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: Plan:
Think of the constant on the right as (distance) times (magnitude of the normal).

Step 2: Steps:
For $\vec r\cdot\vec n = d$, the distance of the plane from the origin is $\frac{|d|}{|\vec n|}$. Here the distance is $5$ and $|\vec n| = 3$, so $d = 5\times3 = 15$.

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