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

Feb 15, 2016

$p r o {j}_{\vec{v}} \vec{u} = \left(- \frac{15}{11} i - \frac{10}{11} j + \frac{15}{11} k\right)$

#### Explanation:

To make it easier to refer to them, let's call the first vector $\vec{u}$ and the second $\vec{v}$. We want the project of $\vec{u}$ onto $\vec{v}$:

$p r o {j}_{\vec{v}} \vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{|} | \vec{v} | {|}^{2}\right) \cdot \vec{v}$

That is, in words, the projection of vector $\vec{u}$ onto vector $\vec{v}$ is the dot product of the two vectors, divided by the square of the length of $\vec{v}$ times vector $\vec{v}$. Note that the piece inside the parentheses is a scalar that tells us how far along the direction of $\vec{v}$ the projection reaches.

First, let's find the length of $\vec{v}$:

$| | \vec{v} | | = \sqrt{{3}^{2} + {2}^{2} + {\left(- 3\right)}^{2}} = \sqrt{22}$

But note that in the expression what we actually want is $| | \vec{v} | {|}^{2}$, so if we square both sides we just get $22$.

Now we need the dot product of $\vec{u}$ and $\vec{v}$:

$\vec{u} \cdot \vec{v} = \left(1 \times 3 + \left(- 2\right) \times 2 + 3 \times \left(- 3\right)\right) = \left(3 - 4 - 9\right) = \left(- 10\right)$

(to find the dot product we multiply the coefficients of $i , j \mathmr{and} k$ and add them)

Now we have everything we need:

proj_vec v vec u = ((vec u*vec v)/||vec v||^2)*vec v = (-10/22)(3i+2j−3k)
$= \left(- \frac{30}{22} i - \frac{20}{22} j + \frac{30}{22} k\right) = \left(- \frac{15}{11} i - \frac{10}{11} j + \frac{15}{11} k\right)$