# What is the projection of  (4 i + 4 j + 2 k) onto (i + j -7k)?

##### 1 Answer
Dec 26, 2016

The vector projection is $< - \frac{2}{17} , - \frac{2}{17} , \frac{14}{17} >$, the scalar projection is $\frac{- 2 \sqrt{51}}{17}$. See below.

#### Explanation:

Given $\vec{a} = \left(4 i + 4 j + 2 k\right)$ and $\vec{b} = \left(i + j - 7 k\right)$, 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, which can be written as $\vec{a} = < 4 , 4 , 2 >$ and $\vec{b} = < 1 , 1 , - 7 >$.

$\vec{a} \cdot \vec{b} = < 4 , 4 , 2 > \cdot < 1 , 1 , - 7 >$

$\implies \left(4 \cdot 1\right) + \left(4 \cdot 1\right) + \left(2 \cdot - 7\right)$

$\implies 4 + 4 - 14 = - 6$

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.

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

$| \vec{b} | = \sqrt{{\left(1\right)}^{2} + {\left(1\right)}^{2} + {\left(- 7\right)}^{2}}$

$\implies \sqrt{1 + 1 + 49} = \sqrt{51}$

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

$p r o {j}_{\vec{b}} \vec{a} = \frac{- 6}{\sqrt{51}} \cdot \frac{< 1 , 1 , - 7 >}{\sqrt{51}}$

$\implies \frac{- 6 < 1 , 1 , - 7 >}{51}$

$\implies - \frac{2}{17} < 1 , 1 , - 7 >$

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

$\implies < - \frac{2}{17} , - \frac{2}{17} , + \frac{14}{17} >$

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 $- \frac{6}{\sqrt{51}}$, which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{- 6 \sqrt{51}}{51} \implies \frac{- 2 \sqrt{51}}{17}$

Hope that helps!