3D Geometry
Line and Plane
Grade 12
Question:
<p>Let \(L\) be a straight line passing through origin. Suppose that all the points on \(L\) are at constant distance from the two planes \(P_1: x + 3y - z + 1 = 0\) and \(P_2: 3x - y + z - 1 = 0\). Then which of the following points lie(s) on the line \(L\)?</p>
<p>(a) \((1, -2, -5)\)</p>
<p>(b) \((1, -2, 5)\)</p>
<p>(c) \((-1, -2, 5)\)</p>
<p>(d) \((-1, 2, 5)\)</p>
Step-by-Step Solution
Key Concept: A line through the origin has all its points equidistant from two planes if and only if the line lies in a plane that is equidistant from both given planes (i.e., lies on one of the two bisector planes of the dihedral angle formed by Pā and Pā).
Step 1: Find the two angle bisector planes of P_1 and P_2. The bisector planes satisfy: $\frac{x + 3y - z + 1}{\sqrt{1+9+1}} = \pm\frac{3x - y + z - 1}{\sqrt{9+1+1}}$ This gives: $\frac{x + 3y - z + 1}{\sqrt{11}} = \pm\frac{3x - y + z - 1}{\sqrt{11}}$ Step 2: Simplify both cases. Case 1: $x + 3y - z + 1 = 3x - y + z - 1$
$4y - 2z + 2 = 2x$
$x - 2y + z = 1$ ... (Bisector 1) Case 2: $x + 3y - z + 1 = -(3x - y + z - 1)$
$x + 3y - z + 1 = -3x + y - z + 1$
$4x + 2y = 0$
$2x + y = 0$ ... (Bisector 2) Step 3: Since L passes through origin, it lies on the bisector plane passing through origin. Bisector 1: $0 - 0 + 0 = 1$ (FALSE - doesn't pass through origin) Bisector 2: $0 + 0 = 0$ (TRUE - passes through origin) So L lies on plane: $2x + y = 0$, giving $y = -2x$ Step 4: The direction vector of L satisfies $y = -2x$ with z arbitrary. Points on L have form $(t, -2t, st)$ for parameters $t, s \in \mathbb{R}$. Check which options satisfy $2x + y = 0$: Any point $(a, -2a, c)$ lies on L. ā“ Answer: BD
Correct Answer: BD