Question

The ( x ) - and ( y )-components of vector ( vec{A} ) are ( 2.80 ) and ( 5.60 ), respectively. The ( x ) and ( y )-components of vector ( vec{B} ) are ( 6.00 ) and ( 2.00 ), respectively. All units are arbitrary. What is the angle between vectors ( vec{A} ) and ( vec{B} ), in degrees? Type in a positive value.

          The ( x ) - and ( y )-components of vector ( vec{A} ) are ( 2.80 ) and ( 5.60 ), respectively. The ( x ) and ( y )-components of vector ( vec{B} ) are ( 6.00 ) and ( 2.00 ), respectively. All units are arbitrary. What is the angle between vectors ( vec{A} ) and ( vec{B} ), in degrees? Type in a positive value.
        
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The ( x ) - and ( y )-components of vector ( vecA ) are ( 2.80 ) and ( 5.60 ), respectively. The ( x ) and ( y )-components of vector ( vecB ) are ( 6.00 ) and ( 2.00 ), respectively. All units are arbitrary. What is the angle between vectors ( vecA ) and ( vecB ), in degrees? Type in a positive value.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
Chapter 1
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Find the angle between each of these pairs of vectors: (a) $\overrightarrow{A}$ = $-$2.00$\hat{\imath}$ $+$ 6.00$\hat{\jmath}$ and $\overrightarrow{B}$ = 2.00$\hat{\imath}$ $-$ 3.00$\hat{\jmath}$ (b) $\overrightarrow{A}$ = 3.00$\hat{\imath}$ $+$ 5.00$\hat{\jmath}$ and $\overrightarrow{B}$ = 10.00$\hat{\imath}$ $+$ 6.00$\hat{\jmath}$ (c) $\overrightarrow{A}$ = $-$4.00$\hat{\imath}$ $+$ 2.00$\hat{\jmath}$ and $\overrightarrow{B}$ = 7.00$\hat{\imath}$ $+$ 14.00$\hat{\jmath}$

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You are given two vectors $\vec{A}=-3.00 \hat{\imath}+6.00 \hat{\jmath} \quad$ and $\vec{B}=7.00 \hat{\imath}+2.00 \hat{\jmath} . \quad$ Let $\quad$ coun- terclockwise angles be positive. (a) What angle does $\vec{A}$ make with the $+x$ -axis? (b) What angle does $\boldsymbol{B}$ make with the $+x$ -axis? (c) Vector $\boldsymbol{C}$ is the sum of $\vec{A}$ and $\vec{B},$ so $\vec{C}=\vec{A}+\vec{B} .$ What angle does $\vec{C}$ make with the $+x$ -axis?

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Transcript

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00:01 Hello students in this question we have given x component and y component for the vector a as 2 .80 x cap and y component is 5 .60s it can be written as y cap and for the vector b this these values are 6 .00 x cap plus 2 .00 y cap so we have to determine the angle between a vector and b vector okay the angle between two vectors theta is given by arc of cause a vector dot b vector divided by modulus of a vector and modulus of b vector okay so we can substitute the values so we can write that theta it is equals to the arc of cause a vector 2 .80 x cap plus 5 .60 y cap and the dot product with the 6 .0 x cap plus 2 .0 y cap okay and divide by the magnitude of a vector can be written as a 2 .80 square plus 5 .60 square and this under root and multiplied with the 6 .00 square plus 2 .00 square and after solving it completely the angle theta will be equals to the 45 degree...
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