# What is the dot product of  (2i -3j + 4k) and (i + j -7k)?

${v}_{1} \cdot {v}_{2} = - 29$

#### Explanation:

I'm going to name these two vectors as ${v}_{1}$ and ${v}_{2}$, where ${v}_{1} = 2 i - 3 j + 4 k = < 2 , - 3 , 4 >$ and ${v}_{2} = i + j - 7 k = < 1 , 1 - 7 >$.

The dot product of two vectors is defined as ${v}_{1} \cdot {v}_{2} = | | {v}_{1} | | | | {v}_{2} | | \cos \left(\theta\right) = \left({i}_{1}\right) \left({i}_{2}\right) + \left({j}_{1}\right) \left({j}_{2}\right) + \left({k}_{1}\right) \left({k}_{2}\right)$.

We don't have an angle to use, so we'll calculate the dot product using by adding the products of the components.

Therefore, ${v}_{1} \cdot {v}_{2} = \left(2\right) \left(1\right) + \left(- 3\right) \left(1\right) + \left(4\right) \left(- 7\right) = 2 - 3 - 28 = - 29$.

Mar 5, 2018

The dot product is $= - 29$

#### Explanation:

The dot product of $2$ vectors

$\vec{a} = < {x}_{1} , {y}_{1} , {z}_{1} >$

and

$\vec{b} = < {x}_{2} , {y}_{2} , {z}_{2} >$

is

$\vec{a} . \vec{b} = < {x}_{1} , {y}_{1} , {z}_{1} > . < {x}_{2} , {y}_{2} , {z}_{2} >$

$= {x}_{1} {x}_{2} + {y}_{1} {y}_{2} + {z}_{1} {z}_{2}$

Here, we have

$< 2 , - 3 , 4 > . < 1 , 1 , - 7 = \left(2\right) \cdot \left(1\right) + \left(- 3\right) \cdot \left(1\right) + \left(4\right) \cdot \left(- 7\right)$

$= 2 - 3 - 28$

$= - 29$