Question

If from a point $Q(a,b,c)$ perpendiculars $QA$ and $QB$ are drawn to the $YZ$ and $ZX$ planes respectively, then find the vector equation of the plane $OAB$.

Answer 1

Point $Q(a,b,c)$ is given.

Perpendiculars from $Q$ are dropped n $YZ$ and $ZX$ planes, having foot as $AandB$.

Coordinates of $A$ are $(0,b,c)$ and $b$ are $(a,0,c)$

Therefore,

$O=(0,0,0)$

$A=(0,b,c)$

$B=(a,0,c)$

$P=(x,y,z)$

$\overrightarrow{OA}=b\hat{j}+c\hat{k}$

$\overrightarrow{OB}=a\hat{i}+c\hat{k}$

Normal vector for the plane $OAB$ is

$\overrightarrow{OA}\times \overrightarrow{OB}=(b\hat{j}+c\hat{k})\times (a\hat{i}+c\hat{k})$

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

Equation of plane:

$\left(\overrightarrow{OP}\right)\cdot \overrightarrow{n}=0$

$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k})\cdot (bc\hat{i}+ac\hat{j}-ab\hat{k})$=0

$\Rightarrow bcx+acy-abz=0$