# How do you find the inner product and state whether the vectors are perpendicular given <4,5,1>*<-1,-2,3>?

Jan 14, 2017

$- 11$, and not perpendicular

#### Explanation:

The inner product ( or scalar dot product) of wo vectors

$\vec{a} = \left(\begin{matrix}{a}_{1} \\ {a}_{2} \\ {a}_{3}\end{matrix}\right)$

$\vec{b} = \left(\begin{matrix}{b}_{1} \\ {b}_{2} \\ {b}_{3}\end{matrix}\right)$

is defined as:

$\vec{a} . \vec{b} = | \vec{a} | | \vec{b} | \cos \theta \text{ }$where $\theta$ is the angle bewteen the vectors.

and can be calculated using the result

$\vec{a} . \vec{b} = {a}_{1} {b}_{1} + {a}_{2} {b}_{2} + {a}_{3} {b}_{3}$

so for the vectors in this question we have;

$\vec{a} = \left(\begin{matrix}4 \\ 5 \\ 1\end{matrix}\right)$

$\vec{b} = \left(\begin{matrix}- 1 \\ - 2 \\ 3\end{matrix}\right)$

$\vec{a} . \vec{b} = \left(\begin{matrix}4 \\ 5 \\ 1\end{matrix}\right) . \left(\begin{matrix}- 1 \\ - 2 \\ 3\end{matrix}\right)$

$\vec{a} . \vec{b} = 4 \times \left(- 1\right) + 5 \times \left(- 2\right) + 1 \times 3$

$\vec{a} . \vec{b} = - 4 - 10 + 3 = - 11$

for the vectors to be perpendicular ie. $\theta = {90}^{0} , \text{ } \cos \theta = 0$

$\implies \vec{a} . \vec{b} = 0$

in this case $\vec{a} . \vec{b} \ne 0 , \therefore \text{ }$not perpendicular