The line $l$ has an equation $r=\begin{pmatrix} 1\\ 2\\ -1 \end{pmatrix}+\lambda \begin{pmatrix} 2\\ 1\\3 \end{pmatrix}$ and equation of plane $p= r.\begin{pmatrix} 2\\ -1\\ -1 \end{pmatrix}$

i) show that line $l$ is parallel to plane $p$

I managed to solve this question by showing scalar product of direction vector of line and normal of plane is $0$

ii)A line $m$ lies in the plane $p$ and is perpendicular to $l$. The line $m$ passes through the point $(5,3,1)$ .Find equation of line $m$.

I know that the equation of the line will be in the form $\begin{pmatrix} 5\\ 3\\ 1 \end{pmatrix}+direction$

How to find the direction , Please assist , also line $m$ lies on plane and $l$ is parallel to plane , how can $m$ be perpendicular to $l$, Ps assist.

thank you, Arif

  • $\begingroup$ You people apparently are considering that lines in space can be perpendicular even if they don't intersect. Not everybody defines it that way. We in school required that the two lines first have to intersect and then their direction vectors must be perpendicular. You guys, it seems to be, require only the second condition. $\endgroup$
    – DonAntonio
    Commented May 25, 2016 at 7:24

1 Answer 1



Let $d$ be the direction of line $m$. Because line $m$ is in the plane, then $d$ is perpendicular on the normal of the plane.

Because line $m$ is perpendicular on $l$ then $d$ is perpendicular on the direction of $l$.

Take into account that perpendicular lines in space may not intersect.

For definitions see this link

  • $\begingroup$ Thanks, I understand cross product will give the direction of the line $m$. I always need help with this kind of sums. Any reference that can be helpful will be great. Arif $\endgroup$
    – Arif
    Commented May 25, 2016 at 9:11

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