Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two vectors, $\boldsymbol{a}$ and $\boldsymbol{b}$:

$\boldsymbol{a} = (3,0,-1)$

$\boldsymbol{b} = (3,7,9)$

I am trying to figure out if $\boldsymbol{a}$ lies on the plane perpendicular to $\boldsymbol{b}$. Not really sure how to approach the problem, any hints are welcomed.


share|cite|improve this question
if $a\cdot b =0$ then $a$ is in a plane perpendicular to $b$ – Maesumi Feb 15 '13 at 21:15
Vectors are perpendicular iff scalar product of them is 0. ( // In your case they are. – zaarcis Feb 16 '13 at 4:03

To find the equation of a plane you need the normal vector and a point on the plane. Because for a given vector there can be infinite number of planes normal to this vector. So I am gonna assume that your plane contains $b$ and $b$ is in the direction of the vector normal to the plane.

Now equation of a plane is given by $p^T x = c$ where $p$ is the unit vector normal to the plane and $x$ is any point in the plane. $c$ is the projection of any vector that has an end point in the plane on $a$. Now since $b$ lies in the plane we can find the projection of $b$ on itself which will be equal to $\frac{b.b}{|b|}$.

In your case: $p = \left[\begin{array}{c} 3/\sqrt{139}\\7/\sqrt{139}\\9/\sqrt{139}\\\end{array}\right]$, $x = \left[\begin{array}{c} x\\y\\z\\\end{array}\right]$ and $c = \sqrt{139}$.

Now if you calculate the equation of plane you will get $3x + 7y + 9z - 139 = 0$.

Substituting $a$ you will get -139 and hence the point $a$ doesn't lie in the plane.

This was the detailed method but if you calculate the dot product of $a$ and $b$ you will get zero which itself indicates that $a$ will not lie in the plane perpendicular to $b$. Although dot product is not a thorough technique and might not always work.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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