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:

In a 3D geometry problem involving a "foot of perpendicular" $N$ and another point $Q$ on the same line, always check if you can use right-triangle trigonometry. It is usually faster than using the dot product formula for the angle between two vectors.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
We need to find two points on line $L$: the foot of the perpendicular $N$ and the intersection point $Q$. The angle $\alpha$ is the angle at vertex $P$ in the triangle $PNQ$. Since $PN \perp L$, $\triangle PNQ$ is a right-angled triangle at $N$.

Step 2: Finding Point N and Point Q
1. Any point on $L$ is $(r, 0, -r)$. 2. For $N$: Vector $\vec{PN} = (r-1, -2, -r+1)$. Since $\vec{PN} \perp (1, 0, -1)$: $(r-1)(1) + (-2)(0) + (-r+1)(-1) = 0 \implies r-1 + r-1 = 0 \implies r=1$. So, $N = (1, 0, -1)$ and $\vec{PN} = (0, -2, 0)$. Length $PN = 2$. 3. For $Q$: $Q$ lies on $L$, so $Q = (k, 0, -k)$. $\vec{PQ} = (k-1, -2, -k+1)$. Since $PQ$ is parallel to the plane $x+y+2z=0$, $\vec{PQ}$ is perpendicular to the normal $(1, 1, 2)$: $(k-1)(1) + (-2)(1) + (-k+1)(2) = 0 \implies k-1 -2 -2k + 2 = 0 \implies -k - 1 = 0 \implies k = -1$. So, $Q = (-1, 0, 1)$ and $\vec{PQ} = (-2, -2, 2)$.

Step 3: Calculating \(\cos \alpha\)
In right-angled $\triangle PNQ$ (right-angled at $N$): $\cos \alpha = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PN}{PQ}$ 1. $PN = 2$ 2. $PQ = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$ 3. $\cos \alpha = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$

Step 4: Final Answer
The value of \( \cos \alpha \) is \( \frac{1}{\sqrt{3}} \).