Three Dimensional Geometry · Class 12 · CBSE · Science Mathematics
Find the equation of the plane passing through the point (1, 2, -1) and perpendicular to the line whose direction ratios are 2, -1 and 2.
Practice question from Three Dimensional Geometry (Class 12 · CBSE · Science Mathematics), with the correct answer and a step-by-step solution.
medium · Subjective · Q-81C23A
Find the equation of the plane passing through the point (1, 2, -1) and perpendicular to the line whose direction ratios are 2, -1 and 2.
Answer: A plane perpendicular to the given line has normal vector (2, -1, 2). Hence its equation through (1, 2, -1) is 2(x - 1) - (y - 2) + 2(z + 1) = 0, or 2x - y + 2z + 2 = 0.
Solution
Since the plane is perpendicular to the line, the line's direction ratios serve as the normal vector of the plane. Substituting the given point in the point-normal form gives , which simplifies to .
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