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 ________.
Show Hint
A vector is parallel to a plane if its dot product with the plane's normal vector is zero. This is a common and vital condition in 3D geometry problems.
Step 1: Understanding the Concept:
We need to find two specific points \(N\) and \(Q\) on the given line \(L\).
\(N\) is found using the perpendicularity condition from \(P\).
\(Q\) is found using the condition that the vector \(\vec{PQ}\) is perpendicular to the normal of a given plane.
The angle is then found using the dot product of vectors \(\vec{PN}\) and \(\vec{PQ}\). Step 2: Key Formula or Approach:
1. General point on line: \((r, 0, -r)\).
2. Perpendicularity: \(\vec{A} \cdot \vec{B} = 0\).
3. \(\cos \alpha = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}| |\vec{v}|}\). Step 3: Detailed Explanation:
Let \(N\) be \((r, 0, -r)\) on line \(L\).
Vector \(\vec{PN} = (r - 1, -2, -r + 1)\).
Direction of line \(L\) is \((1, 0, -1)\).
Since \(\vec{PN} \perp L\): \((r - 1)(1) + (-2)(0) + (-r + 1)(-1) = 0\).
\(r - 1 + r - 1 = 0 \Rightarrow 2r = 2 \Rightarrow r = 1\).
So, \(N = (1, 0, -1)\) and \(\vec{PN} = (0, -2, 0)\). Magnitude \(|\vec{PN}| = 2\).
Let \(Q\) be \((\lambda, 0, -\lambda)\) on line \(L\).
Vector \(\vec{PQ} = (\lambda - 1, -2, -\lambda + 1)\).
The line \(PQ\) is parallel to the plane \(x + y + 2z = 0\), so \(\vec{PQ}\) is perpendicular to the normal \(\vec{n} = (1, 1, 2)\).
\((\lambda - 1)(1) + (-2)(1) + (-\lambda + 1)(2) = 0\).
\(\lambda - 1 - 2 - 2\lambda + 2 = 0 \Rightarrow -\lambda - 1 = 0 \Rightarrow \lambda = -1\).
So, \(Q = (-1, 0, 1)\) and \(\vec{PQ} = (-2, -2, 2)\). Magnitude \(|\vec{PQ}| = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}\).
