Let L be the line passing through the point P = (1, −2, −1) with direction vector →d = <5, 1, 3>, and let T be the plane defined by x + 4y − 2z = −28. Find the point Q where L and T intersect.
I'm not sure how to solve this problem. Would I find an arbitrary point on the plane first? e.g. (-28, 0, 0)
Any point on your line will be of the form $\left( 1 + 5 \lambda, -2 + \lambda, -1 + 3 \lambda \right)$. Now, this also satisfies the equation of the plane (and hence is in the intersection. So we have
$$1 + 5 \lambda + 4 \left( -2 + \lambda \right) -2 \left( -1 + 3 \lambda \right) = -28$$
On solving, we get $\lambda = - \dfrac{23}{3}$ and hence the point of intersection will be $\left( - \dfrac{112}{3}, - \dfrac{29}{3}, -24 \right)$.