# How do you find the inner product and state whether the vectors are perpendicular given <-5,3>*<2,-3>?

Jan 6, 2017

Use the fact that the dot product of two vectors that are perpendicular is equal to zero.

#### Explanation:

The inner product (more commonly called the dot product) of two vectors of dimension n:

$\vec{A} = < {A}_{1} , {A}_{2} , \ldots , {A}_{n} > \mathmr{and} \vec{B} = < {B}_{1} , {B}_{2} , \ldots , {B}_{n} >$

Is computed as follows:

$\vec{A} \cdot \vec{B} = \left({A}_{1}\right) \left({B}_{1}\right) + \left({A}_{2}\right) \left({B}_{2}\right) + , \ldots , + \left({A}_{n}\right) \left({B}_{n}\right)$

The given vectors are of dimension 2:

Let $\vec{A} = < - 5 , 3 >$
Let $\vec{B} = < 2 , - 3 >$

$\vec{A} \cdot \vec{B} = \left({A}_{1}\right) \left({B}_{1}\right) + \left({A}_{2}\right) \left({B}_{2}\right)$

$\vec{A} \cdot \vec{B} = \left(- 5\right) \left(2\right) + \left(3\right) \left(- 3\right)$

$\vec{A} \cdot \vec{B} = - 19$

Another way to compute the dot product is:

$\vec{A} \cdot \vec{B} = | \vec{A} | | \vec{B} | \cos \left(\theta\right)$

where:

$| \vec{A} | = \sqrt{{A}_{1}^{2} + {A}_{2}^{2} + , \ldots , + {A}_{n}^{2}}$,

$| \vec{B} | = \sqrt{{B}_{1}^{2} + {B}_{2}^{2} + , \ldots , + {B}_{n}^{2}}$,

and $\theta$ is the angle between the two vectors.

If all that you care about is whether the vector are perpendicular, then you do not care about the magnitudes; you only care whether (or not) the dot product is zero.

If the two vectors were perpendicular, then $\theta = \frac{\pi}{2} \left(\mathmr{and} {90}^{\circ}\right)$ and the $\cos \left(\theta\right) = 0$. This means that the dot product would be zero.

Because the dot product of the given vectors is NOT zero, one can say that the vectors are not perpendicular.