Question

The vector equation of the plane which is at a distance of \(5\) units from the origin and normal to the vector \(2\hat{i}+\hat{j}-2\hat{k}\) is

• A. \(\overset{⃗}{r}\cdot (2\hat{i}+\hat{j}-2\hat{k}) = 12\)
• B. \(\overset{⃗}{r}\cdot (2\hat{i}+\hat{j}-2\hat{k}) = 15\)
• C. \(\overset{⃗}{r}\cdot (2\hat{i}+\hat{j}-2\hat{k}) = 9\)
• D. \(\overset{⃗}{r}\cdot (2\hat{i}+\hat{j}-2\hat{k}) = 18\)

Hint:

Divide the normal by its magnitude to get a unit normal, then \(\vec r\cdot\hat n = 5\).

Answer 1

The Correct Option is B

Step 1: Key Formula:
A plane at distance \(p\) from the origin with unit normal \(\hat n\) is \(\vec r\cdot\hat n = p\).

Step 2: Calculate:
\(\vec n = 2\hat i + \hat j - 2\hat k\), \(|\vec n| = \sqrt{4+1+4} = 3\). So \(\hat n = \frac{\vec n}{3}\).
\(\vec r\cdot\frac{\vec n}{3} = 5\), so \(\vec r\cdot\vec n = 15\).

Step 3: Other options:
The values \(12\), \(9\) and \(18\) correspond to distances \(4\), \(3\) and \(6\) from the origin, not \(5\).

Final Answer:
The plane is \(\vec r\cdot(2\hat i+\hat j-2\hat k) = 15\), option (B). \[ \boxed{\vec r\cdot(2\hat i+\hat j-2\hat k) = 15} \]