Define cross product of two vectors. If \(\vec{a} = \vec{i} + 3\vec{j} + 4\vec{k}\)  and \(\vec{b} = 2\vec{i} + 7\vec{j} – 5\vec{k}\) find the vector \(\vec{b} \times \vec{a} \enspace and \enspace \vec{a} \times \vec{b}\)

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Let \(\vec{a}\) and \(\vec{b}\) are two non vectors. Then the cross product of \(\vec{a}\) and \(\vec{b}\) is denoted by \(\vec{b} \times \vec{a} \) and is defined as

\(\vec{a} \times \vec{b} = |\vec{a}| \enspace |\vec{b}| sin\theta \)

where \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\) and \(\hat{n}\) is unit vector along \(( \vec{a} \times \vec{b} )\)

 

Problem Part:

Let

\(\vec{a} = \vec{i} + 3\vec{j} + 4\vec{k}\)  and \(\vec{b} = 2\vec{i} + 7\vec{j} – 5\vec{k}\)

Then

\(\vec{a} \times \vec{b} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
1 & 3 & 4\\
2 & 7 & 5
\end{vmatrix}\)

\( = (-15-28)\vec{i} – (-5-8)\vec{j} + (7-6)\vec{k}\)

\( = (-43)\vec{i} + (13)\vec{j} + \vec{k}\)

and we know \(\vec{a} \times \vec{b} \enspace =\enspace – \vec{b} \times \vec{a}\)

so

\(\vec{b} \times \vec{a} \enspace =\enspace – \vec{a} \times \vec{b}\)

\( = 43\vec{i} – 13\vec{j} – \vec{k}\)

 

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