Question

If A = 3i -2j +5k and B = 7i -3j +2k, find A×B and direction of the resultant vector

Answer 1

$\large{ \tt{Answer : }}$

$\large\tt \vec{A} = 3 \hat{i} - 2\hat{j} + 5\hat{k}$

$\large\tt \vec{B}= 7 \hat{i} - 3\hat{j} + 2\hat{k}$

$\tt \vec{A } \times \vec{B} = \begin{array}{|l|}\hat{i} \: \: \: \: \: \: \: \: \hat{j} \: \: \: \: \: \: \hat{k} \: \: \\ 3\: \: - 2 \: \: \: \: \: \: 5 \\ 7\: \: - 3 \: \: \: \: \: \: 2 \end{array}$

$\tt \vec{A } \times \vec{B} = ( (- 2 \times 2) -( 5 \times - 3))\hat{i} + ( (5 \times 7 ) -( 2 \times 3))\hat{j} + ( (3 \times - 3 ) -( - 2 \times 7))\hat{k}$

$\tt \vec{A } \times \vec{B} = ( - 4 -( - 15))\hat{i} + ( 35 -6)\hat{j} + ( - 9 -( - 14))\hat{k}$

$\tt \vec{A } \times \vec{B} = ( - 4 + 15)\hat{i} + 29\hat{j} + ( - 9 + 14)\hat{k}$

$\tt \vec{A } \times \vec{B} = 11\hat{i} + 29\hat{j} + 5\hat{k}$

$\large \tt \vec{A}. \vec{B}= | A | |B| \cos \theta$

$21 + 6 + 10 =( \sqrt{ {(3)}^{2} + { (- 2)}^{2} + {(5)}^{2} } )( \sqrt{ {(7)}^{2} + { (- 3)}^{2} + {(2)}^{2}} ) \ \cos \theta$

$37=( \sqrt{ 9 + 4+ 25 } )( \sqrt{ 49+ 9 + 4}) \cos \theta$

$37=( \sqrt{ 38 } )( \sqrt{ 62}) \cos \theta$

$\displaystyle \tt\cos \theta = \frac{37}{( \sqrt{ 38 } )( \sqrt{ 62})}$

$\displaystyle \tt \theta = { \cos }^{ - 1} \frac{37}{( \sqrt{ 38 } )( \sqrt{ 62})}$

$\displaystyle \tt \theta \approx40 \degree$