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a_sh-v [17]
3 years ago
8

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 .
Mathematics
1 answer:
Genrish500 [490]3 years ago
8 0
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=53

Check:
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
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For equations like this, we want to isolate x in order to get the value of it.

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