Question

The distance of the point \( O(0,0,0) \) from the plane \( \overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 \) measured parallel to \( 2\hat{i} + 3\hat{j} - 6\hat{k} \) is?

• A. \( 35 \)
• B. \( 30 \)
• C. \( 25 \)
• D. \( 42 \)

Hint:

When measuring distance from a point to a plane along a given direction, parameterize the line and solve for \( \lambda \) using the plane equation. The final distance is obtained by scaling the magnitude of the direction vector.

Answer 1

The Correct Option is A

We are given the equation of the plane: \[ \overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 \] and we need to find the distance of the point \( O(0,0,0) \) from the plane, measured parallel to the direction \( 2\hat{i} + 3\hat{j} - 6\hat{k} \). Step 1: Find the normal to the plane
The normal to the given plane is: \[ \overrightarrow{N} = \hat{i} + \hat{j} + \hat{k}. \] Step 2: Find the equation of the line
The given direction \( \overrightarrow{D} = 2\hat{i} + 3\hat{j} - 6\hat{k} \) represents the direction along which we measure the distance. A general point on this line through \( O(0,0,0) \) is given by: \[ \overrightarrow{r} = \lambda (2\hat{i} + 3\hat{j} - 6\hat{k}). \] Step 3: Find intersection of line with plane
Substituting this point into the equation of the plane: \[ (2\lambda) + (3\lambda) + (-6\lambda) = 5 \] \[ 2\lambda + 3\lambda - 6\lambda = 5 \] \[ - \lambda = 5 \quad \Rightarrow \quad \lambda = -5. \] Step 4: Find the required distance
Since the displacement along the given direction is \( \lambda \) times the magnitude of the direction vector, \[ \text{Distance} = |\lambda| \times |\overrightarrow{D}|. \] \[ |\overrightarrow{D}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7. \] Thus, the required distance is: \[ |-5| \times 7 = 35. \]