Question

Find the angle between the vectors. (first find an exact expression and then approximate to the nearest degree.) a = 1, −4, 1 , b = 0, 3, −3

Answer 1

The formula of the dot product of two vectors:

$\overrightarrow{a}\circ\overrightarrow{b}=|\overrightarrow{a}|\cdot|\overrightarrow{b}|\cos\alpha\to\cos\alpha=\dfrac{\overrightarrow{a}\circ\overrightarrow{b}}{|\overrightarrow{a}|\cdot|\overrightarrow{b}|}$

We have:

$\overrightarrow{a}=[1,\ -4,\ 1]\\\\\overrightarrow{b}=[0,\ 3,\ -3]$

Calculate the dot product:

$\overrightarrow{a}\circ\overrightarrow{b}=1\cdot0+(-4)\cdot3+1\cdot(-3)=-15$

Calculate the lengths of vectors:

$|\overrightarrow{a}|=\sqrt{1^2+(-4)^2+1^2}=\sqrt{18}=\sqrt{9\cdot2}=\sqrt9\cdot\sqrt2=3\sqrt2\\\\|\overrightarrow{b}|=\sqrt{0^2+3^2+(-3)^2}=\sqrt{18}=\sqrt{9\cdot2}=\sqrt9\cdot\sqrt2=3\sqrt2$

Substitute:

$\cos\alpha=\dfrac{-15}{3\sqrt2\cdot3\sqrt2}=\dfrac{-15}{9\sqrt4}=-\dfrac{15}{18}\approx-0.8333\to\alpha\approx146^o$

Answer: Measure of the angle between the vectors a and b is equal 146°.