Question

Let L be the line given by < 3, −1, 2 > + t < 1, 1, −1 >, for t ∈ R.

(4.1) Show that the above line L lies on the plane −2x + 3y − 4z + 1 = 0.

(4.2) Find an equation for the plane through the point P = (3, −2, 4) that is perpendicular to the (4) line < −8, 2, 0 > + t < −3, 2, −7 >

Answer 1

$<3,-1,2>+t<1,1,-1>\\ x=3+t, y=-1+t,\\ z=2-t$

(4.1)

To show that the Line lies on the plane −2x + 3y − 4z + 1 = 0

$-2(3+t)+3(-1+t)-4(2-t)+1\\ -6-2t-3+3t-8+4t+1=5t-16=0\\ t=\frac{16}{5}\\ x=3+t=\frac{31}{5}\\ y=-1+t=\frac{11}{5}\\ z=2-t=\frac{-6}{5}\\ -2x+3y-4z-1=-2\times \frac{31}{5}+3\times \frac{11}{5}-4\times \frac{-6}{5}\\+1=0$

Hence it lies on the plane -2x+3y-4z+1=0

(4.2)

$r_o=<3,-2,4>\\ x=-8-3t,\\ y=2+2t,\\ z=-7t\\ a=-3\\ b=2\\ c=-7\\ a(x-x_o)+b(y-y_o)+c(z-z_o)=0\\ -3(x-3)+2(y-(-2))-7(z-4)=0\\ 3x-2y+7z=41$