2t) = 0
4 + t + 4 + 4t + 4t = 0
9t = -8
t = -8/9

Q = (3+t, 2+2t, 1+2t), t = -8/9
Q = (19/9, 2/9, -7/9)

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P = (-1, 0, 1)
Q = (19/9, 2/9, -7/9)

PQ = < 28/9, 2/9, -16/9 > = 2/9 < 14, 1, -8 >

Line through points PQ has equation:
r = (-1, 0, 1) + s(14, 1, -8)

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|PQ| = 2/9 √(14²+1²+8²) = 2/9 √261 = 2/9 * 3√29 = 2(√29)/3

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Line PQ has direction vector < 14, 1, -8 >
Line r = (3,2,1) + t(1,2,2) has direction vector < 1, 2, 2 >

< 14, 1, -8 > x < 1, 2, 2 > = < 18, -36, 27 > = 9 < 2, -4, 3 >

Vector < 2, -4, 3 > is normal to plane containing the two lines

Equation of plane with normal < 2, -4, 3 > and passing through point (-1, 0, 1):
2(x + 1) - 4 (y - 0) + 3 (z - 1) = 0
2x + 2 - 4y + 3z - 3 = 0
2x - 4y + 3z = 1

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