The planes 1 : 2x - 3y + 2z = 4 2: 12x - 12y + 4z = 36 in R 3 are given. In this

Question

The planes 1 : 2x - 3y + 2z = 4 2: 12x - 12y + 4z = 36 in R 3 are given. In this question you must determine if the planes intersect and how. If they intersect in a line, then you are to find the equations of the line of intersection. Select which of the following statements is correct and provide the additional information where appropriate. The planes 1 and 2 are parallel to each other and never intersect. The planes 1 and 2 are the same plane. The planes 1 and 2 intersect in a line having parametric equations x = y = z =

Explanation / Answer

We first need to find a point on the line. P1 = 2x - 3y + 2z = 4 P2 = 12x - 12y + 4z = 36    = 3x - 3y + z = 9 We can set x = 0 in both equations 2x - 3y = 4 3x - 3y = 9 Solving for these: x = 13/5 and y = 2/5 So the point (13/5 , 2/5 , 0) is on the line. Since the line lies on both planes it is perpendicular to bothnormal vectors. Thus a vector v parallel to the line is givenby the cross product: v = n1 xn2 solving this v = 3i + 4j+0k The lines parametric equationsare: x = 13/5 + 3t y = 2/5 + 4t z = 0

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