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Q6E

Expert-verifiedFound in: Page 556

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**To find a dot product between \({\rm{a}}\) and \({\rm{b}}\).**

The dot product \(a \cdot b\) is \( - pq\)

**The minimum of two vectors are required to perform a dot product. The resultant dot product of two vectors is scalar. hence, the dot product is also known as a scalar product.**

The given two vectors are\({\rm{a}} = \langle p, - p,2p\rangle \) and \({\rm{b}} = \langle 2q,q, - q\rangle \)

Consider a general expression to find the dot product between two three-dimensional vectors.

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= \left\langle {{a_1},{a_2},{a_3}} \right\rangle \cdot \left\langle {{b_1},{b_2},{b_3}} \right\rangle \\{\rm{a}} \cdot {\rm{b}} &= {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\end{aligned}\)

Substitute \(p\) for \({a_1}, - p\) for \({a_2},2p\) for \({a_3},2q\) for \({b_1},q\) for \({b_2}\) and \( - q\) for \({b_3}\).

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (p)(2q) + ( - p)(q) + (2p)( - q)\\{\rm{a}} \cdot {\rm{b}} &= 2pq - pq - 2pq\\{\rm{a}} \cdot {\rm{b}} &= - pq\end{aligned}\)

Thus, \(a \cdot b\) is \( - pq\).

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