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

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 .

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Answers ( No )

  1. Emma
    0
    2021-09-14T22:38:57+00:00

    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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