Question

If a= 2i +j -k, b= i + 2j + 3k and c= 6i - 2j - 6k then the angle between a+b and c will be

Answer 1

Step-by-step explanation:

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Answer 2

The angle between a+b and c will be 1.5707 radians.

Step-by-step explanation:

We are given the following in the question:

$a= 2i +j -k\\b= i + 2j + 3k\\c= 6i - 2j - 6k$

a + b =

$a + b = 2i +j -k + ( i + 2j + 3k)\\d = 3i+3j+2k$

The magnitude can be found as:

$|d| = \sqrt{3^2 + 3^2 + 2^2} = \sqrt{22}\\|c| = \sqrt{6^2 + (-2)^2 + (-6)^2} = \sqrt{76}$

Formula:

$\cos\theta = \dfrac{c.d}{|c||d|}\\\\\cos\theta = \dfrac{(3i+3j+2k)(6i - 2j - 6k)}{\sqrt{22}\sqrt{76}}\\\\\cos\theta = \dfrac{18-6-12}{\sqrt{22}\sqrt{76}} = 0\\\\\theta = \arccos(0)\\\\\theta = \dfrac{\pi}{2}\\\\\theta \approx 1.5707\text{ radians}$

Thus, the vector a+b and c are perpendicular.

The angle between a+b and c will be 1.5707 radians.

#LearnMore

Show that vectors a=2i+3j+6k,b=3i-6j+2k and c= 6i+2j-3k are mutually perpendiular

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