Problem

Given two planes in 3D space, Plane 1: 3x−4y+2z=12 and Plane 2: x+2y−z=5. Find the intersection of the line that passes through the origin and is perpendicular to Plane 1 with Plane 2.

Answer

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Answer

Step 4: Substitute t=−53 back into the equations of the line to get the intersection point. x=3t=−5, y=−4t=203, z=2t=−103.

Steps

Step 1 ：Step 1: Find the normal vector of Plane 1. Since the coefficients of x, y, and z in the equation of a plane give the normal vector, the normal vector of Plane 1 is n→1=(3,−4,2).

Step 2 ：Step 2: Since the line is perpendicular to Plane 1 and passes through the origin, its parametric equations are x=3t, y=−4t, and z=2t.

Step 3 ：Step 3: Substitute the equations of the line into the equation of Plane 2 to find the intersection point. 3t+2(−4t)−2t=5 => −3t=5 => t=−53.

Step 4 ：Step 4: Substitute t=−53 back into the equations of the line to get the intersection point. x=3t=−5, y=−4t=203, z=2t=−103.

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