# What is the projection of <2,7,1 > onto <3,-5,7 >?

Aug 18, 2017

The vector projection is $< - \frac{66}{83} , \frac{110}{83} , - \frac{154}{83} >$ and the scalar projection is $- \frac{22}{\sqrt{83}}$.

#### Explanation:

Given $\vec{a} = < 2 , 7 , 1 >$ and $\vec{b} = < 3 , - 5 , 7 > ,$ we can find $p r o {j}_{\vec{b}} \vec{a}$, the vector projection of $\vec{a}$ onto $\vec{b}$ using the following formula:

$\textcolor{\mathrm{da} r k b l u e}{p r o {j}_{\vec{b}} \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{\left\mid \vec{b} \right\mid}\right) \frac{\vec{b}}{\left\mid \vec{b} \right\mid}}$

• 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 (a 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 given by:

color(darkblue)(comp_(vecb)veca=(a*b)/(abs(b))

Which may also written as $\left\mid p r o {j}_{\vec{b}} \vec{a} \right\mid$.

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

$\vec{a} \cdot \vec{b} = < 2 , 7 , 1 > \cdot < 3 , - 5 , 7 >$

$\implies \left(2 \cdot 3\right) + \left(7 \cdot - 5\right) + \left(1 \cdot 7\right)$

$\implies 6 - 35 + 7 = - 22$

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.

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

$\left\mid \vec{b} \right\mid = \sqrt{{\left(3\right)}^{2} + {\left(- 5\right)}^{2} + {\left(7\right)}^{2}}$

$\implies \sqrt{9 + 25 + 49} = \textcolor{\mathrm{da} r k b l u e}{\sqrt{83}}$

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

$p r o {j}_{\vec{b}} \vec{a} = \frac{- 22}{\sqrt{83}} \cdot \frac{< 3 , - 5 , 7 >}{\sqrt{83}}$

$\implies \frac{- 22 < 3 , - 5 , 7 >}{83}$

$\implies < - \frac{66}{83} , \frac{110}{83} , - \frac{154}{83} >$

The scalar projection of $\vec{a}$ onto $\vec{b}$ is just the first half of the formula, as described above. Therefore, the scalar projection is $- \frac{22}{\sqrt{83}}$.

Hope that helps!