# What is the projection of <0, 1, 3> onto <0, 4, 4>?

Dec 24, 2016

The vector projection is $< 0 , 2 , 2 >$, the scalar projection is $2 \sqrt{2}$. See below.

#### Explanation:

Given $\vec{a} = < 0 , 1 , 3 >$ and $\vec{b} = < 0 , 4 , 4 >$, 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:

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

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

$\implies 0 + 4 + 12 = 16$

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.

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

$| \vec{b} | = \sqrt{{\left(0\right)}^{2} + {\left(4\right)}^{2} + {\left(4\right)}^{2}}$

$\implies \sqrt{0 + 16 + 16} = \sqrt{32}$

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

$p r o {j}_{\vec{b}} \vec{a} = \frac{16}{\sqrt{32}} \cdot \frac{< 0 , 4 , 4 >}{\sqrt{32}}$

$\implies \frac{16 < 0 , 4 , 4 >}{32}$

$\implies \frac{< 0 , 4 , 4 >}{2}$

$\implies < 0 , 2 , 2 >$

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 $\frac{16}{\sqrt{32}}$, which further simplifies to $2 \sqrt{2}$. I've shown the simplification below.

$\frac{16}{\sqrt{32}}$

$\implies \frac{16}{\sqrt{16 \cdot 2}}$

$\implies \frac{16}{4 \cdot \sqrt{2}}$

$\implies \frac{4}{\sqrt{2}}$

$\implies \frac{4 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$

$\implies \frac{4 \sqrt{2}}{2}$

$\implies 2 \sqrt{2}$

Hope that helps!