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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the distance of a point(2,4,-1) from the line $\large \frac{x-5}{1}\;=\frac{y-3}{4}\;=\frac{z-6}{-9}$

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  • To get distance between a point and line get foot of $ \perp$ of the point on the line and get the distance.
  • If two lines are $\perp$ then the dot product of their d.r. =0
Step 1
Let $ P(2, 4,-1)$ be the given point and the Line $ L \rightarrow \large\frac{x-5}{1}=\large\frac{y-3}{4}=\large\frac{z-6}{-9}= \lambda$
$\Rightarrow\:d.r.$ of the line $ L=(1,4,-9)$
Any point $Q(x,y,z)$ on the line L is given by $Q (\lambda+5, 4\lambda+3, -9\lambda+6)$
Let this point $Q$ be the foot of $\perp$ of the point P on the line $L$.
Step 2
d.r of $\overline{ PQ}$ the $ \perp$
$ = (\lambda+3, 4\lambda-1, -9\lambda+7)$
Step 3
PQ is $ \perp$ to the line L
$ \therefore\: \overline {PQ} . (1, 4, -9)=0$
$(\lambda+3, 4\lambda-1, -9\lambda+7).(1,4,-9)=0$
$ \Rightarrow ( \lambda+3)1+(4\lambda-1)4-9(-9\lambda+7)=0$
$\Rightarrow\:98 \lambda=64$
Step 4
Substituting $ \lambda$ in Q we get the foot of $ \perp Q$ as
Step 5
$ \perp$ distance $ \equiv \overline{PQ}=7$
answered Mar 6, 2013 by thanvigandhi_1
edited May 31, 2013 by rvidyagovindarajan_1

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