Question 1.4: Given vectors A = 3ax + 4ay + az and B = 2ay - 5az, find the...

The angle \theta_{AB} can be found by using either dot product or cross product.

A \cdot B=\left(3, 4, 1\right)\cdot \left(0, 2, -5\right)= 0 + 8 -5=3

\left|A\right|=\sqrt{3^{2}+4^{2}+1^{2}}=\sqrt{26}

\left|B\right|=\sqrt{0^{2}+2^{2}+\left(-5\right) ^{2}}=\sqrt{29}

\cos \theta_{AB}=\frac{A\cdot B}{\left|A\right|\left|B\right|}=\frac{3}{\sqrt{\left(26\right)\left(29\right) }}=0.1092

\theta_{AB}=\cos^{-1}0.1092=83.73^{\circ}

Alternatively:

A\times B=\left | \begin{matrix} a_{x} & a_{y} & a_{z} \\ 3 & 4 & 1 \\ 0 & 2 & -5 \end{matrix} \right | =\left(-20-2\right)a_{x}+\left(0+15\right)a_{y}+\left(6-0\right)a_{z} =\left(-22,15,6\right)

\left|A\times B\right|=\sqrt{\left(-22\right)^{2}+15^{2}+6^{2} }=\sqrt{745}

\sin \theta_{AB}=\frac{\left|A\times B\right| }{\left|A\right| \left|B\right| }=\frac{\sqrt{745}}{\sqrt{\left(26\right)\left(29\right) }}=0.994

\theta_{AB}=\sin^{-1}0.994=83.73^{\circ}