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: Find A and B:
The foot of the perpendicular from \(P(a,b,c)\) to the XY plane is \(A(a,b,0)\). The foot on the ZX plane is \(B(a,0,c)\).

Step 2: Normal Vector:
The plane passes through \(O\), so its normal is \(\overrightarrow{OA}\times\overrightarrow{OB}\):
\[ (a,b,0)\times(a,0,c)=(bc-0,\ 0\cdot a-ac,\ 0-ab)=(bc,\,-ac,\,-ab) \]

Step 3: Equation:
\(bc\,x-ac\,y-ab\,z=0\). Divide by \(abc\):
\[ \frac xa-\frac yb-\frac zc=0 \]

Step 4: Check:
Substitute \(A(a,b,0)\): \(1-1-0=0\). Substitute \(B(a,0,c)\): \(1-0-1=0\). Both lie on the plane, so option (A) is correct. The other options fail one of these checks.

Final Answer:
The plane is \(\dfrac xa-\dfrac yb-\dfrac zc=0\), option (A). \[ \boxed{\text{(A) } \frac{x}{a}-\frac{y}{b}-\frac{z}{c}=0} \]