Question

the angle between the vectors -2i+3j-k and i+2j+4k is

Answer 1

Answer :

We have been provided two vectors.

▪ A = -2i + 3j - k

▪ B = i + 2j + 4k

We have to find angle between both vectors$.$

Whenever we have been provided two different vectors and asked to find angle b/w them then the concept by which we can solve this problem is Dot product of both vectors.

Dot product of two vectors A and B is given by

◈ A ▪ B = AB cosΦ

where Φ denotes angle b/w both vectors.

Magnitude of A = √14 unit

Magnitude of B = √21 unit

A ▪ B = -2 + 6 - 4 = 0 unit

⇒ A ▪ B = AB cosΦ

⇒ 0 = (√14×√21) cosΦ

⇒ cosΦ = 0

⇒ Φ = 90°

☃ God Bless :)

Answer 2

Let ,

A = -2i + 3j - k

B = i + 2j + 4k

We know that , the angle between two vectors is given by

$\boxed{ \sf{\cos( \theta) = \frac{ \vec{x} .\vec{y}}{ |x| |y| } }}$

Thus ,

$\sf \mapsto \cos( \theta) = \frac{ - 2 + 6 + ( - 4)}{ \sqrt{ {( - 2)}^{2} + {(3)}^{2} + {( - 1)}^{2} } \sqrt{ {(1)}^{2} + {(2)}^{2} } + {(4)}^{2} } \\ \\ \sf \mapsto \cos( \theta) = \frac{0}{ \sqrt{14} \sqrt{21} } \\ \\ \sf \mapsto \cos( \theta) =0 \\ \\ \sf \mapsto \theta = 90$

Therefore ,

• The angle between two given vectors is 90