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Physics 6 Online
OpenStudy (anonymous):

Using the definition of the scalar product, find the angles between vectors A=-2i + 4j + 1k and B=3i - 4j + 2k.

OpenStudy (anonymous):

\[\huge \text{Definition of Dot/Scalar Product:}\]\[\huge \vec{\text{A}}\cdot\vec{\text{B}}=|\vec{\text{A}}||\vec{\text{B}}| \cos(\theta)\]

OpenStudy (anonymous):

\[\huge \theta=\cos^{-1} \left(\frac{\vec{\text{A}}\cdot\vec{\text{B}}}{|\vec{\text{A}}||\vec{\text{B}}|}\right)\]

OpenStudy (unklerhaukus):

\[\mathbf A\cdot \mathbf B=(a_1\hat\imath +a_2\hat\jmath+a_3\hat k)\cdot(b_1\hat\imath +b_2\hat\jmath+b_3\hat k)\\ \qquad=a_1b_2+a_2b_2+a_3b_3\] \[|\mathbf A|=\sqrt{a_1^2+a_2^2+a_3^2}\]

OpenStudy (anonymous):

Fixing a small typo here: \[\mathbf A\cdot \mathbf B=(a_1\hat\imath +a_2\hat\jmath+a_3\hat k)\cdot(b_1\hat\imath +b_2\hat\jmath+b_3\hat k)\\ \qquad=a_1b_\color{red}{1}+a_2b_2+a_3b_3\]

OpenStudy (unklerhaukus):

*thankyou

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