Question

Find the angle between the vectors
α
Ì
…
=
2
i
+
2
j
−
k
αÌ…=2i+2j−k and
\
e
t
a
=
6
i
−
3
j
+
2
k
\eta=6i−3j+2k.

Answer 1

Find the angle between the vectors  $\bar{\alpha}=2i+2j-k$ and $\bar{\beta}=6i-3j+2k$

The cosine of the angle $\theta$ between two vectors is equal to the dot product of this vectors divided by the product of vector magnitude

$$\cos \theta={\bar{\alpha}\cdot\bar{\beta} \over |\bar{\alpha}|\cdot|\bar{\beta}|}$$

$\bar{\alpha}=(2, 2, -1), \bar{\beta}=(6, -3, 2)$

$$\bar{\alpha}\cdot\bar{\beta}=2(6)+2(-3)+(-1)(2)=4$$

$$|\bar{\alpha}|=\sqrt{(2)^2+(2)^2+(-1)^2}=3$$

$$|\bar{\beta}|=\sqrt{(6)^2+(-3)^2+(2)^2}=7$$

Hence

$$\cos \theta={4 \over 3\cdot7}={4 \over 21}$$

$$\theta=\arccos{{4 \over 21}}\approx79.02\degree$$