Question

In a quadrilateral $ABCD$, $\overrightarrow{AC}$ is the bisector of the $(\overrightarrow{AB}\wedge \overrightarrow{AD})$ which is $\frac{2\pi }{3},$
$15\left|\overrightarrow{AC}\right|=3\left|\overrightarrow{AB}\right|=5\left|\overrightarrow{AD}\right|$ then $\cos (\overrightarrow{BA}\wedge \overrightarrow{CD})$ is :

• A. $-\frac{\sqrt{14}}{7\sqrt{2}}$
• B. $-\frac{\sqrt{21}}{7\sqrt{3}}$
• C. $\frac{2}{\sqrt{7}}$
• D. $\frac{2\sqrt{7}}{14}$

Answer 1

The correct option is C $\frac{2}{\sqrt{7}}$

Let $A$ be the origin, $\overrightarrow{AB}=\overrightarrow{b},\overrightarrow{AC}=\overrightarrow{c}$ and $\overrightarrow{AD}=\overrightarrow{d}$

Now, according to the conditions given, we have $c=\frac{b}{5}=\frac{d}{3}$
Also, $\frac{\overrightarrow{c}\cdot \overrightarrow{d}}{cd}=\cos {60}^o=\frac{1}{2}$
$\frac{\overrightarrow{c}.\overrightarrow{b}}{cb}=\cos {60}^o=\frac{1}{2}$

$\Rightarrow \overrightarrow{c}\cdot \overrightarrow{d}=1.5c^2$ and $\overrightarrow{c}.\overrightarrow{b}=2.5c^2$
We want cosine of the angle between $BA$ and $CD$, i.e. $\frac{\overrightarrow{b}.(\overrightarrow{d}-\overrightarrow{c})}{b|\overrightarrow{d}-\overrightarrow{c}|}$
$=-\frac{\overrightarrow{b}.\overrightarrow{d}-\overrightarrow{b}.\overrightarrow{c}}{b\sqrt{d^2+c^2-2\overrightarrow{c}.\overrightarrow{d}}}$
$=-\frac{-7.5c^2-2.5c^2}{5c\times \sqrt{9c^2+c^2-3c^2}}$
$=-\frac{-10c^2}{5\sqrt{7}{c}^2}$

$=\frac{2}{\sqrt{7}}$