Question

Find the cosine of the angle between the vectors $\overrightarrow{a}=i+j+k$ and $\overrightarrow{b}=2i-3j+2k$

Answer 1

We have

$\overrightarrow{a}\cdot \overrightarrow{b}=ab\cos \theta$

$\cos \theta =\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab}\right)$

$\overrightarrow{a}\cdot \overrightarrow{b}=a_x{b}_x+a_y{b}_y+a_c{b}_c$

$=\left(1\right)\left(2\right)+\left(1\right)(-3)+\left(1\right)\left(2\right)=1$

$a=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$b=\sqrt{2^2+(-3{)}^2+2^2}=\sqrt{1}7$

Therefore, we have

$\cos \theta =\left(\frac{1}{\sqrt{3}\sqrt{1}7}\right)=0.14$