Question

If $A$ and $B$ are foot of perpendicular drawn from point $Q(a,b,c)$ to the planes $yz$ and $zc$, then equation of plane through the points $A,B$ and $O$ is ............

• A. $\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$
• B. $\frac{x}{a}-\frac{y}{b}+\frac{z}{c}=0$
• C. $\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=0$
• D. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$

Answer 1

The correct option is A $\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$

$A$ and $B$ are the foot of perpendiculars drawn from $Q(a,b,c)$ to the planes $yz$ and $xz$

$\therefore A=(0,b,c)$ and $B=(a,0,c)$

Also, $O=(0,0,0)$ is the origin

To determine the equation of plane, let's determine the normal vector

$OA\times OB=\left|\begin{array}{ccc}i& j& k\\ 0& b& c\\ a& 0& c\end{array}\right|$

$=i(bc-0)-j(0-ac)+k(0-ab)$

$=bc\overrightarrow{i}+ac\overrightarrow{j}-ab\overrightarrow{k}$

Equation of the plane is $bcx+acy-abz+d=0$

Plug in any point to find the value of $d$

Let's substitute $O=(0,0,0)$ to find $d$

$⟹d=0$

$\therefore$ The equation becomes $bcx+acy-abz=0$

$⟹\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$ is the required equation of plane.