3D Geometry
Line of intersection of planes
Grade 12
Question:
<p>The line of intersection of the planes, \(\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1\) and \(\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2\), is</p>
<p>\(\dfrac{x - \dfrac{4}{7}}{2} = \dfrac{y}{7} = \dfrac{z - \dfrac{5}{7}}{13}\)</p>
<p>\(\dfrac{x - \dfrac{6}{13}}{2} = \dfrac{y - \dfrac{5}{13}}{7} = \dfrac{z}{-13}\)</p>
<p>\(\dfrac{x - \dfrac{4}{7}}{2} = \dfrac{y}{-7} = \dfrac{z + \dfrac{5}{7}}{13}\)</p>
<p>\(\dfrac{x - \dfrac{6}{13}}{2} = \dfrac{y - \dfrac{5}{13}}{-7} = \dfrac{z}{-13}\)</p>
Step-by-Step Solution
Key Concept: The line of intersection of two planes is perpendicular to both normal vectors; find the direction vector using the cross product of the normal vectors, then find a point on the line by solving the plane equations.
Step 1: Find the direction vector The normal vectors are n_1 = 3 î - ĵ + k̂ and n_2 = î + 4 ĵ - 2 k̂ Direction vector d = n_1 × n_2 : d = | î ĵ k̂ | |3 -1 1| |1 4 -2| = î (2 - 4) - ĵ (-6 - 1) + k̂ (12 + 1) = -2 î + 7 ĵ + 13 k̂ Step 2: Find a point on the line Set z = 0 and solve: 3x - y = 1 ... (1) x + 4y = 2 ... (2) From (2): x = 2 - 4y Substitute in (1): 3(2 - 4y) - y = 1 6 - 12y - y = 1 → y = 5/13 x = 2 - 20/13 = 6/13 Point: (6/13, 5/13, 0) Step 3: Write the line equation Parametric form: r = (6/13) î + (5/13) ĵ + 0 k̂ + t(-2 î + 7 ĵ + 13 k̂ ) Or Cartesian form: (x - 6/13)/(-2) = (y - 5/13)/7 = z/13 ∴ Answer: A
Correct Answer: A