# What is the projection of <3, -6, 2> onto <1, 1, 1>?

Mar 5, 2017

The vector projection is $< - \frac{1}{3} , - \frac{1}{3} , - \frac{1}{3} >$, the scalar projection is $- \frac{\sqrt{3}}{3}$.

#### Explanation:

Given $\vec{a} = < 3 , - 6 , 2 >$ and $\vec{b} = < 1 , 1 , 1 > ,$ 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.

$\vec{a} \cdot \vec{b} = < 3 , - 6 , 2 > \cdot < 1 , 1 , 1 >$

$\implies \left(3 \cdot 1\right) + \left(- 6 \cdot 1\right) + \left(2 \cdot 1\right)$

$\implies 3 - 6 + 2 = - 1$

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.

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

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

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

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

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

$\implies \frac{- < 1 , 1 , 1 >}{3}$

$\implies < - \frac{1}{3} , - \frac{1}{3} , - \frac{1}{3} >$

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 $- \frac{1}{\sqrt{3}}$. You can rationalize the denominator to yield $- \frac{\sqrt{3}}{3}$ equivalently.

Hope that helps!