Question

In the isosceles triangle ABC $\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=8$, a point E divides AB internally in the ratio $1:3$, then the cosine of the angle between $\overrightarrow{CE}$ & $\overrightarrow{CA}$ is? (where $\left|\overrightarrow{CA}\right|=12$)

• A. $\frac{3\sqrt{7}}{8}$
• B. $\frac{3\sqrt{8}}{17}$
• C. $\frac{-3\sqrt{7}}{8}$
• D. $\frac{-3\sqrt{8}}{17}$

Answer 1

The correct option is A $\frac{3\sqrt{7}}{8}$

Consider the problem

Given ,

$|AB|=|BC|=8$

Let,

$|\overrightarrow{b}|=8$

$|\overrightarrow{c}|=12$

$|\overrightarrow{b}-\overrightarrow{c}|=8$

$\overrightarrow{CE}=\overrightarrow{E}-\overrightarrow{C}$

$=\frac{\overrightarrow{b}}{4}-\overrightarrow{c}$

$|\overrightarrow{CE}|=|\frac{\overrightarrow{b}}{4}-\overrightarrow{c}|=\sqrt{\frac{|\overrightarrow{b}{|}^2}{16}+|\overrightarrow{c}{|}^2-2.\frac{\overrightarrow{b}.\overrightarrow{c}}{4}}$

$|\overrightarrow{CE}|=|\frac{\overrightarrow{b}}{4}-\overrightarrow{c}|=\sqrt{\frac{64}{16}+144-\frac{\overrightarrow{b}.\overrightarrow{c}}{2}}$

$=\sqrt{148-\frac{\overrightarrow{b}.\overrightarrow{c}}{2}}$ ---- (i)

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\cos A=\frac{8^2+{12}^2-8^2}{\mathrm{2.8.12}}$

$=\frac{3}{4}$

$\overrightarrow{b}.\overrightarrow{c}=|\overrightarrow{b}||\overrightarrow{c}|\cos A$

$=8\times 12\times \frac{3}{4}=72$

from (i)

$=\sqrt{148-\frac{72}{2}}=\sqrt{112}$

Now in triangle $ABC$

$\frac{CE}{\sin A}=\frac{AE}{\sin \alpha }$ $(\cos A=\frac{3}{4},\sin A=\frac{\sqrt{7}}{4})$

$\frac{\sqrt{112}}{\frac{\sqrt{7}}{4}}=\frac{2}{\sin \alpha }$

$\sin \alpha =\frac{2\times \sqrt{7}}{\sqrt{112}.4}$

$=\frac{1}{8}$

then,

$\cos \alpha =\frac{\sqrt{63}}{8}=\frac{3\sqrt{7}}{8}$