Question

if a=2i+3j+4k and b=4i+3j+2k find the angle betwen them

Answer 1

Given : if A=2i+3j+4k and B=4i+3j+2k find the angle betwen them

To find : Angle between the vectors A=2i+3j+4k and B=4i+3j+2k

A bar . B bar  = AB Cos x

=> Cos x = A bar . B bar / AB

=> x = $cos^{-1}(A bar . B bar)/ AB$

Where A bar. B bar = Ax.Bx + Ay.By + Az.Bz

=> A bar . B bar = 2*4 + 3*3 + 4*2

=> A bar . B bar = 8 + 9 + 8

=> A bar . B bar = 25

A  and  B  are the magnitudes of vectors  

A =

       $\sqrt{2^{2} +3^{2} +4^{2} }\\=> \sqrt{4+9+16}\\ =>\sqrt{29}$

B =

        $\sqrt{4^{2} +3^{2} +2^{2} }\\=> \sqrt{16+9+4}\\ =>\sqrt{29}$

Now substitute these values in main formula  x = $cos^{-1}(A bar . B bar)/ AB$

=> x = $cos^{-1}( 25 )/ (\sqrt{29} * \sqrt{29} )\\=> cos^{-1}( 25 )/ (29 )\\=> acos(2529)=acos(2529)rad ≈ 0.531458237938851rad ≈ 30.45031402135560$