Question

A line \(L_1\) passing through the point \(A\) with position vector \( \vec{a} = 4\hat{i} + 2\hat{j} + 2\hat{k} \) is parallel to the vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \). The length of the perpendicular drawn from a point \(P\) with position vector \( \vec{p} = \hat{i} + 2\hat{j} + 3\hat{k} \) to \(L_1\) is

• A. 0
• B. \( \sqrt{15} \)
• C. \( 2\sqrt{3} \)
• D. \( \sqrt{10} \)

Hint:

Distance from a point to a line in vector form is given by \( \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|} \), where \( \vec{d} \) is the direction vector.

Answer 1

The Correct Option is D

Step 1: Write position vectors.
Point on line:
\[ \vec{A} = (4,2,2) \]
Direction vector of line:
\[ \vec{b} = (2,3,6) \]
Point \(P\):
\[ \vec{P} = (1,2,3) \]

Step 2: Find vector \( \vec{AP} \).
\[ \vec{AP} = \vec{P} - \vec{A} = (1-4,\ 2-2,\ 3-2) \]
\[ \vec{AP} = (-3,0,1) \]

Step 3: Use formula for perpendicular distance.
Distance from point to line is:
\[ \text{Distance} = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}. \]

Step 4: Find cross product \( \vec{AP} \times \vec{b} \).
\[ \vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -3 & 0 & 1 2 & 3 & 6 \end{vmatrix} \]
\[ = \hat{i}(0\cdot6 - 1\cdot3) - \hat{j}(-3\cdot6 - 1\cdot2) + \hat{k}(-3\cdot3 - 0\cdot2) \]
\[ = \hat{i}(-3) - \hat{j}(-18 - 2) + \hat{k}(-9) \]
\[ = -3\hat{i} + 20\hat{j} - 9\hat{k}. \]

Step 5: Find magnitude of cross product.
\[ |\vec{AP} \times \vec{b}| = \sqrt{(-3)^2 + 20^2 + (-9)^2} \]
\[ = \sqrt{9 + 400 + 81} = \sqrt{490}. \]

Step 6: Find magnitude of direction vector.
\[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7. \]

Step 7: Compute distance.
\[ \text{Distance} = \frac{\sqrt{490}}{7} = \sqrt{10}. \]
Final Answer:
\[ \boxed{\sqrt{10}} \]