Question #1f88b

1 Answer
Mar 30, 2017

The vector projection is $= < \frac{2}{9} , \frac{4}{9} , \frac{4}{9} >$
The scalar projection is $= \frac{2}{3}$

Explanation:

The vector projection of $\vec{a}$ over $\vec{b}$ is

$= \frac{\vec{a} . \vec{b}}{| | \vec{b} | |} ^ 2 \cdot \vec{b}$

$\vec{b} = < 1 , 2 , 2 >$

$\vec{a} = < 2 , - 1 , 1 >$

The dot product is

$\vec{a} . \vec{b} = < 2 , - 1 , 1 > . < 1 , 2 , 2 \ge 2 - 2 + 2 = 2$

The modulus of $\vec{b}$ is

$= | | < 1 , 2 , 2 > | | = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

The vector projection is

$= \frac{2}{9} \cdot < 1 , 2 , 2 >$

The scalar projection is

$= \frac{\vec{a} . \vec{b}}{|} | \vec{b} | |$

$= \frac{2}{3}$