Question

If $\overline{a},\overline{b}$ are vectors in space given by $\overline{a}=\frac{\overline{i}-2\overline{j}}{\sqrt{5}},\overline{b}=\frac{2\overline{i}+\overline{j}+3\overline{k}}{\sqrt{14}}$ then value of $(2\overline{a}+\overline{b})[(\overline{a}\times \overline{b})\times (\overline{a}-2\overline{b})]$ is

Answer 1

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1/\sqrt{5}& -2/\sqrt{5}& 0\\ 2/\sqrt{14}& \frac{1}{\sqrt{14}}& \frac{3}{\sqrt{14}}\end{array}\right|=\frac{1}{\sqrt{5}\sqrt{14}}\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& -2& 0\\ 2& -1& 3\end{array}\right|$

$=\frac{1}{\sqrt{70}}\left(\overrightarrow{i}\right(-6)+(3)\overrightarrow{j}+(5\left)\overrightarrow{k}\right)$

$=\frac{-6\overrightarrow{i}+3\overrightarrow{j}+5\overrightarrow{k}}{\sqrt{70}}$

let $y=(2\overrightarrow{a}+\overrightarrow{b}).\left[\right(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{a}-2\overrightarrow{b}\left)\right]$

$=\left[\begin{array}{ccc}2\overrightarrow{a}+\overrightarrow{b}& a^{-1}\times \overrightarrow{b}& \overrightarrow{a}-2\overrightarrow{b}\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}\overrightarrow{i}-\frac{4}{\sqrt{5}}\overrightarrow{j}\frac{2}{\sqrt{14}}\overrightarrow{i}+\frac{1}{\sqrt{14}}\overrightarrow{j}+\frac{3}{\sqrt{14}}\overrightarrow{k}+& \frac{-6\overrightarrow{i}+3\overrightarrow{j}5\overrightarrow{k}}{\sqrt{70}}& \overrightarrow{a}-2\overrightarrow{b}\end{array}\right]$

$=\left|\begin{array}{ccc}\frac{2}{\sqrt{5}}+\frac{2}{\sqrt{14}}& \frac{-4}{\sqrt{5}}+\frac{1}{\sqrt{14}}& \frac{3}{\sqrt{14}}\\ \frac{-6}{\sqrt{70}}& \frac{3}{\sqrt{70}}& \frac{5}{\sqrt{70}}\\ \frac{1}{\sqrt{5}}-\frac{4}{\sqrt{14}}& \frac{-2}{\sqrt{5}}-\frac{2}{\sqrt{14}}& \frac{-6}{\sqrt{14}}\end{array}\right|$

$=(\sqrt{70}{)}^3\left|\begin{array}{ccc}2(\sqrt{5}+\sqrt{14})& -4\sqrt{14}+\sqrt{5}& 3\sqrt{5}\\ -6& 3& 5\\ \sqrt{14}-4\sqrt{5}& -2\sqrt{14}-2\sqrt{5}& -6\sqrt{5}\end{array}\right|$

$R_1\to R-2R_3$

$y=(\sqrt{70}{)}^3\left|\begin{array}{ccc}10\sqrt{5}& 5\sqrt{5}& 15\sqrt{5}\\ -6& 3& 5\\ \sqrt{14}-4\sqrt{5}& -2\sqrt{14}-2\sqrt{5}& -6\sqrt{5}\end{array}\right|$

$=(\sqrt{70}{)}^3\times 5\sqrt{5}\left|\begin{array}{ccc}2& 1& 3\\ -6& 3& 5\\ -& -& -\end{array}\right|$

$c_1\to c_1-2c_2{c}_3\to c_3-3c_1$

$=(\sqrt{70}{)}^3\times 5\sqrt{5}\left|\begin{array}{ccc}0& 1& 0\\ -12& 3& -4\\ 5\sqrt{4}& -2\sqrt{14}-2\sqrt{5}& 6\sqrt{14}\end{array}\right|$

expanding along $c_2$

$=(\sqrt{70}{)}^3\times 5\sqrt{5}\times (-1\left)\right(-72\sqrt{14}+80\sqrt{14})$

$=\left(70\sqrt{70}\right)5\sqrt{5}(-8\sqrt{14})$

$=70\times 70(-40)$

$=-196000$

No option.