Question

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) a = 6i − 9j + k, b = 8i − k

Answer 1

Answer:

The angle between given vectors is 58 degrees.          

Step-by-step explanation:

We are given the following vectors in the question:

$a = 6i- 9j + k\\b = 8i - k$

We have to find the angle between the two vectors.

Formula:

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

Evaluating,

$a.b\\= ( 6i- 9j + k)(8i - k)\\=6.8 -9.0 +1(-1) = 47\\|a| = \sqrt{6^2 + (-9)^2+1^2} = \sqrt{118}\\|b| = \sqrt{8^2 + (-1)^2} = \sqrt{65}\\\text{Putting values}\\\\\cos \theta = \displaystyle\frac{a.b}{|a||b|}\\\\\cos \theta = \frac{47}{\sqrt{118}\sqrt{65}} = \frac{47}{\sqrt{7670}}\\\\\theta = \arccos \frac{47}{\sqrt{7670}}\\\\\theta = 57.54336^\circ \approx 58\text{ degrees}$

The angle between given vectors is 58 degrees.