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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the coordinates of the foot of the perpendicular drawn from the point $(0,2,3)$ on the line \(\large\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3} \). Also, find length of perpendicular.

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  • Length of the $\perp$ from point on the given line=$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Let $L$ be the foot of the $\perp$ drawn from the point $P(0,2,3)$ to the given line.
The coordinates of a general point on $\large\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ are given by
$\large\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=$$\lambda$
(i.e) $x=5\lambda-3$----(1)
$y=2\lambda+1$-----(1)
$z=3\lambda-4$------(1)
This is represented by $L$ in the figure.
Step 2:
$\therefore$ Direction ratios of PL are proportional to $5\lambda-3-0,2\lambda-1-2,3\lambda-4-3$
(i.e) $5\lambda-3,2\lambda-1,3\lambda-7$
The direction ratios of the given line are proportional to $5,2,3$
Since $PL$ is $\perp$ to the given line
$a_1a_2+b_1b_2+c_1c_2=0$
$5(5\lambda-3)+2(2\lambda-1)+3(3\lambda-7)=0$
On simplifying we get,
$\lambda=1$
Step 3:
Put $\lambda=1$ in equ(1) we get
$x=2,y=3$ and $z=-1$
$\therefore$ Length of the $\perp$ from $P$ on the given line is
$PL=\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$\;\;\;\;=\sqrt {(2-0)^2+(3-2)^2+(-1-3)^2}$
$\;\;\;\;=\sqrt {21}$units
answered Sep 27, 2013 by sreemathi.v
 

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