Question

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

विकल्प

• ${\displaystyle \frac{x}{a}+\frac{y}{b}-\frac{z}{c}}$ = 0

• ${\displaystyle \frac{x}{a}-\frac{y}{b}+\frac{z}{c}}$ = 0

• ${\displaystyle \frac{x}{a}-\frac{y}{b}-\frac{z}{c}}$ = 0

• ${\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}}$ = 0

Answer 1

If A and B are foot of perpendicular drawn from point Q(a, b, c) to the planes yz and zx, then equation of plane through the points A, B and O is ${\displaystyle \underline{\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0}}$.

Explanation:

It is given that, the foot of perpendicular from point Q(a, b, c) to the yz plane is A(0, b, c) and the foot of perpendicular from point Q to the zx plane is B(a, 0, c).

Let the equation of plane passing through the point (0, 0, 0) be Ax + By + Cz = 0 ......(i)

As it is passing through the point A(0, b, c) and B(a, 0, c).

So, 0 + Bb + Cc = 0 and Aa + 0 + Cc = 0

${\displaystyle \Rightarrow }$ Cc = Bb and Cc = –Aa

Therefore, – Aa = – Bb = Cc = k

${\displaystyle \Rightarrow }$  A = ${\displaystyle -\frac{k}{a}}$, B = ${\displaystyle -\frac{k}{b}}$ and C = ${\displaystyle \frac{k}{c}}$

${\displaystyle -\frac{k}{a}x-\frac{k}{b}y+\frac{k}{c}z}$ = 0  ...[From equation (i)]

${\displaystyle \Rightarrow -\frac{x}{a}-\frac{y}{b}+\frac{z}{c}}$ = 0 or ${\displaystyle \frac{x}{a}+\frac{y}{b}-\frac{z}{c}}$ = 0