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Let $\left| {{{\vec A}_1}} \right| = 3,\,\left| {\vec A_2} \right| = 5$, and $\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5$. The value of $\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right)\cdot \left( {3{{\vec A}_1} - 2{{\vec A}_2}} \right)$ is
$-106.5$
$-112.5$
$-118.5$
$-99.5$
Solution
$\begin{array}{l}
\left| {{{\vec A}_1}} \right| = 3,\left| {{{\vec A}_2}} \right| = 5,\,and\,\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5.\\
\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = {\left| {{{\vec A}_1}} \right|^2} + {\left| {{{\vec A}_2}} \right|^2} + 2\left| {{{\vec A}_1}} \right|\left| {{{\vec A}_2}} \right|\,\cos \,\theta \\
\cos \,\theta \, = – \frac{3}{{10}}\\
\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right).\left( {3{{\vec A}_1} – 2{{\vec A}_2}} \right)\\
= 6{\left| {{{\vec A}_1}} \right|^2} + 9{{\vec A}_1}.{{\vec A}_2} – 4{{\vec A}_1}.{{\vec A}_2} – 6{\left| {{{\vec A}_2}} \right|^2}\\
= – 118.5
\end{array}$
Similar Questions
If $\theta$ is the angle between two vectors $A$ and $B$, then match the following two columns.
| colum $I$ | colum $II$ |
| $(A)$ $A \cdot B =| A \times B |$ | $(p)$ $\theta=90^{\circ}$ |
| $(B)$ $A \cdot B = B ^2$ | $(q)$ $\theta=0^{\circ}$ or $180^{\circ}$ |
| $(C)$ $|A+B|=|A-B|$ | $(r)$ $A=B$ |
| $(D)$ $|A \times B|=A B$ | $(s)$ None |
