Question

Find the projection of u onto v
URGENT HELP

Answer 1

Given:

Two vectors are:

$u=\left <0,6\right>$

$v=\left <2,17\right>$

To find:

The projection of u onto v.

Solution:

Magnitude of a vector $v=\left <a,b\right>$ is:

$|v|=\sqrt{a^2+b^2}$

Dot product of two vector $v_1=\left <a_1,b_1\right>$ and $v_2=\left <a_2,b_2\right>$ is:

$v_1\cdot v_2=a_1a_2+b_1b_2$

Formula for projection of u onto v is:

$Proj_vu=\dfrac{u\cdot v}{|v|^2}v$

$Proj_vu=\dfrac{\left <0,6\right>\cdot \left <2,17\right>}{(\sqrt{2^2+17^2)^2}}\left <2,17\right>$

$Proj_vu=\dfrac{0\cdot 2+6\cdot 17}{4+289}\left <2,17\right>$

$Proj_vu=\dfrac{0+102}{293}\left <2,17\right>$

On further simplification, we get

$Proj_vu=\dfrac{102}{293}\left <2,17\right>$

$Proj_vu=\left <\dfrac{102}{293}\cdot 2,\dfrac{102}{293}\cdot 17\right>$

$Proj_vu=\left <\dfrac{204}{293},\dfrac{1734}{293}\right>$

Therefore, the projection of u onto v is $Proj_vu=\left <\dfrac{204}{293},\dfrac{1734}{293}\right>$.