Find the angle θθ between the vectors v=2i+kv=2i+k, w=j−3kw=j−3k.

1 answer:

4 0

Hello :
v . w = | | v | | × | | w | | cos(θ) ....(1)
v(2,0,1) w(0,1,-3)

v . w = (2)(0)+(0)(1)+(1)(-3) = - 3
| | v | | = √((2)²+(0)²+(1)²) = √5
| | w | = √((0)²+(1)²+(-3)²) = √10
by (1) :
cos(θ) = (v . w ) / | | v | | × | | w | |

cos(θ) = = - 3/√50
θ =..... calculate by 2ind function ( calculator)

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