# What is the projection of <6,5,3 > onto <2,-1,8 >?

Dec 30, 2016

#### Answer:

The vector projection is $< \frac{62}{69} , - \frac{31}{69} , \frac{248}{69} >$, the scalar projection is $\frac{31 \sqrt{69}}{69}$.

#### Explanation:

Given $\vec{a} = < 6 , 5 , 3 >$ and $\vec{b} = < 2 , - 1 , 8 >$, 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.

$\vec{a} \cdot \vec{b} = < 6 , 5 , 3 > \cdot < 2 , - 1 , 8 >$

$\implies \left(6 \cdot 2\right) + \left(5 \cdot - 1\right) + \left(3 \cdot 8\right)$

$\implies 12 - 5 + 24 = 31$

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.

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

$| \vec{b} | = \sqrt{{\left(2\right)}^{2} + {\left(- 1\right)}^{2} + {\left(8\right)}^{2}}$

$\implies \sqrt{4 + 1 + 64} = \sqrt{69}$

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

$p r o {j}_{\vec{b}} \vec{a} = \frac{31}{\sqrt{69}} \cdot \frac{< 2 , - 1 , 8 >}{\sqrt{69}}$

$\implies \frac{31 < 2 , - 1 , 8 >}{69}$

$= \frac{31}{69} < 2 , - 1 , 8 >$

You can distribute the coefficient to each component of the vector and write as:

$\implies < \frac{62}{69} , - \frac{31}{69} , \frac{248}{69} >$

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 $\frac{31}{\sqrt{69}}$, which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{31 \sqrt{69}}{69}$.

Hope that helps!