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

Angle=70°

Step-by-step 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=(a.b)/(|a||b|)\\\alpha=cos^(-1) ((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) ((a.b)/(|a||b|))\\\alpha=cos^(-1) ((17)/((5.477)*(9.055)))\\\alpha =70^(o)`