a) Give parametric equation (point-direction form) of the line which lies on both of the planes:
x + y + z = 1 and −x + 2y + 10z = 2. What is the direction d of this line?
b) Let n1 and n2 be the normal vectors to the two given planes. Without actual computation,
describe the relationship between d and n1 × n2.
1
Expert's answer
2021-09-27T17:29:14-0400
ANSWER: a)The parametric equation of the line which lies on both of the planes is the line passing through P0=(0,1,0) and parallel to the vector d :x=1948t,y=1−19411t,z=1943t .d=1941⟨8,−11,3⟩,∥∥d∥∥=1
b)the vector d is perpendicular to both normal vector , d∥n1×n2
EXPLANATION:
a)
The point (a,b,c) lies on both of the planes, then {a+b+c=1−a+2b+10c=2 . Let c=0 , we have
{a+b=1−a+2b=2⇔{a+b=1−a+a+b+2b=3⇔{a=0b=1 . Let a=1 , then b+c=0 and −1+8c=2 . Thus the points (0,1,0),(1,−3/8,3/8) lie on both planes . Therefore the vector D=(1,−83,83)−(0,1,0)=⟨1,−811,83⟩ determines the direction
of the line of intersection of these planes .Since ∥∥D∥∥=1+64121+649=64194 , then d=∥∥D∥∥D=1948⟨1,−811,83⟩=1941⟨8,−11,3⟩,∥∥d∥∥=1. So, the parametric equations of the line passing through the point P0=(0,1,0) and parallel to d the vector d:x=1948t,y=1−19411t,z=1943t.
b)
The plane x+y+z=1 have normal vector n1=(1,1,1) and the plane−x+2y+10z=2 have normal vector n2=(−1,2,10). The line is on both planes and thus the line (and the vector d) is perpendicular to both normal vectors.
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