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

Answer 1

The correct option is B $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(P)}$

$\text{(A)}$
Given, $\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|=1$
$\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=2$
$\therefore \frac{1}{2}|(3\overrightarrow{a}+4\overrightarrow{b})\times (\overrightarrow{a}-3\overrightarrow{b})|$
$=\frac{1}{2}|13(\overrightarrow{a}\times \overrightarrow{b})|=13$

$\text{(B)}$
Given, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\frac{1}{4}$
$\therefore \left[3\right(\overrightarrow{a}+\overrightarrow{b}\left)\right(\overrightarrow{b}+\overrightarrow{c}\left)4\right(\overrightarrow{c}+\overrightarrow{a}\left)\right]$
$=12\left(2\right)\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
$=\frac{24}{4}=6$