Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 03/Solution 1

Consider connecting the two vectors into a right triangle similar the picture to the right

. To get the component of $\mathbf {x}$ onto $\mathbf {y}$, we follow like in the picture and take,

${\textrm {proj}}_{\mathbf {y} }\mathbf {x} =|\mathbf {x} |\cos(\theta ){\hat {\mathbf {y} }}$

where $\theta$ is the angle between the vectors $\mathbf {x}$ and $\mathbf {y}$ and ${\hat {\mathbf {y} }}$ is a unit vector in the direction of $\mathbf {y}$. We can rewrite $\cos(\theta )$ in terms of the dot (scalar) product as,

$\cos(\theta )={\frac {\mathbf {x} \cdot \mathbf {y} }{|\mathbf {x} ||\mathbf {y} |}}.$

Therefore, the projection of $\mathbf {x}$ onto $\mathbf {y}$ is

${\textrm {proj}}_{\mathbf {y} }\mathbf {x} ={\frac {\mathbf {x} \cdot \mathbf {y} }{|\mathbf {y} |}}{\hat {\mathbf {y} }}={\frac {\mathbf {x} \cdot \mathbf {y} }{|\mathbf {y} |^{2}}}\mathbf {y}$

where we have recognized that

${\hat {\mathbf {y} }}={\frac {\mathbf {y} }{|\mathbf {y} |}}.$

Computing the dot product we get,

$\mathbf {x} \cdot \mathbf {y} =0-3+2=-1$

and for the magnitude of $\mathbf {y}$,

$|\mathbf {y} |={\sqrt {1^{2}+1^{2}+1^{2}}}={\sqrt {3}}.$

Therefore we get that the projection of $\mathbf {x}$ onto $\mathbf {y}$ is,

${\textrm {proj}}_{\mathbf {y} }\mathbf {x} ={\frac {\mathbf {x} \cdot \mathbf {y} }{|\mathbf {y} |^{2}}}\mathbf {y} =-{\frac {1}{3}}\mathbf {y} =\left[-{\frac {1}{3}},-{\frac {1}{3}},-{\frac {1}{3}}\right].$