## Find the equation of the plane with the given description. contains the lines r1(t) = 2, 1, 0 + t, 2t, 4t and r2(t) = 2, 1, 0 + 4t, t, 15t .

Question

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## Answers ( No )

The intersection point of the two lines is (2,1,0).

The respective direction vectors are

L1: <1,2,4>

L2: <4,1,15>

Since the normal vector of the required plane is perpendicular to both direction vectors, the normal vector of the plane is obtained by the cross product of L1 and L2.

i j k

1 2 4

4 1 15

=<30-4, 16-15, 1-8>

=<26, 1, -7>

We know that the plane must pass through (2,1,0), the equation of the plane is

26(x-2)+1(y-1)-7(z-0)=0

simplifying,

26x+y-7z=52+1+0=53

or

26x+y-7z=53Check:

Put points on L1 in the plane

26(t+2)+(2t+1)-7(4t+0)=53 ok

For L2,

26(4t+2)+(t+1)-7(15t+0)=53 ok