Question

Find the intersection of the line and plane:

3y−x+z=−4,r(t)=⟨−1,2,1⟩+t⟨−2,−3,1⟩

3y−x+z=−4,r(t)=⟨−1,2,1⟩+t⟨−2,−3,1⟩

P=(

P=(  , , )

Answer 1

$r(t)$

$=⟨-1,2,1⟩+t⟨-2,-3,1⟩$

$=⟨-1,2,1⟩+⟨-2t,-3t,t⟩$

Adding, this gives

$=⟨-1-2t,2-3t,1+t⟩$

$\implies x=-1-2t$

$y=2-3t$

$z=1+t$

Substituting $x,y,z$ into $3y-x+z$, we have

$3(2-3t)-(-1-2t)+(1+t)=-4$

$\implies6-9t+1+2t+1+t=-4$

$\implies-9t+2t+t+6+1+1=-4$

$\implies-6t+8=-4$

$\implies-6t=-4-8$

$\implies-6t=-12$

Dividing both sides by $-6$ .This gives

$t=\frac{-12}{-6}=2$

Substituting $t$ into $⟨-1-2t,2-3t,1+t⟩$

We have

$⟨−1−2(2),2−3(2),1+2⟩$

$P=⟨3,-4,3⟩$