Question

Find plane at distance 5 from origin perpendicular to \(2\hat{i}+\hat{j}+2\hat{k}\)

• A. \( 2x+y+2z=10 \)
• B. \( 2x+y+2z=5 \)
• C. \( 2x+y+2z=15 \)
• D. \( x+y+z=5 \)
• E. \( 2x+2y+z=15 \)

Hint:

Distance from origin to plane uses \( \frac{|d|}{\sqrt{a^2+b^2+c^2}} \).

Answer 1

The Correct Option is C

Concept: Plane equation using normal vector.

Step 1: Normal vector: \[ \vec{n} = (2,1,2) \]

Step 2: Magnitude: \[ |\vec{n}| = \sqrt{4+1+4}=3 \]

Step 3: Distance formula: \[ \frac{|d|}{|\vec{n}|} = 5 \Rightarrow |d|=15 \]

Step 4: Plane: \[ 2x+y+2z = \pm 15 \]

Step 5: Choose positive: \[ 2x+y+2z=15 \]