Physics, asked by drikithroshan1642, 1 year ago

The angle between vector a=3i-4j and b= -2I+3k

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Answered by AR17
17
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Answered by SushmitaAhluwalia
6

The angle between the vectors is \alpha = cos^{-1}(\frac{-6}{5\sqrt{13} } )

  • Given vectors are

                   a = 3i - 4j, b = -2i + 3k

  • The angle between vectors can be calculated using the formula

                   cos\alpha = \frac{a_{1}b_{1}+a_{2}b_{2} +a_{3}b_{3} }{\sqrt{(a_{1}^{2}+ a_{2}^{2}+a_{3}^{2})((b_{1}^{2}+ b_{2}^{2}+b_{3}^{2}} )}

  • Here,

                  a_{1} = 3, a_{2} = -4, a_{3} = 0

                  b_{1} = -2, b_{2} = 0, b_{3} = 3

                  cos\alpha = \frac{-6+0+0}{\sqrt{(9+16)(4+9)} }

                           =\frac{-6}{5\sqrt{13} }

                   \alpha = cos^{-1}(\frac{-6}{5\sqrt{13} } )

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