Question

From a point P \((a,b,c)\), perpendiculars PA and PB are drawn to XY plane and ZX plane respectively. If O is the origin, then the equation of plane OAB 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. \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 0\)

Hint:

Find the coordinates of A and B, then use the cross product as the plane normal.

Answer 1

The Correct Option is A

Step 1: Test the Options:
The plane contains $O$, $A(a,b,0)$ and $B(a,0,c)$. Plug each option.

Step 2: Point A:
For $A$: option (A) gives $1-1-0=0$. Option (B) gives $1-1+0=0$. Option (C) gives $1+1-0=2$. Option (D) gives $2$. So (C) and (D) are out.

Step 3: Point B:
For $B$: option (A) gives $1-0-1=0$. Option (B) gives $1-0+1=2$, so (B) is out. Only option (A) survives.

Final Answer:
Option (A). \[ \boxed{\text{(A) } \frac{x}{a}-\frac{y}{b}-\frac{z}{c}=0} \]