Question

The coordinates of the foot of the perpendicular from the points $(1,-2,1)$ on the plane containing the lines, $\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}$ and $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}$ is:

• A. $(2,-4,2)$
• B. $(-1,2,-1)$
• C. $(0,0,0)$
• D. $(1,1,1)$

Answer 1

Answer: (C)

$(0,0,0)$

$\text{give the Point}P(1,-2,1)\text{,}$

$L=\frac{x+1}{6}=\frac{y-1}{7}=\frac{2-3}{8}$

$L_2=\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}$

$\therefore$The $DRs$ of $L_1$ is $(6,7,8)$ and $L_2$

is $(3,5,7)$

Equation of plane $=\left[\begin{array}{ccc}x+1& y-1& z-3\\ 6& 7& 8\\ 3& 5& 7\end{array}\right]$

$\Rightarrow x+1(49-40)-(y-1)(42-24)+2-3(30-21)$

$\Rightarrow 9x-18y+9z=0$

$9(x-2y+z)=0[\because ax+by+cz=0]$

Now, we know, the given point is

$P(1,-2,1)$

$\because Q(x,y,z)=\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

$=\frac{-a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$

$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-1}{1}=\frac{-(1+4+1)}{6}$

$\begin{array}{cc}x-1& =-1,y+2=-(-2),z=-1+1\\ x=& 0y=0z=0\end{array}$

$\therefore$ Option $c=0,0,0$