Question

Find the distance of point -2i+3j-4k from the line r=i+2j-k+a(i+3j-9k) measured parallel to the plane: x-y+2z-3=0

Answer 1

let Q(x1,y1,z1) be any point on the line

$\frac{x - 1}{1} = \frac{y - 2}{3} = \frac{z + 1}{ - 9} = \alpha$

therefore,

$x1 = \alpha + 1$

$y1 = 3 \alpha + 2$

$z1 = - 9 \alpha - 1$

the given point is P(-2,3,-4)

as we have to find the distance of P from line parallel to the plane

$x - y + 2z - 3 = 0$

therefore, the join of P and Q must be parallel to the plane

so dot product of d.r. of PQ with the d.r. of plane is equal to zero

d.r. of plane =<1,-1,2>

d.r. pf PQ =

$< \alpha + 3. \: \: 3 \alpha - 1 .\: \: - 9 \alpha + 3 >$

therefore,

$1( \alpha + 3) + (- 1)(3 \alpha - 1) + 2( - 9 \alpha + 3) = 0 \\ \alpha + 3 - 3 \alpha + 1 - 18 \alpha + 6 = 0 \\ - 20 \alpha + 10 = 0 \\ \alpha = \frac{1}{2}$

so, coordinates of point Q is

$( \frac{3}{2} . \: \: \: \frac{7}{2} . \: \: \: \frac{ - 11}{2} )$

hence required distance, PQ=

$\sqrt{ {( \frac{7}{2} )}^{2} + {( \frac{1}{2}) }^{2} + {( \frac{3}{2}) }^{2} } \\ = \sqrt{ \frac{49 + 1 + 9}{4} } = \frac{ \sqrt{59} }{2} units$