JEE Main 2023 — Vector Algebra Question with Solution

Given that $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ are coplanar.

$\Rightarrow \overrightarrow{b}-\overrightarrow{a},\overrightarrow{c}-\overrightarrow{a},\overrightarrow{d}-\overrightarrow{a}$ are also coplanar.

$\Rightarrow \left[\begin{array}{ccc}\overrightarrow{b}-\overrightarrow{a}& \overrightarrow{c}-\overrightarrow{a}& \overrightarrow{d}-\overrightarrow{a}\end{array}\right]=0$

$\Rightarrow \left(\overrightarrow{b}-\overrightarrow{a}\right).\left(\left(\overrightarrow{c}-\overrightarrow{a}\right)\times \left(\overrightarrow{d}-\overrightarrow{a}\right)\right)=0$

$\Rightarrow \left(\overrightarrow{b}-\overrightarrow{a}\right).\left(\overrightarrow{c}\times \overrightarrow{d}-\overrightarrow{c}\times \overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{d}\right)=0 (\mathrm{Since} \overrightarrow{a}\times \overrightarrow{a}=0)$

$\Rightarrow \overrightarrow{b}.\left(\overrightarrow{c}\times \overrightarrow{d}\right)-\overrightarrow{b}.\left(\overrightarrow{c}\times \overrightarrow{a}\right)-\overrightarrow{b}.\left(\overrightarrow{a}\times \overrightarrow{d}\right)-\overrightarrow{a}.\left(\overrightarrow{c}\times \overrightarrow{d}\right)+\overrightarrow{a}.\left(\overrightarrow{c}\times \overrightarrow{a}\right)+\overrightarrow{a}.\left(\overrightarrow{a}\times \overrightarrow{d}\right)=0$

$\Rightarrow \left[\begin{array}{ccc}\overrightarrow{b}& \overrightarrow{c}& \overrightarrow{d}\end{array}\right]-\left[\begin{array}{ccc}\overrightarrow{b}& \overrightarrow{c}& \overrightarrow{a}\end{array}\right]-\left[\begin{array}{ccc}\overrightarrow{b}& \overrightarrow{a}& \overrightarrow{d}\end{array}\right]-\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{c}& \overrightarrow{d}\end{array}\right]=0$

We know that $\left[\begin{array}{ccc}\overrightarrow{x}& \overrightarrow{y}& \overrightarrow{z}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{y}& \overrightarrow{z}& \overrightarrow{x}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{z}& \overrightarrow{x}& \overrightarrow{y}\end{array}\right]$ and $\left[\begin{array}{ccc}\overrightarrow{x}& \overrightarrow{y}& \overrightarrow{z}\end{array}\right]=-\left[\begin{array}{ccc}\overrightarrow{x}& \overrightarrow{z}& \overrightarrow{y}\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}\overrightarrow{b}& \overrightarrow{c}& \overrightarrow{d}\end{array}\right]+\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]+\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{d}& \overrightarrow{c}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]$

$\Rightarrow \left[\begin{array}{lll}\overrightarrow{c}& \overrightarrow{d}& \overrightarrow{b}\end{array}\right]+\left[\begin{array}{lll}\overrightarrow{b}& \overrightarrow{d}& \overrightarrow{a}\end{array}\right]+\left[\begin{array}{lll}\overrightarrow{d}& \overrightarrow{c}& \overrightarrow{a}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]$

Hence, the required answer is $\left[\begin{array}{lll}\overrightarrow{d}& \overrightarrow{c}& \overrightarrow{a}\end{array}\right]+\left[\begin{array}{lll}\overrightarrow{b}& \overrightarrow{d}& \overrightarrow{a}\end{array}\right]+\left[\begin{array}{lll}\overrightarrow{c}& \overrightarrow{d}& \overrightarrow{b}\end{array}\right]$