91Ó°ÊÓ

u and v are vectors

$\text{Consider the vector}\mathbf{u}=⟨-2,3,5âŸ\copyright ,\mathbf{v}=⟨13,-5,8âŸ\copyright$

$\text{given by}\mathbf{u}\mathbf{.}\mathbf{v}=u_1{v}_1+u_2{v}_2+w_1{w}_2$

$\mathbf{u}·\mathbf{v}=-2\left(13\right)+3(-5)+5\left(8\right)$

$=-26-15+40$$=-1$

$\text{the vector}\mathbf{u}=⟨-2,3,5âŸ\copyright ,\mathbf{v}=⟨13,-5,8âŸ\copyright$

$u.v=-1$

$\text{Also,}\mathbf{u}=⟨-2,3,5âŸ\copyright$

$‖\mathbf{u}‖=\sqrt{(-2{)}^2+3^2+5^2}$

$=\sqrt{38}$

$\text{Also,}\mathbf{v}=⟨13,-5,8âŸ\copyright$

$‖\mathbf{v}‖=\sqrt{{13}^2+(-5{)}^2+8^2}$

$\sqrt{258}$

now

$\cos \mathrm{θ}=\frac{-1}{\sqrt{38}\sqrt{358}}$

$=\frac{-1}{2\sqrt{2451}}$

Therefore

$\mathrm{θ}={\cos }^{-1}\left(\frac{-1}{2\sqrt{2451}}\right)$

Hence, the angle between the vectors is$\mathrm{θ}={\cos }^{-1}\left(\frac{-1}{2\sqrt{2451}}\right)$

$\text{the vector}\mathbf{u}=⟨-2,3,5âŸ\copyright ,\mathbf{v}=⟨13,-5,8âŸ\copyright$

$u.v=-1$(calculated)

$‖\mathbf{u}‖=\sqrt{38}$(calculated)

$\text{Let}\mathbf{u}\text{be any non-zero vector, then the vector projection of}\mathbf{v}\text{onto}\mathbf{u}\text{is given by}$

${proj}_{\mathbf{u}}\mathbf{v}=\frac{\mathbf{u}·\mathbf{v}}{‖\mathbf{u}{‖}^2}\mathbf{u}\text{.Now, substitute the values in}{proj}_{\mathbf{u}}\mathbf{v}=\frac{\mathbf{u}·\mathbf{v}}{‖\mathbf{u}{‖}^2}\mathbf{u}$

${proj}_{\mathrm{u}}\mathbf{v}=\frac{-1}{(\sqrt{38}{)}^2}⟨-2,35âŸ\copyright$

$=\frac{-1}{38}⟨-2,3,5âŸ\copyright$

${\text{Hence, value ofproj}}_u\mathrm{v}\text{is}\frac{-1}{38}(-2,3,5)\text{.}$