⇒Vectors

Correct answer option is (C)

$\begin{array}{rl}& (\overline{a}+2\overline{b}+\overline{c})\cdot [(\overline{a}-\overline{b})\times (\overline{a}-\overline{b}-\overline{c})]\\ & =(\overline{a}+2\overline{b}+\overline{c})\cdot (\overline{a}\times \overline{a}-\overline{a}\times \overline{b}-\overline{a}\times \overline{c}-\overline{b}\times \overline{a}+\overline{b}\times \overline{b}+\overline{b}\times \overline{c})\\ & =(\overline{a}+2\overline{b}+\overline{c})\cdot (\bar{0}-\overline{a}\times \overline{b}-\overline{a}\times \overline{c}+\overline{a}\times \overline{b}+\bar{0}+\overline{b}\times \overline{c})\\ & =(\overline{a}+2\overline{b}+\overline{c})\cdot (\overline{c}\times \overline{a}+\overline{b}\times \overline{c})\\ & =\overline{a}\cdot (\overline{c}\times \overline{a})+\overline{a}\cdot (\overline{b}\times \overline{c})+2\overline{b}\cdot (\overline{c}\times \overline{a})+2\overline{b}\cdot (\overline{b}\times \overline{c})\\ & +\overline{c}\cdot (\overline{c}\times \overline{a})+\overline{c}\cdot (\overline{b}\times \overline{c})\end{array}$

$\begin{array}{rl}& =\bar{0}+\overline{a}\cdot (\overline{b}\times \overline{c})+2\overline{b}\cdot (\overline{c}\times \overline{a})+2\times 0-0-0\\ & =\left[\begin{array}{lll}\overline{a}& \overline{b}& \overline{c}\end{array}\right]+2\left[\begin{array}{lll}\overline{b}& \overline{c}& \overline{a}\end{array}\right]\\ & =\left[\begin{array}{lll}\overline{a}& \overline{b}& \overline{c}\end{array}\right]+2\left[\begin{array}{lll}\overline{a}& \overline{b}& \overline{c}\end{array}\right]\\ & =3\left[\begin{array}{lll}\overline{a}& \overline{b}& \overline{c}\end{array}\right]\end{array}$

Correct answer option is (C)

R.H.S. of the given equality can be written as

$\begin{array}{rl}& (2\bar{\mathrm{a}}+3\bar{\mathrm{b}}+\bar{\mathrm{c}})\cdot [(\bar{\mathrm{a}}+2\bar{\mathrm{b}}+3\bar{\mathrm{c}})\times (3\bar{\mathrm{a}}+\bar{\mathrm{b}}+2\bar{\mathrm{c}})]\\ & =(2\overline{a}+3\overline{b}+\overline{c})\cdot [3(\overline{a}\times \overline{a})+(\overline{a}\times \overline{b})+2(\overline{a}\times \overline{c})\\ & +6(\overline{b}\times \overline{a})+2(\overline{b}\times \overline{b})+4(\overline{b}\times \overline{c})\\ & +9(\bar{\mathrm{c}}\times \bar{\mathrm{a}})+3(\bar{\mathrm{c}}\times \bar{\mathrm{b}})+6(\bar{\mathrm{c}}\times \bar{\mathrm{c}})]\\ & =(2\bar{\mathrm{a}}+3\bar{\mathrm{b}}+\bar{\mathrm{c}})[0+(\bar{\mathrm{a}}\times \bar{\mathrm{b}})+2(\bar{\mathrm{a}}\times \bar{\mathrm{c}})\\ & -6(\overline{a}\times \overline{b})+0+4(\overline{b}\times \overline{c})\\ & -9(\bar{\mathrm{a}}\times \bar{\mathrm{c}})-3(\bar{\mathrm{b}}\times \bar{\mathrm{c}})+0]\\ & =(2\bar{\mathrm{a}}+3\bar{\mathrm{b}}+\bar{\mathrm{c}})[-5(\bar{\mathrm{a}}\times \bar{\mathrm{b}})+(\bar{\mathrm{b}}\times \bar{\mathrm{c}})-7(\bar{\mathrm{a}}\times \bar{\mathrm{c}})]\\ & =-10[\bar{\mathrm{a}}\cdot (\bar{\mathrm{a}}\times \bar{\mathrm{b}})]+2[\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})]-14[\bar{\mathrm{a}}\cdot (\bar{\mathrm{a}}\times \bar{\mathrm{c}})]\\ & -15[\bar{\mathrm{b}}\cdot (\bar{\mathrm{a}}\times \bar{\mathrm{b}})]+3[\bar{\mathrm{b}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})]-21[\bar{\mathrm{b}}\cdot (\bar{\mathrm{a}}\times \bar{\mathrm{c}})]\\ & -5[\bar{\mathrm{c}}\cdot (\bar{\mathrm{a}}\times \bar{\mathrm{b}})]+[\bar{\mathrm{c}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})]-7[\bar{\mathrm{c}}\cdot (\bar{\mathrm{a}}\times \bar{\mathrm{c}})]\\ & =0+2[\bar{\mathrm{a}}\bar{\mathrm{b}}\mathrm{c}]+0\\ & +0+0+21[\bar{\mathrm{a}}\bar{\mathrm{b}}\bar{\mathrm{c}}]\\ & -5[\bar{\mathrm{a}}\bar{\mathrm{b}}\bar{\mathrm{c}}]+0+0\\ & =18[\bar{\mathrm{a}}\bar{\mathrm{b}}\bar{\mathrm{c}}]\\ \therefore & 18[\bar{\mathrm{a}}\bar{\mathrm{b}}\bar{\mathrm{c}}]=54\\ & \Rightarrow [\bar{\mathrm{a}}\bar{\mathrm{b}}\bar{\mathrm{c}}]=3\end{array}$

Correct answer option is (B)

$\begin{array}{rl}& \text{ Volume of parallelepiped }=\left[\begin{array}{lll}\overrightarrow{\mathrm{u}}& \overrightarrow{\mathrm{v}}& \overrightarrow{\mathrm{w}}\end{array}\right]\\ & \left|\begin{array}{lll}1& 1& \lambda \\ 1& 1& 3\\ 2& 1& 1\end{array}\right|=1\\ & \Rightarrow \lambda =2\\ & \therefore \cos \theta =\frac{2+1+2}{\sqrt{6}\cdot \sqrt{6}}=\frac{5}{6}\end{array}$

Correct answer option is (D)

We have $\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})=4$

Volume of required parallelopiped is

$\begin{array}{rl}& (\bar{\mathrm{a}}+2\bar{\mathrm{b}})\cdot [(\bar{\mathrm{b}}+2\bar{\mathrm{c}})\times (\bar{\mathrm{c}}+2\bar{\mathrm{a}})]\\ & =(\bar{\mathrm{a}}+2\bar{\mathrm{b}})\cdot [(\bar{\mathrm{b}}\times \bar{\mathrm{c}})+2(\bar{\mathrm{c}}\times \bar{\mathrm{c}})+2(\bar{\mathrm{b}}\times \bar{\mathrm{a}})+4(\bar{\mathrm{c}}\times \bar{\mathrm{a}})]\\ & =\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})+\bar{\mathrm{a}}(0)+2\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{a}})+4\bar{\mathrm{a}}\cdot (\bar{\mathrm{c}}\times \bar{\mathrm{a}})+2\bar{\mathrm{b}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})+4\bar{\mathrm{b}}(0)+4\bar{\mathrm{b}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{a}})+8\bar{\mathrm{b}}\cdot (\bar{\mathrm{c}}\times \bar{\mathrm{a}})\\ & =\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})+0+0+0+0+0+0+8\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})=9[\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})]=9(4)=36\end{array}$

Correct answer option is (C)

As per data given, we write

$24=\frac{1}{6}\left\{\left(\bar{a}+\bar{b}\right).\left[\left(\bar{b}+\bar{c}\right)\times \left(\bar{c}+\bar{a}\right)\right]\right\}$

$=\frac{1}{6}\left\{\left(\bar{a}+\bar{b}\right).\left[\left(\bar{b}+\bar{c}\right)+\left(\bar{b}+\bar{a}\right)+\left(\bar{c}+\bar{a}\right)\right]\right\}$

... [$\because$ $\bar{c}\times \bar{c}=0$]

$\therefore$ $144=\left[\bar{a}.\left(\bar{b}\times \bar{c}\right)\right]+\left[\bar{b}.\left(\bar{b}\times \bar{c}\right)\right]+\left[\bar{a}.\left(\bar{b}\times \bar{a}\right)\right]+\left[\bar{b}.\left(\bar{b}\times \bar{a}\right)\right]+\left[\bar{a}.\left(\bar{c}\times \bar{a}\right)\right]+\left[\bar{b}.\left(\bar{c}\times \bar{a}\right)\right]$

$=\left[\bar{a}.\left(\bar{b}\times \bar{c}\right)\right]+0+0+0+0+\left[\bar{b}.\left(\bar{c}\times \bar{a}\right)\right]$

$=2\left[\bar{a}.\left(\bar{b}\times \bar{c}\right)\right]$ ...... [$\because$ $\bar{b}.\left(\bar{c}\times \bar{a}\right)=\bar{a}.\left(\bar{b}\times \bar{c}\right)$]

$\therefore$ $\bar{a}.\left(\bar{b}\times \bar{c}\right)=72$, i.e. volume of parallelopiped with conterminous edges $\bar{\mathrm{a}},\bar{\mathrm{b}},\bar{\mathrm{c}}$.