Find the line through $P = (2,4,7) $ that intersects and is perpendicular to the line : $x = -1 + t, y = -2 + 3t, z = 4 $

Question

Find the line through $P = (2,4,7) $ that intersects and is perpendicular to the line : $x = -1 + t, y = -2 + 3t, z = 4 $

I set up a system of equations and obtained :

$-1 + t = x_0 +2 \\ -2 + 3t = y_0 + 4 \\ 4 = z_0 + 7$

Solving for each and putting them I obtain the vector $ v_1 = <t -3, 3t-6, -3>$ I then set the direction vector as $ v_2 =<1,3,4>$

Since $v_1 \cdot v_2 = 0 $ means they are orthogonal I obtain :

$<t -3, 3t-6, -3> \cdot <1,3,4> = 0 $

Multiplying out and solving for t I get $t = 10/21$

Would I now just substitute my given value for t back into $t_1$ ?

Answer 1

You are almost done! You just need to find the right value for $t$.

Let $Q_t:=(-1 + t,-2 + 3t,4)$ then the vector $Q_t-P=(t−3,3t−6,−3)$ (that you have already found) should be perpendicular to the direction vector of the given line $Q_1-Q_0=(1,3,0)$. Find such $t$. Then the desired line is $$s\to P+s\cdot (Q_t-P).$$

Answer 2

1. The plane $P_1$ which includes the line $$\ell_0: x = -1 + t, y = -2 + 3t, z = 4 $$ and the point $p=(2,4,7)$ is $$\boxed{3x-y-z=-5}$$ 2. The plane $P_2$ perpendicular to the line $\ell_0$ psses through point $p=(2,4,7)$ is $$\boxed{x+3y=14}$$ 3. Intersection of planes $P_1$ and $P_2$ is desired line that is $$\boxed{\dfrac{x-14}{-3}=y=\dfrac{-z+47}{10}}$$