0
$\begingroup$

I'm attempting one of the optional problems in my linear algebra textbook and I am very confused as I have never seen a problem in this form. Question is:

Find the vector form of the equation of the line in $\mathbb R^3$ that passes through the point $P = (-1,1,3)$ and is perpendicular to the plane with general equation $x-3y+2z=5$.

I know the normal vector $\overrightarrow n = [1,-3,2]$ from the general equation but i'm not sure how it helps me. Based on the question it would seem this would be parallel to line i'm trying to find?

My thought was to find 2 other points on the plane, say Q and R, by satisfying the general equation and use those points to find the direction vectors $\overrightarrow u = \vec PQ$ and $\overrightarrow v = \vec PR$. Then I would use those to form my vector equation $$\overrightarrow x = \overrightarrow p + s\overrightarrow u + t\overrightarrow v$$

Is this the proper way to solve this type of problem? Or would this only work if the line is parallel to the plane? The parallel/perpendicular to the plane concept has me really confused.

I just started this course and i'm struggling, so i'm doing my best to try and understand problems outside of the assigned homework (one which I cannot look up an answer for). If someone could give me the proper advice for how to approach these types of problems, I would be extremely grateful. Thanks.

$\endgroup$
  • $\begingroup$ The vector equation of a line passing through point $P(\overrightarrow a)$ and parallel to a vector $\overrightarrow b$ is given by $$\overrightarrow r = \overrightarrow a + k \overrightarrow b$$ $\endgroup$ – Shreyas S May 15 '17 at 5:45
1
$\begingroup$

the solution is: $$\vec{r}=\vec{v}t+\vec{OP}$$ in your case is: $$(x,y,z)=(1,-3,2)t+(-1,1,3)$$ god bless us

$\endgroup$
1
$\begingroup$

Hint. The parametric equation of a line passing through $(x_0,y_0,z_0)$ with direction $(v_x,v_y,v_z)$ is $$t\to(x_0+v_xt,y_0+v_y t,z_0+v_zt)$$

$\endgroup$
  • $\begingroup$ In this case would the normal vector also be my direction vector since the line I am looking for is parallel to normal vector? Or am I way off here? $\endgroup$ – FuegoJohnson May 15 '17 at 5:58
  • $\begingroup$ Yes, you are right! $\endgroup$ – Robert Z May 15 '17 at 6:00
  • $\begingroup$ Thanks! I think I understand now. So the equation I was trying to use above is for planes only and not lines I take it? $\endgroup$ – FuegoJohnson May 15 '17 at 6:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.