1.
x = 3 + t , y = − 1 + t , z = 2 − t x=3+t, y=-1+t, z=2-t x = 3 + t , y = − 1 + t , z = 2 − t
− 2 x + 3 y − 4 z + 1 -2x+3y-4z+1 − 2 x + 3 y − 4 z + 1
= − 2 ( 3 + t ) + 3 ( − 1 + t ) − 4 ( 2 − t ) + 1 =-2(3+t)+3(-1+t)-4(2-t)+1 = − 2 ( 3 + t ) + 3 ( − 1 + t ) − 4 ( 2 − t ) + 1
= − 2 ( 3 + t ) + 3 ( − 1 + t ) − 4 ( 2 − t ) + 1 =-2(3+t)+3(-1+t)-4(2-t)+1 = − 2 ( 3 + t ) + 3 ( − 1 + t ) − 4 ( 2 − t ) + 1
= 5 t − 16 =5t-16 = 5 t − 16 Consider t = 0 t=0 t = 0
5 t − 16 = 5 ( 0 ) − 16 = − 16 ≠ 0 5t-16=5(0)-16=-16\not=0 5 t − 16 = 5 ( 0 ) − 16 = − 16 = 0 Therefore the line L does not lie on the plane − 2 x + 3 y − 4 z + 1 = 0. -2x+3y-4z+1=0. − 2 x + 3 y − 4 z + 1 = 0.
2.
n ⃗ = ⟨ − 3 , 2 , − 7 ⟩ \vec n=\langle-3, 2, -7\rangle n = ⟨ − 3 , 2 , − 7 ⟩ Point P = ( 3 , − 2 , 4 ) . P=(3,-2,4). P = ( 3 , − 2 , 4 ) .
The equation for the plane through the point P = ( 3 , − 2 , 4 ) P=(3,-2,4) P = ( 3 , − 2 , 4 ) that is perpendicular to the line ⟨ − 8 , 2 , 0 ⟩ + t ⟨ − 3 , 2 , 7 ⟩ \langle-8, 2, 0\rangle+t\langle-3, 2, 7\rangle ⟨ − 8 , 2 , 0 ⟩ + t ⟨ − 3 , 2 , 7 ⟩ is
− 3 ( x − 3 ) + 2 ( y − ( − 2 ) ) − 7 ( z − 4 ) = 0 -3(x-3)+2(y-(-2))-7(z-4)=0 − 3 ( x − 3 ) + 2 ( y − ( − 2 )) − 7 ( z − 4 ) = 0 or
3 x − 2 y + 7 z − 41 = 0 3x-2y+7z-41=0 3 x − 2 y + 7 z − 41 = 0
Comments
Dear Nhlanhla Muthwam the direction vector of the straight line <-8,2,0>+t<-3,2,-7> is <-3,2,-7>, the straight line is perpendicular to the plane, that is why n=<-3,2,-7>.
How to did you get n=⟨−3,2,−7⟩?