# Perpendicular Vector Plane

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.

Thanks

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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. (en.wikipedia.org/wiki/Scalar_product) // 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.

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