Question

Let \( \overrightarrow{OB} = \hat{i} + 2\hat{j} + 2\hat{k} \) and \( \overrightarrow{OA} = 4\hat{i} + 2\hat{j} + 2\hat{k} \). The distance of the point \( B \) from the straight line passing through \( A \) and parallel to the vector \( 2\hat{i} + 3\hat{j} + 6\hat{k} \) is

• A. \( \frac{7\sqrt{5}}{9} \)
• B. \( \frac{5\sqrt{7}}{9} \)
• C. \( \frac{3\sqrt{5}}{7} \)
• D. \( \frac{9\sqrt{5}}{7} \)
• E. \( \frac{9\sqrt{7}}{5} \)

Hint:

Distance from point to line in 3D = cross product magnitude divided by direction magnitude.

Answer 1

The Correct Option is B

Concept:
• Distance from point to line: \[ d = \frac{|\overrightarrow{AB} \times \vec{v}|}{|\vec{v}|} \] where \( \vec{v} \) is direction vector of line.

Step 1: Write coordinates.
\[ A = (4,2,2), \quad B = (1,2,2) \]

Step 2: Find vector \( \overrightarrow{AB} \).
\[ \overrightarrow{AB} = B - A = (-3,0,0) \]

Step 3: Direction vector of line.
\[ \vec{v} = (2,3,6) \]

Step 4: Compute cross product.
\[ \overrightarrow{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 0 \\ 2 & 3 & 6 \end{vmatrix} \] \[ = \hat{i}(0 - 0) - \hat{j}(-18 - 0) + \hat{k}(-9 - 0) \] \[ = (0,18,-9) \]

Step 5: Find magnitude.
\[ |\overrightarrow{AB} \times \vec{v}| = \sqrt{0^2 + 18^2 + (-9)^2} \] \[ = \sqrt{324 + 81} = \sqrt{405} = 9\sqrt{5} \]

Step 6: Magnitude of direction vector.
\[ |\vec{v}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Step 7: Distance.
\[ d = \frac{9\sqrt{5}}{7} \]

Step 8: Final Answer.
\[ \boxed{\frac{9\sqrt{5}}{7}} \]