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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 $C(2,-3,-1)$.

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}$