Answer
(a) $11i-4j-k$
(b) $\sqrt 6$
(c) $-1$
(d) $-3i-7j-5k$
(e) $3\sqrt {35}$
(f) $18$
(g) $96^{\circ}$
Work Step by Step
We are given $\vec u=i+j-2k$, $\vec v=3i-2j+k$, and $\vec w=0i+j-5k$, use the corresponding formulas:
(a) $2\vec u+3\vec v=(2\times1+3\times3)i+(2\times1+3(-2))j+(2(-2)+3\times1)k=11i-4j-k$
(b) $|\vec u|=\sqrt {(1)^2+(1)^2+(-2)^2}=\sqrt 6$
(c) $\vec u\cdot\vec v=1\times3+1(-2)-2\times1=-1$
(d) $\vec u\times\vec v=(1\times1-(-2)(-2))i+((-2)3-1\times1)j+(1(-2)-1\times3)k=-3i-7j-5k$
(e) $|\vec v\times\vec w|=|(10-1)i+(0+15)j+(3-0)k|=|(9)i+(15)j+(3)k|=\sqrt {(9)^2+(15)^2+(3)^2}=3\sqrt {35}$
(f) $\vec u\cdot(\vec v\times\vec w)=\vec u\cdot(9i+15j+3k)=1\times9+1\times15-2\times3=18$
(g) The angle between $\vec u$ and $\vec v$ can be found as $cos\theta=\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}=\frac{-1}{\sqrt 6 \sqrt {3^2+(-2)^2+1^2}}=\frac{-1}{\sqrt 6\sqrt {14}}=-\frac{\sqrt {21}}{42}$ and $\theta\approx96^{\circ}$
