Question

Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:

• A. \( \frac{1}{\sqrt{5}} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{1}{\sqrt{3}} \)
• D. \( \frac{1}{2\sqrt{3}} \)

Hint:

For finding the angle between two lines, use the dot product formula and ensure to calculate the direction ratios carefully.

Answer 1

The Correct Option is C

Let the direction ratios of the line \( L \) be \( \hat{i} - \hat{k} \). 
Step 1: The foot of the perpendicular \( N \) lies on the line \( L \), so let \( \overrightarrow{N} = (\lambda, 0, -\lambda) \). 
Step 2: The vector \( \overrightarrow{PN} = (\lambda - 1, 0 - 2, -\lambda + 1) = (\lambda - 1, -2, -\lambda + 1) \). The vector \( \overrightarrow{PQ} \) is parallel to the plane, so its direction ratios are proportional to the normal vector \( \mathbf{n} = \hat{i} + \hat{j} + 2\hat{k} \). 
Step 3: Now, calculate the cosine of the angle between the two lines \( PN \) and \( PQ \) using the dot product formula: \[ \cos \alpha = \frac{\overrightarrow{PN} \cdot \overrightarrow{PQ}}{|\overrightarrow{PN}| |\overrightarrow{PQ}|} \] After calculation, we find: \[ \cos \alpha = \frac{1}{\sqrt{3}} \]