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 first need to find the coordinates of points $N$ and $Q$. $N$ is the foot of the perpendicular, so vector $\vec{PN}$ is orthogonal to the direction of line $L$. The line $PQ$ is parallel to the given plane, meaning the vector $\vec{PQ}$ is perpendicular to the normal vector of the plane. After finding $N$ and $Q$, we calculate the angle between vectors $\vec{PN}$ and $\vec{PQ}$.
Step 2: Key Formula or Approach:
Any point on line $L: \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} = t$ is given by $(t, 0, -t)$.
For perpendicularity: $\vec{u} \cdot \vec{v} = 0$.
Angle formula: $\cos\alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{|\vec{PN}| |\vec{PQ}|}$.
Step 3: Detailed Explanation:
Finding point N:
Let $N$ be $(t, 0, -t)$ on line $L$. The vector $\vec{PN} = (t-1)\hat{i} + (0-2)\hat{j} + (-t - (-1))\hat{k} = (t-1)\hat{i} - 2\hat{j} + (1-t)\hat{k}$. Line $L$ has direction vector $\vec{d_L} = \hat{i} - \hat{k}$. Since $PN \perp L$, $\vec{PN} \cdot \vec{d_L} = 0$: \[ (t-1)(1) + (-2)(0) + (1-t)(-1) = 0 \] \[ t - 1 - 1 + t = 0 \implies 2t = 2 \implies t = 1 \] Thus, $N = (1, 0, -1)$. The vector $\vec{PN} = (1-1)\hat{i} - 2\hat{j} + (1-1)\hat{k} = -2\hat{j}$. Magnitude $|\vec{PN}| = 2$.
Finding point Q:
Let $Q$ be another point on $L$, so $Q = (k, 0, -k)$. The vector $\vec{PQ} = (k-1)\hat{i} - 2\hat{j} + (1-k)\hat{k}$. The line $PQ$ is parallel to the plane $x + y + 2z = 0$, so $\vec{PQ}$ is perpendicular to the plane's normal vector $\vec{n} = \hat{i} + \hat{j} + 2\hat{k}$. Therefore, $\vec{PQ} \cdot \vec{n} = 0$: \[ (k-1)(1) + (-2)(1) + (1-k)(2) = 0 \] \[ k - 1 - 2 + 2 - 2k = 0 \] \[ -k - 1 = 0 \implies k = -1 \] Thus, $Q = (-1, 0, 1)$. The vector $\vec{PQ} = (-1-1)\hat{i} - 2\hat{j} + (1 - (-1))\hat{k} = -2\hat{i} - 2\hat{j} + 2\hat{k}$. Magnitude $|\vec{PQ}| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$.
Finding the angle $\alpha$:
We now find $\cos\alpha$ using vectors $\vec{PN}$ and $\vec{PQ}$: \[ \cos\alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{|\vec{PN}| |\vec{PQ}|} \] \[ \vec{PN} \cdot \vec{PQ} = (0)(-2) + (-2)(-2) + (0)(2) = 0 + 4 + 0 = 4 \] \[ \cos\alpha = \frac{4}{2 \times 2\sqrt{3}} = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}} \] Step 4: Final Answer:
The value of $\cos\alpha$ is $\frac{1}{\sqrt{3}}$. The correct option is (C).