# What is the projection of <5,8,3 > onto <2,4,-2 >?

Feb 28, 2017

The vector projection is $< 3 , 6 , - 3 > ,$ the scalar projection is $3 \sqrt{6}$.

#### Explanation:

Given $\vec{a} = < 5 , 8 , 3 >$ and $\vec{b} = < 2 , 4 , - 2 > ,$ 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.

$\vec{a} \cdot \vec{b} = < 5 , 8 , 3 > \cdot < 2 , 4 , - 2 >$

$\implies \left(5 \cdot 2\right) + \left(8 \cdot 4\right) + \left(3 \cdot - 2\right)$

$\implies 10 + 32 - 6 = 36$

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.

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

$| \vec{b} | = \sqrt{{\left(2\right)}^{2} + {\left(4\right)}^{2} + {\left(2\right)}^{2}}$

$\implies \sqrt{4 + 16 + 4} = \sqrt{24}$

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

$p r o {j}_{\vec{b}} \vec{a} = \frac{36}{\sqrt{24}} \cdot \frac{< 2 , 4 , - 2 >}{\sqrt{24}}$

$\implies \frac{36 < 2 , 4 , - 2 >}{24}$

$= \frac{3}{2} < 2 , 4 , - 2 >$

$\implies < 3 , 6 , - 3 >$

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 $\frac{36}{\sqrt{24}}$. This could be simplified further if desired:

$\implies \frac{36}{2 \sqrt{6}}$

$\implies \frac{18}{\sqrt{6}}$

$\implies \frac{18 \sqrt{6}}{6}$

$\implies 3 \sqrt{6}$

Hope that helps!