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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the perpendicular distance of the points (2,3,4) from the line \( \large\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}\), Also, find the coordinates of the foot of the perpendicular.

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  • Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
The given line $l$ is
$\large\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
$\Rightarrow\large\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$
Any point on $Q$ on the line $l$ is given by
$\Rightarrow\large\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$$=\lambda$
$x=-2\lambda+4,y=6\lambda,z=-3\lambda+1$
Step 2:
Direction ratios of PQ are
$-2\lambda+4-2,6\lambda-3,-3\lambda+1-4$
Or $-2\lambda+2,6\lambda-3,-3\lambda-3$
Since $PQ\perp l$
$a_1a_2+b_1b_2+c_1c_2=0$
$-2(-2\lambda+2)+6(6\lambda-3)-3(-3\lambda-3)=0$
$\Rightarrow 49\lambda-13=0$
$\Rightarrow \lambda=\large\frac{13}{49}$
Step 3:
Hence the coordinates of $Q$ are
$x=(-2)(\large\frac{13}{49}$$)+4$
$\;\;=\large\frac{170}{49}$
$y=(6)(\large\frac{13}{49})$
$\;\;=\large\frac{78}{49}$
$x=(-3)(\large\frac{13}{49}$$)+4$
$\;\;=\large\frac{10}{49}$
Step 4:
The distance PQ is given by
$\sqrt{(2-\large\frac{170}{49})^2+(3-\large\frac{78}{49})^2+(4-\large\frac{10}{49})^2}$
$\sqrt{(\large\frac{72}{49})^2+(\large\frac{115}{49})^2+(\large\frac{186}{49})^2}$
On simplifying we get,
$\Rightarrow 4$units
answered Oct 2, 2013 by sreemathi.v
 

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