Question

$|a|=|b|=5$ and the angle between a and b is $\frac{\pi }{4}$, The area of the triangle constructed on the vectors $a-2b$ and $3a+2b$ is.

• A. $560$
• B. $50\sqrt{2}$
• C. $\frac{50}{\sqrt{2}}$
• D. $100$

Answer 1

The correct option is B $50\sqrt{2}$

Let $\overrightarrow{A}=\overrightarrow{(a-2b)}$ and $\overline{B}=\overrightarrow{(3a+2b)}$

Then, $\frac{1}{2}\overline{A}\times \overline{B}=\frac{1}{2}\overrightarrow{(a-2b)}\times \overrightarrow{(3a+2b)}$

$\Rightarrow (-6(b\times a)+2(a\times b))$

$\frac{8(a\times b)}{2}=4\times \left|a\right|\left|b\right|\sin {45}^0$

$=4\times 5\times 5\frac{1}{\sqrt{2}}$$=50\sqrt{2}$