# What is the projection of  (-4i + 3k) onto (-2i -j + 2k)?

Mar 4, 2017

The vector projection is $< - \frac{28}{9} , - \frac{14}{9} , \frac{28}{9} > ,$ the scalar projection is $\frac{14}{3}$.

#### Explanation:

Given $\vec{a} = < - 4 , 0 , 3 >$ and $\vec{b} = < - 2 , - 1 , 2 > ,$ we can find $p r o {j}_{\vec{b}} \vec{a}$, the vector projection of $\vec{a}$ onto $\vec{b}$ using the following formula:

$p r o {j}_{\vec{b}} \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}\right) \frac{\vec{b}}{|} \vec{b} |$

That is, the dot product of the two vectors divided by the magnitude of $\vec{b}$, multiplied by $\vec{b}$ divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide $\vec{b}$ by its magnitude in order to obtain a unit vector (vector with magnitude of $1$). You might notice that the first quantity is scalar, as we know that when we take the dot product of two vectors, the resultant is a scalar.

Therefore, the scalar projection of $a$ onto $b$ is $c o m {p}_{\vec{b}} \vec{a} = \frac{a \cdot b}{| b |}$, also written $| p r o {j}_{\vec{b}} \vec{a} |$.

We can start by taking the dot product of the two vectors.

$\vec{a} \cdot \vec{b} = < - 4 , 0 , 3 > \cdot < - 2 , - 1 , 2 >$

$\implies \left(- 4 \cdot - 2\right) + \left(0 \cdot - 1\right) + \left(3 \cdot 2\right)$

$\implies 8 + 0 + 6 = 14$

Then we can find the magnitude of $\vec{b}$ by taking the square root of the sum of the squares of each of the components.

$| \vec{b} | = \sqrt{{\left({b}_{x}\right)}^{2} + {\left({b}_{y}\right)}^{2} + {\left({b}_{z}\right)}^{2}}$

$| \vec{b} | = \sqrt{{\left(- 2\right)}^{2} + {\left(- 1\right)}^{2} + {\left(2\right)}^{2}}$

$\implies \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

And now we have everything we need to find the vector projection of $\vec{a}$ onto $\vec{b}$.

$p r o {j}_{\vec{b}} \vec{a} = \frac{14}{3} \cdot \frac{< - 2 , - 1 , 2 >}{3}$

$\implies \frac{14 < - 2 , - 1 , 2 >}{9}$

$\implies < - \frac{28}{9} , - \frac{14}{9} , \frac{28}{9} >$

The scalar projection of $\vec{a}$ onto $\vec{b}$ is just the first half of the formula, where $c o m {p}_{\vec{b}} \vec{a} = \frac{a \cdot b}{| b |}$. Therefore, the scalar projection is $\frac{14}{3}$.

Hope that helps!