Evaluate the product (3 a-5 b) (2 a+7 b)

Question

Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$

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Answer

According to question,
$(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$
$\Rightarrow\binom{3}{3} \cdot(2 \vec{b})+\binom{3}{3} \cdot(7 \vec{b})+(-5 \vec{b}) \cdot(2 \vec{a})+(-5 \vec{b}) \cdot(7 \vec{b})$
$=6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{b} \cdot \vec{a}-35 \vec{b} \cdot \vec{b}$
$=6|\vec{a}|^2+21(\vec{a} \cdot \vec{b})-10(\vec{b} \cdot \vec{a})-35|\vec{b}|^2$
$\because \vec{a} \cdot \vec{a}=|\vec{a}|^2, \vec{b} \cdot \vec{b}=|\vec{b}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$
So, $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$
$=6|\vec{a}|^2+21(\vec{a} \cdot \vec{b})_{-10} \vec{a} \cdot \vec{b}-35|\vec{b}|^2$
$=6|\vec{a}|^2+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^2$

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