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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra2 Marks
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,
$
\begin{array}{l}
\begin{array}{r}
(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b}) \\
\begin{aligned}
\Rightarrow\binom{3}{3} \cdot(2 \vec{b})+\binom{3}{3} \cdot(7 \vec{b})+ \\
\quad(-5 \vec{b}) \cdot(2 \vec{a})+(-5 \vec{b}) \cdot(7 \vec{b})
\end{aligned} \\
\quad=6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{b} \cdot \vec{a}-35 \vec{b} \cdot \vec{b} \\
\quad=6|\vec{a}|^2+21(\vec{a} \cdot \vec{b})-10(\vec{b} \cdot \vec{a})-35|\vec{b}|^2
\end{array} \\
\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}
\end{array}
$
So, $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$
$
\begin{array}{l}
=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
\end{array}
$

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