Question

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

Answer 1

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Answer 2

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

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

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

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