Question

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 - proj_wu) and w are orthogonal.

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

Answer 1

a)

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

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

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

$$=277$$

$V=277$ cubic units

b)

$$proj_{\vec{w}}\vec{u}=\dfrac{\vec{u}\cdot\vec{w}}{|\vec{w}|^2}\vec{w}$$

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

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

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

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

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

c)

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

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

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

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

$$=\sqrt{5129}$$

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

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

$$\sin \theta=\dfrac{|\vec{u}\times\vec{w}|}{|\vec{u}||\vec{w}|}=\sqrt{\dfrac{5129}{47(114)}}\approx0.9784$$

$$\theta=\sin^{-1}{\sqrt{\dfrac{5129}{47(114)}}}\approx78\degree$$