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3D Geometry - Lines and Planes
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The cartesian equation of the plane which is at a distance of 10 unite from the original and perpendicular to the vector i + 2j -2k is
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A
x+2y-2z = 30
💡 Explanation:
The equation of a plane at distance d from origin with normal vector n is r dot (n/|n|) = d. Here n = i + 2j - 2k, so |n| = sqrt(1+4+4) = 3. The equation is x + 2y - 2z = 10 * 3 = 30.