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Worked solution

Evaluate one scalar product in two ways

Given a=(121)\mathbf a=\begin{pmatrix}1\\2\\1\end{pmatrix} and b=(23−1)\mathbf b=\begin{pmatrix}2\\3\\-1\end{pmatrix}, find (3a−b)⋅(2a+b)(3\mathbf a-\mathbf b)\cdot(2\mathbf a+\mathbf b).

CJC · Vectors Basic · Example 9

Answer

2929.

Compare your reasoning

Method 1: combine the coordinates first

(3a−b)⋅(2a+b)=[3(121)−(23−1)]⋅[2(121)+(23−1)]=(134)⋅(471)=(1)(4)+(3)(7)+(4)(1)=29.\begin{aligned}(3\mathbf a-\mathbf b)\cdot(2\mathbf a+\mathbf b)&=\left[3\begin{pmatrix}1\\2\\1\end{pmatrix}-\begin{pmatrix}2\\3\\-1\end{pmatrix}\right]\cdot\left[2\begin{pmatrix}1\\2\\1\end{pmatrix}+\begin{pmatrix}2\\3\\-1\end{pmatrix}\right]\\&=\begin{pmatrix}1\\3\\4\end{pmatrix}\cdot\begin{pmatrix}4\\7\\1\end{pmatrix}\\&=(1)(4)+(3)(7)+(4)(1)=29.\end{aligned}

Subtract and add the columns before taking the componentwise scalar product.

Method 2: distribute the dot product

(3a−b)⋅(2a+b)=6a⋅a+3a⋅b−2b⋅a−b⋅b=6∣a∣2+a⋅b−∣b∣2=6(12+22+12)+(121)⋅(23−1)−[22+32+(−1)2]=36+[(1×2)+(2×3)+(1×(−1))]−14=36+7−14=29.\begin{aligned}(3\mathbf a-\mathbf b)\cdot(2\mathbf a+\mathbf b)&=6\mathbf a\cdot\mathbf a+3\mathbf a\cdot\mathbf b-2\mathbf b\cdot\mathbf a-\mathbf b\cdot\mathbf b\\&=6|\mathbf a|^2+\mathbf a\cdot\mathbf b-|\mathbf b|^2\\&=6(1^2+2^2+1^2)+\begin{pmatrix}1\\2\\1\end{pmatrix}\cdot\begin{pmatrix}2\\3\\-1\end{pmatrix}-[2^2+3^2+(-1)^2]\\&=36+[(1\times2)+(2\times3)+(1\times(-1))]-14\\&=36+7-14=29.\end{aligned}

Use commutativity to collect the two mixed products. Self-products are squared magnitudes.

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Solution notes

Method 2 is printed in both copies; the Teacher additionally completes Method 1.