# Let L1 be the line passing through the points Q1=(4, 2, −3) and Q2=(0, −2, 3). Find a value of k so the line L2 passing through the point P1 = P1(−11, 2, k) with direction vector →d=[3, −2, −3]T intersects with L1

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1. Direction of L1 = <4,4,-6> or reduced to <2,2,-3>
direction of L2 = <3,-2,-3> , that was given

If L1 and L2 intersect they must lie on the same plane.
The normal of that plane is the cross product of <2,2,-3> with <3,-2,-3>

Using whatever method you learned, that normal is <12,3,10>

so the equation of the plane is 12x + 3y + 10z = k
but the point (4, 2, −3) lies on it, so 48 + 6 - 30 = k ----> k = 24
( I could have used the point (0, −2, 3) to get 0 -6 + 30 = k , k = 24)

So the equation of the plane containing our two lines is
12x + 3y + 10z = 24
That also contains any point on either line, and (-11,2,k) is supposed to be on it, so
-132 + 6 + 10k = 24
10k = 150
k = 15

check my arithmetic

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Reiny
2. don't reduce, the rest should be fine

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