Question
If $\vec{\text{A}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}})$ and $\vec{\text{B}}=(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$ find $\vec{\text{A}}\times\vec{\text{B}}$ and $\vec{\text{A}}.\vec{\text{B}}$

Answer

  1. $\vec{\text{A}}\times\vec{\text{B}}=\begin{vmatrix}\hat{\text{i}} & \hat{\text{j}}&\hat{\text{k}} \\-2 & 3&-4\\3&-4&5 \end{vmatrix}$

$=\hat{\text{i}}\begin{vmatrix}3&-4\\-4&5\end{vmatrix}-\hat{\text{j}}\begin{vmatrix}-2&-4\\3&5\end{vmatrix}+\hat{\text{k}}\begin{vmatrix}-2&3\\3&-4\end{vmatrix}$

$=\hat{\text{i}}(15-16)-\hat{\text{j}}(-10+12)+\hat{\text{k}}(8-9)$

$=-\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$

  1. $\vec{\text{A}}.\vec{\text{B}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$

$=-6-12-20=-38$

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