# What is the projection of (3i + 2j - 6k) onto  (3i – 4j + 4k)?

Dec 26, 2016

The vector projection is $< - \frac{69}{41} , \frac{92}{41} , - \frac{92}{41} >$, the scalar projection is $\frac{- 23 \sqrt{41}}{41}$.

#### Explanation:

Given $\vec{a} = \left(3 i + 2 j - 6 k\right)$ and $\vec{b} = \left(3 i - 4 j + 4 k\right)$, 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 $\vec{a} = < 3 , 2 , - 6 >$ and $\vec{b} = < 3 , - 4 , 4 >$.

$\vec{a} \cdot \vec{b} = < 3 , 2 , - 6 > \cdot < 3 , - 4 , 4 >$

$\implies \left(3 \cdot 3\right) + \left(2 \cdot - 4\right) + \left(- 6 \cdot 4\right)$

$\implies 9 - 8 - 24 = - 23$

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.

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

$| \vec{b} | = \sqrt{{\left(3\right)}^{2} + {\left(- 4\right)}^{2} + {\left(4\right)}^{2}}$

$\implies \sqrt{9 + 16 + 16} = \sqrt{41}$

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

$p r o {j}_{\vec{b}} \vec{a} = \frac{- 23}{\sqrt{41}} \cdot \frac{< 3 , - 4 , 4 >}{\sqrt{41}}$

$\implies \frac{- 23 < 3 , - 4 , 4 >}{41}$

$\implies - \frac{23}{41} < 3 , - 4 , 4 >$

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

$\implies < - \frac{69}{41} , \frac{92}{41} , - \frac{92}{41} >$

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 $- \frac{23}{\sqrt{41}}$, which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{- 23 \sqrt{41}}{41}$.

Hope that helps!