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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the foot of perpendicular from the point(2,3,-8)to the line $\Large \frac{4-x}{2}\;=\frac{y}{6}\;=\frac{1-z}{3}$.Also,find the perpendicular distance from the given point to the line.

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1 Answer

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  • If two lines are $ \perp$ then $a_1a_2+b_1b_2+c_1c_2=0$, where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are the direction ratios of the two lines.
Step 1:
Let the given equation be $\large\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\lambda$.
This can be written as
$\large\frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3}$$=\lambda$-----(1)
Therefore the direction ratios of the line is $(-2,6,-3)$
Therefore the coordinates of any point on the line is
$x=4-2 \lambda,y=6 \lambda, z=1-3 \lambda$
Step 2:
Let $ Q(4-2\lambda, 6\lambda, 1-3\lambda)$ be the foot of $ \perp$ from the point $ P(2, 3, -8)$ on line (1)
We know the direction ratios of any line segement $PQ$ is given by $(x_2-x_1),(y_2-y_1),(z_2-z_1)$
The direction cosines of $ \overline{PQ} $ is given by
$=(-2\lambda+4-2), (6\lambda-3), (-3\lambda+1+8)$
$=( -2\lambda+2, 6 \lambda-3, -3\lambda+9)$
Now $Q$ is the foot of the perpendicular of the line (1)
(ie) $\overrightarrow {PQ}$ is the perpendicular to the line (i)
hence the sum of the product of this direction ratios is 0
$=(-2\lambda+2).(-2)+ (6\lambda-3).6+ (-3\lambda+9).(-3)=0$
$=4\lambda-4+ 36\lambda-18+9\lambda-27=0$
On simplifying we get,
$49\lambda-49=0$
Therefore $\lambda=1$
Step 3:
Substituting $\lambda$ in $Q$ we get the
$Q(0,3,6)$
Therefore the perpendicular distance $=\sqrt {0^2+3^2+6^2}$
$=\sqrt {45}$
answered Jun 11, 2013 by meena.p
 

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