Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Area of a triangle with adjacent sides determined by vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{is}1\text{. Then the area of}& \text{(P)}& 6\\ & \text{the triangle with adjacent sides determined by}(3\overrightarrow{a}+4\overrightarrow{b})\text{and}(\overrightarrow{a}-3\overrightarrow{b})\text{is}& & \\ \text{(B)}& \text{Volume of parallelopiped determined by vectors}\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{is}\frac{1}{4}\text{. Then the volume of the}& \text{(Q)}& 9\\ & \text{parallelopiped determined by vectors}3(\overrightarrow{a}+\overrightarrow{b}),(\overrightarrow{b}+\overrightarrow{c}),4(\overrightarrow{c}+\overrightarrow{a})\text{is}& & \\ \text{(C)}& \text{Area of a parallelogram with adjacent sides determined by vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{is}8\text{. Then the}& \text{(R)}& 13\\ & \text{area of the parallelogram with adjacent sides determined by vectors}(2\overrightarrow{a}-\overrightarrow{b})\text{and}\overrightarrow{b}\text{is}& & \\ \text{(D)}& \text{Volume of tetrahedron determined by vectors}\overrightarrow{a},\overrightarrow{b}\text{and}\overrightarrow{c}\text{is}\frac{1}{2}\text{. Then the volume of the}& \text{(S)}& 16\\ & \text{tetrahedron determined by vectors}2(\overrightarrow{a}\times \overrightarrow{b})\text{,}3(\overrightarrow{b}\times \overrightarrow{c})\text{and}(\overrightarrow{c}\times \overrightarrow{a})\text{is}& & \end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(P)}$
• B. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(P)}$
• C. $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(S)}$
• D. $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(Q)}$

Answer 1

The correct option is D $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(Q)}$

$\text{(C)}$
Given, $|\overrightarrow{a}\times \overrightarrow{b}|=8$
$\therefore |(2\overrightarrow{a}-\overrightarrow{b})\times \overrightarrow{b}|$
$=2|\overrightarrow{a}\times \overrightarrow{b}|=16$

$\text{(D)}$
Given, $\frac{1}{6}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\frac{1}{2}$
$\Rightarrow \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=3$
$\therefore \frac{1}{6}\left[2\right(\overrightarrow{a}\times \overrightarrow{b}\left)3\right(\overrightarrow{b}\times \overrightarrow{c}\left)\right(\overrightarrow{c}\times \overrightarrow{a}\left)\right]$
$=[\overrightarrow{a}\times \overrightarrow{b}\overrightarrow{b}\times \overrightarrow{c}\overrightarrow{c}\times \overrightarrow{a}]$
$={\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2=9$