Question

find the angle between A vector=4i+3j-4k and B vector 3i+4j+6k​

Answer 1

Angle between A vector=4i+3j-4k and B vector 3i+4j+6k :

• Let the angle between them be $\theta$.

Performing Scalar (Dot) Product :

$\vec{A}. \vec{B} = | \vec{A} || \vec{ B}| \cos( \theta)$

$\rm\implies(4 \hat{i} + 3 \hat{j} - 4 \hat{k}).(3 \hat{i} + 4 \hat{k} + 6 \hat{k}) = ( \sqrt{ {4}^{2} + {3}^{2} + {4}^{2} } )( \sqrt{ {3}^{2} + {4}^{2} + {6}^{2} } ) \cos( \theta)$

$\rm\implies 12 + 12 - 24= ( \sqrt{ 41 } )( \sqrt{ 61} ) \cos( \theta)$

$\rm\implies 0= ( \sqrt{ 41 } )( \sqrt{ 61} ) \cos( \theta)$

$\rm\implies \cos( \theta) = 0$

$\rm\implies \theta = {90}^{ \circ}$

So, angle between the vectors is 90°.

Answer 2

Explanation:

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