Question

The distance of the point having position vector $\hat{i}-2\hat{j}-6\hat{k}$ from the straight line passing through the point $(2,-3,-4)$ and parallel to the vector $6\hat{i}+3\hat{j}-4\hat{k}$ is

• A. $\sqrt{\frac{340}{61$
• B. $\frac{341}{61}$
• C. $\frac{\sqrt{341{61}$
• D. $\sqrt{\frac{341}{61$

Hint:

Logic Tip: Alternatively, you can use the Pythagorean theorem method: $d = \sqrt{|\vec{\alpha} - \vec{a}|^2 - \left(\frac{(\vec{\alpha} - \vec{a}) \cdot \vec{b{|\vec{b}|}\right)^2}$. This avoids computing the cross product determinant, but requires calculating the dot product and two magnitudes instead!

Answer 1

The Correct Option is D

Concept:
The perpendicular distance of a point $\vec{\alpha}$ from a line passing through $\vec{a}$ and parallel to $\vec{b}$ is: \[ d = \frac{|(\vec{\alpha} - \vec{a}) \times \vec{b}|}{|\vec{b}|} \] 
Step 1: Identify the vectors.
\[ \vec{\alpha} = \hat{i} - 2\hat{j} - 6\hat{k}, \quad \vec{a} = 2\hat{i} - 3\hat{j} - 4\hat{k}, \quad \vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k} \] 
Step 2: Compute $\vec{\alpha} - \vec{a}$.
\[ \vec{\alpha} - \vec{a} = (1-2)\hat{i} + (-2+3)\hat{j} + (-6+4)\hat{k} \] \[ \vec{\alpha} - \vec{a} = -\hat{i} + \hat{j} - 2\hat{k} \] 
Step 3: Compute cross product.
\[ (\vec{\alpha} - \vec{a}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 6 & 3 & -4 \end{vmatrix} \] \[ = \hat{i}(1\cdot -4 - (-2)\cdot 3) - \hat{j}((-1)\cdot -4 - (-2)\cdot 6) + \hat{k}((-1)\cdot 3 - 1\cdot 6) \] \[ = \hat{i}(-4 + 6) - \hat{j}(4 + 12) + \hat{k}(-3 - 6) \] \[ = 2\hat{i} - 16\hat{j} - 9\hat{k} \] 
Step 4: Find magnitudes.
\[ |(\vec{\alpha} - \vec{a}) \times \vec{b}| = \sqrt{2^2 + (-16)^2 + (-9)^2} = \sqrt{4 + 256 + 81} = \sqrt{341} \] \[ |\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61} \] 
Step 5: Final distance.
\[ d = \frac{\sqrt{341}}{\sqrt{61}} = \sqrt{\frac{341}{61}} \] 
Final Answer:
\[ \boxed{\sqrt{\frac{341}{61}}} \]