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Q5E

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 \(19\)

**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\(a = \left\langle {4,1,\frac{1}{4}} \right\rangle \) and \(b = \langle 6, - 3, - 8\rangle \)

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 4 for \({a_1},1\) for \({a_2},\frac{1}{4}\) for \({a_3},6\) for \({b_1}, - 3\) for \({b_2}\) and \( - 8\) for \({b_3}\).

\(\begin{aligned}{l}a \cdot b &= (4)(6) + (1)( - 3) + \left( {\frac{1}{4}} \right)( - 8)\\a \cdot b &= 24 - 3 - 2\\a \cdot b &= 19\end{aligned}\)

Thus, \(a \cdot b\) is \(19\).

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