Question

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) a = 2i − 5j + k, b = 9i − k exact

Answer 1

Answer:

Angle=70°

Explanation:

Given data

a=2i-5j+k

b=9i-k

To find

Angle Θ

Solution

As we know from dot product rule that

$a.b=|a||b|cos\alpha \\cos\alpha=\frac{a.b}{|a||b|}\\\alpha=cos^{-1} (\frac{a.b}{|a||b|})$

First we need to find a.b

As

i.i=j.j=k.k=1

i.j=i.k=j.k=0

So

$a.b=(18)+(-1)\\a.b=17$

Now for |a| and |b|

$|a|=\sqrt{x^{2} +y^{2}+z^{2} }\\|a|=\sqrt{(2)^{2} +(-5)^{2}+(1)^{2} }\\ |a|=5.477\\And\\|b|=\sqrt{x^{2} +y^{2}+z^{2} }\\|b|=\sqrt{(9)^{2} +(0)^{2}+(-1)^{2} }\\|b|=9.055$

So the angle is given as

$\alpha=cos^{-1} (\frac{a.b}{|a||b|})\\\alpha=cos^{-1} (\frac{17}{(5.477)*(9.055)})\\\alpha =70^{o}$