Question

The angle between two vectors -2i+3j+k and i+2j-4i

Answer 1

Answer: CONSULTATION : Here must be i+2j - 4k

α = 90  

Explanation: Suppose the angle be ' α '

Let P and Q be the two vectors respectively ;

vector P = - 2i +3j +k

vector Q = i + 2j - 4k

Now, by using resolution method

Px = -2 , Py = 3 , Pz = 1

Qx =  1 , Qy = 2 , Qz = -4

Magnitude of P = $\sqrt{Px ^2 +Py^2 + Pz^2$

                           =$\sqrt{(-2)^2 + (3)^2 + (1)^2$

                           =$\sqrt{4 + 9 +1}$

                           = $\sqrt{14}$

Magnitude of Q  = $\sqrt{Qx^2 + Qy^2 + Qz^2}$

                            =$\sqrt{(1)^2 + (2)^2 + (-4)^2$

                            =$\sqrt{1 + 4+ 16}$

                            = $\sqrt{21}$

We have the formula of dot product  :

Vector P . Vector Q =  magnitude of P . magnitude pf Q . cos α

( -2i + 3j + k ) . ( i + 2j - 4k) =  $\sqrt{14}. \sqrt{21}$ .cosα

-2 +6-4 =$\sqrt{14}. \sqrt{21}$ .cosα     [ i .i = 1 , i .j = 0 , i.k =0 ] , [ j.j = 1 , j .i = 0 , j.k = 0]

0 = cosα                             [ k.k =1 , k.i =0 , k.j =0]

cos 90 = cosα

Therefore , α = 90   ans .