Question

If from a point $P(a,b,c)$ perpendiculars are drawn to the $YZ$ and $ZX$ planes, prove that the equation of the planes through $OAB$ is $x/a+y/b-z/c=0$.

Answer 1

Given that $A$ is the foot of the perpendicular from $P(a,b,c)$ on $YZ$ plane.

$\therefore A=(0,b,c)$

and $B$ is the foot of the perpendicular from $P(a,b,c)$ on $ZX$ plane.

$\therefore A=(a,0,c)$

Normal vector $\overrightarrow{n}$ to the plane $OAB$ is $(0,b,c)\times (a,0,c)$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& b& c\\ a& 0& c\end{array}\right|$

$=(bc,ac,-ab)$

$\Rightarrow \overrightarrow{n}=(\frac{1}{a},\frac{1}{b},\frac{-1}{c})$

Since the plane passes through origin $O$, its vector eqn. is given by

$\overrightarrow{r}(\frac{\hat{i}}{a}+\frac{\hat{j}}{b}-\frac{\hat{k}}{c})=0$

$(x\hat{i}+y\hat{j}+\hat{k}).(\frac{\hat{i}}{a}+\frac{\hat{j}}{b}-\frac{\hat{k}}{c})=0$

$\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$ isthe equation of the planes through $OAB$