<3,−1,2>+t<1,1,−1>
x=3+t,y=−1+t,z=2−t Show that the Line lies on the plane
−2(3+t)+3(−1+t)−4(2−t)+1=0
−6−2t−3+3t−8+4t+1=0
5t=16
t=3.2
x=6.2,y=2.2,z=−1.2 Let t=0
x=3,y=−1,z=2 Check whether the point (3,−1,2) lies on the plane
−2(3)+3(−1)+4(2)+1=0,True Therefore the line L lies on the plane -2x + 3y -4z + 1 = 0.
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