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Given two planes:

3x + 1y + 2z = 1

3x + 2z = 0

A) Find the unique point P on the Y axis where the planes intersect

B) Find a unit vector u with positive first coordinate that is parallel to both planes: ___i + ___j + ___k

C) Use parts A and B to find a vector equation for the line of intersection of both planes: r(t) = ___i + ___j + ___k

Part A is the easy part. To see where each plane intersects the Y axis, substitute 0 for X and 0 for Z.

This gives you y = 1 for one of the planes, and 0 = 0 for the other plane. The y = 1 tells you that the first plane intersects the Y axis at (0,1,0). The 0 = 0 tells you that the second plane covers the entire Y axis, so it also passes through (0,1,0).

For part B, we take the coefficients and use them to set up a cross product:

i j k

3 1 2

3 0 2

Taking the determinant, we get i * (1*2-0*2) + j * (3*2-3*2) + k * (3*0-3*1) = 2i + 0j -3k. This vector has a length of √((2*2)+(-3*-3)) = √(13). So we have to divide by that to normalize it to a unit vector.

2 √(13) / 13 i + 0 j + 3 √(13) / 13 k

Putting the two together for the vector equation of the line, we get:

r(t) =

(0 + (2 √13/13)t) i

+

(1 + 0t)j

+

(0 + (3 √13/13)t) k

You can leave off the 0+ and the +0t if you want simpler expressions in the boxes.