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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the co-ordinates of the foot of perpendicular drawn from the point $A(1,8,4)$ to the line joining the points $B(0,-1,3)$ and $C(2,-3,-1)$.

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Toolbox:
  • The cartesian equation of a line passing through the points $B(x_1,y_1,z_1)$ and $C(x_2,y_2,z_2)$ is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}=\large\frac{z-z_1}{z_2-z_1}$
Step 1:
The cartesian equation of a line passing through the points $B(0,-1,3)$ and $C(2,-3,-1)$ is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}=\large\frac{z-z_1}{z_2-z_1}$
Now substituting the values we get
$\large\frac{x-0}{2-0}=\frac{y+1}{-3+1}=\frac{z-3}{-1-3}$
$\Rightarrow \large\frac{x}{2}=\frac{x+1}{-2}=\frac{z-3}{-4}$
Step 2:
Let $L$ be the foot of the perpendicular from the point $A(1,8,4)$ to the given line.
The coordinates of the point $L$ on the line BC is given by
$\large\frac{x}{2}=\frac{x+1}{-2}=\frac{z-3}{-4}$$=\lambda$
(i.e) $x=2\lambda,y=-2\lambda-1,z=-4\lambda+3$
Step 3:
The direction ratios of AL is $(x_2-x_1),(y_2-y_1),(z_2-z_1)$
(i.e) $(2\lambda-1),(-2\lambda-1-8),(-4\lambda+3-4)$
(i.e) $(2\lambda-1),(-2\lambda-9),(-4\lambda-1)$
Direction ratios of the given lines are proportional to $(2,-2,-4)$
Since $AL$ is perpendicular to the given line BC
$\therefore a_1a_2+b_1b_2+c_1c_2=0$
(i.e) $2(2\lambda-1)+(-2)(-2\lambda-9)+(-4)(-4\lambda-1)=0$
Step 4:
On simplifying we get,
$24\lambda=20$
$\lambda=\large\frac{20}{24}$
$\lambda=\large\frac{5}{6}$
Step 5:
Hence the coordinates are
$x=2(\large\frac{5}{6})$
$y=-2(\large\frac{5}{6})$$-1$
$Z=-4(\large\frac{5}{6})$$+3$
$\Rightarrow x=\large\frac{5}{3}$$,y=\large\frac{-8}{3},$$,z=-\large\frac{1}{3}$
answered Oct 1, 2013 by sreemathi.v
 

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