Question

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.)

Answer 1

Answer:

$a.b = (1*0) + (-4*2) +(1*-2)= -10$

And replacing we got:

$cos \theta = \frac{-10}{\sqrt{18} \sqrt{8}}= -\frac{10}{\sqrt{144}}= -\frac{10}{12}= -\frac{5}{6}$

And we can find the angle with the inverse cosine function and we got:

$\theta =cos^{-1} (-\frac{5}{6})= 146.44°$

Step-by-step explanation:

for this case we can use the following identity:

$cos \theta = \frac{a.b}{|a| |b|}$

We can begin finding the norm for each vector and we got:

$|a| =\sqrt{(1)^2 +(-4)^2 +(1)^2}= \sqrt{18}$

$|b| =\sqrt{(0)^2 +(2)^2 +(-2)^2}= \sqrt{8}$

Now we can find the dot product and we got:

$a.b = (1*0) + (-4*2) +(1*-2)= -10$

And replacing we got:

$cos \theta = \frac{-10}{\sqrt{18} \sqrt{8}}= -\frac{10}{\sqrt{144}}= -\frac{10}{12}= -\frac{5}{6}$

And we can find the angle with the inverse cosine function and we got:

$\theta =cos^{-1} (-\frac{5}{6})= 146.44°$