Question

Find the angle between each of the following pairs of vectors A^rightarrow = A_x i^cap + A_y j^cap and B^cap = B_x i^cap + B_y j^cap. A_x1 = -200, A_y1 = 5.20, B_x1 = 1.10, B_y1 = -3.80. A_x2 = 3.20, A_y2 = 4.20; B_x1 = 11.8, B_yx = 6.20. A_x1 = -4.00, A_y2 = 2.00, B_x2 = 7.00, B_y1 = 14.00

Answer 1

As we know that- Cos $\Theta$ = $\underset{(A}{\rightarrow}\underset{.B)}{\rightarrow}$ / $(\left | A \right |.\left | B \right |)$

For Part a)- $\underset{(A}{\rightarrow}\underset{.B)}{\rightarrow}$ = (-2 )(1.10 ) + ( 5.20)(-3.80 ) = -21.96

And $(\left | A \right |.\left | B \right |)$ = (5.57)( 3.96) = 22.06

so Cos $\Theta$ = -21.96 / 22.06 = -0.89

than $\Theta$ = cos^-1( -0.89) = 152.87^o

Now the same formula will use for the rest of the parts of question.

Part b)- Cos $\Theta$ = 63.80 / 70.38 = 0.91

So $\Theta$ = 24.97^o

Part C)-  Cos $\Theta$ = 0

so, $\Theta$ = 90^o

Thank you !!! Have fun