Question #321485

Let u=(-5, -2, 4), v= (3, 6, -5) and w = (-7, 1, -8)

a) Calculate (u × v).w and hence find the volume of the parallelepiped with adjacent sides u, v, and w.

b) Show that vector (u - projwu) and w are orthogonal.

c) Use the cross product to find the angle between u and w.

1
Expert's answer
2022-05-24T23:22:14-0400

a)


(u×v)w=524365718(\vec {u}\times\vec {v})\vec {w}=\begin{vmatrix} -5 & -2 & 4 \\ 3 & 6 & -5 \\ -7 & 1 & -8 \\ \end{vmatrix}

=56518+23578+43671=-5\begin{vmatrix} 6 & -5 \\ 1 & -8 \end{vmatrix}+2\begin{vmatrix} 3 & -5 \\ -7 & -8 \end{vmatrix}+4\begin{vmatrix} 3 & 6 \\ -7 & 1 \end{vmatrix}

=5(48+5)+2(2435)+4(3+42)=-5(-48+5)+2(-24-35)+4(3+42)

=277=277

V=277V=277 cubic units


b)


projwu=uww2wproj_{\vec{w}}\vec{u}=\dfrac{\vec{u}\cdot\vec{w}}{|\vec{w}|^2}\vec{w}

uprojwu=uuww2w\vec {u}-proj_{\vec{w}}\vec{u}=\vec {u}-\dfrac{\vec{u}\cdot\vec{w}}{|\vec{w}|^2}\vec{w}

(uprojwu)w=(uuww2w)w(\vec {u}-proj_{\vec{w}}\vec{u})\cdot\vec {w}=(\vec {u}-\dfrac{\vec{u}\cdot\vec{w}}{|\vec{w}|^2}\vec{w})\cdot\vec {w}

=uwuww2w2=\vec{u}\cdot\vec{w}-\dfrac{\vec{u}\cdot\vec{w}}{|\vec{w}|^2}|\vec{w}|^2




=uwuw=0=\vec{u}\cdot\vec{w}-\vec{u}\cdot\vec{w}=0

Therefore (uprojwu)(\vec {u}-proj_{\vec{w}}\vec{u}) and w\vec{w} are orthogonal.


c)



u×w=ijk524718\vec {u}\times\vec {w}=\begin{vmatrix} \vec {i} & \vec {j} & \vec {k} \\ -5 & -2 & 4 \\ -7 & 1 & -8 \\ \end{vmatrix}

=i2418j5478+k5271=\vec {i}\begin{vmatrix} -2 & 4 \\ 1 & -8 \end{vmatrix}-\vec {j}\begin{vmatrix} -5 & 4 \\ -7 & -8 \end{vmatrix}+\vec {k}\begin{vmatrix} -5 & -2 \\ -7 & 1 \end{vmatrix}

=12i68j19k=12\vec {i}-68\vec {j}-19\vec {k}


u×w=(12)2+(68)2+(19)2|\vec{u}\times\vec{w}|=\sqrt{(12)^2+(-68)^2+(-19)^2}

=5129=\sqrt{5129}

u=(5)2+(2)2+(4)2=47|\vec{u}|=\sqrt{(-5)^2+(-2)^2+(4)^2}=\sqrt{47}

w=(7)2+(1)2+(8)2=114|\vec{w}|=\sqrt{(-7)^2+(1)^2+(-8)^2}=\sqrt{114}

sinθ=u×wuw=512947(114)0.9784\sin \theta=\dfrac{|\vec{u}\times\vec{w}|}{|\vec{u}||\vec{w}|}=\sqrt{\dfrac{5129}{47(114)}}\approx0.9784

θ=sin1512947(114)78°\theta=\sin^{-1}{\sqrt{\dfrac{5129}{47(114)}}}\approx78\degree


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