Question

What is the product of  ${\displaystyle (3\overrightarrow{a}\cdot 5\overrightarrow{b})\cdot (2\overrightarrow{a}+7\overrightarrow{b})}$

Options

• ${\displaystyle 6{\left|\overrightarrow{a}\right|}^2}$

• ${\displaystyle 11\overrightarrow{a}\cdot \overrightarrow{b}}$

• ${\displaystyle -35{\left|\overrightarrow{b}\right|}^2}$

• ${\displaystyle 6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}\cdot \overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2}$

Answer 1

${\displaystyle 6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}\cdot \overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2}$

Explanation:

${\displaystyle (3\overrightarrow{a}-5\overrightarrow{b})\cdot (2\overrightarrow{a}-7\overrightarrow{b})3\overrightarrow{a}\cdot (2\overrightarrow{a}+7\overrightarrow{b})-5\overrightarrow{b}(2\overrightarrow{a}+7\overrightarrow{b})}$

= ${\displaystyle 6\overrightarrow{a}\cdot \overrightarrow{a}+21\overrightarrow{a}\cdot \overrightarrow{b}-10\overrightarrow{b}\cdot \overrightarrow{a}-35\overrightarrow{b}\cdot \overrightarrow{b}}$

= ${\displaystyle 6{\left|\overrightarrow{a}\right|}^2+21\overrightarrow{a}\cdot \overrightarrow{b}-10\overrightarrow{b}\cdot \overrightarrow{a}-35{\left|\overrightarrow{b}\right|}^2}$  ......${\displaystyle [\because \overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}]}$

= ${\displaystyle 6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}\cdot \overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2}$