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If $\overline{e_1}, \overline{e_2}$ are two non-collinear unit vectors such that $\left|\bar{e}_1+\bar{e}_2\right|=\sqrt{3}$ then $\left(2 \bar{e}_1-5 \bar{e}_2\right),\left(3 \bar{e}_1+\bar{e}_2\right)=$
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Verified Answer
The correct answer is:
$\frac{-11}{2}$
Given $\overrightarrow{\mathrm{e}_1} \cdot \& \overrightarrow{\mathrm{e}_2}$ are units vectors, then $\left|\overrightarrow{\mathrm{e}_{\mathrm{i}}}\right|=1=\left|\overrightarrow{\mathrm{e}_2}\right|$
Now, $\left|\overrightarrow{\mathrm{e}_{\mathrm{i}}}+\overrightarrow{\mathrm{e}_2}\right|=\sqrt{3}$
$\begin{aligned} & \left|\overrightarrow{\mathrm{e}_{\mathrm{i}}}+\overrightarrow{\mathrm{e}_2}\right|^2=3 \\ & \left(\overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2}\right) \cdot\left(\overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2}\right)=3 \\ & \overrightarrow{\mathrm{e}_1} \cdot \overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_1} \cdot \overrightarrow{\mathrm{e}_2}+\overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_2}=3 \\ & \left|\overrightarrow{\mathrm{e}_1}\right|^2+\left|\overrightarrow{\mathrm{e}_2}\right|^2+2 \overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=3 \\ & 1+1+2 \overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=3 \\ & 2 \overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=1\end{aligned}$
$\overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=\frac{1}{2}$...(i)
$\left(2 \overrightarrow{\mathrm{e}_1}-5 \overrightarrow{\mathrm{e}_2}\right) \cdot\left(3 \overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2}\right)=6\left|\overrightarrow{\mathrm{e}_1}\right|^2+2 \overrightarrow{\mathrm{e}_1} \cdot \overrightarrow{\mathrm{e}_2}-1-5\left|\overrightarrow{\mathrm{e}_2}\right|^2$
Now,
$\begin{aligned} & 6 \times 1+13 \times \frac{1}{2}-5 \times 1 \\ & 1-\frac{13}{2}=\frac{-11}{2}\end{aligned}$
Now, $\left|\overrightarrow{\mathrm{e}_{\mathrm{i}}}+\overrightarrow{\mathrm{e}_2}\right|=\sqrt{3}$
$\begin{aligned} & \left|\overrightarrow{\mathrm{e}_{\mathrm{i}}}+\overrightarrow{\mathrm{e}_2}\right|^2=3 \\ & \left(\overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2}\right) \cdot\left(\overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2}\right)=3 \\ & \overrightarrow{\mathrm{e}_1} \cdot \overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_1} \cdot \overrightarrow{\mathrm{e}_2}+\overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_2}=3 \\ & \left|\overrightarrow{\mathrm{e}_1}\right|^2+\left|\overrightarrow{\mathrm{e}_2}\right|^2+2 \overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=3 \\ & 1+1+2 \overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=3 \\ & 2 \overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=1\end{aligned}$
$\overrightarrow{\mathrm{e}_2} \cdot \overrightarrow{\mathrm{e}_1}=\frac{1}{2}$...(i)
$\left(2 \overrightarrow{\mathrm{e}_1}-5 \overrightarrow{\mathrm{e}_2}\right) \cdot\left(3 \overrightarrow{\mathrm{e}_1}+\overrightarrow{\mathrm{e}_2}\right)=6\left|\overrightarrow{\mathrm{e}_1}\right|^2+2 \overrightarrow{\mathrm{e}_1} \cdot \overrightarrow{\mathrm{e}_2}-1-5\left|\overrightarrow{\mathrm{e}_2}\right|^2$
Now,
$\begin{aligned} & 6 \times 1+13 \times \frac{1}{2}-5 \times 1 \\ & 1-\frac{13}{2}=\frac{-11}{2}\end{aligned}$
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