Question

# Area of a rectangle having vertices A, B, C, D with position vectors $- \hat i + \dfrac{1}{2}\hat j + 4\hat k,{\text{ }}\hat i + \dfrac{1}{2}\hat j + 4\hat k,{\text{ }}\hat i - \dfrac{1}{2}\hat j + 4\hat k,{\text{ }} - \hat i - \dfrac{1}{2}\hat j + 4\hat k$ respectively is,$a.{\text{ }}\dfrac{1}{2} \\ b.{\text{ }}1 \\ c.{\text{ }}2 \\ d.{\text{ }}4 \\$

Hint: In this question first find out the position vectors of the adjacent sides, later on apply the property of cross product, so, using these concepts we can reach the solution of the question.

Position vectors of the rectangle having vertices A, B, C, D are given as
$\vec O\vec A = - \hat i + \dfrac{1}{2}\hat j + 4\hat k \\ \vec O\vec B = \hat i + \dfrac{1}{2}\hat j + 4\hat k \\ \vec O\vec C = \hat i - \dfrac{1}{2}\hat j + 4\hat k \\ \vec O\vec D = - \hat i - \dfrac{1}{2}\hat j + 4\hat k \\$
The adjacent sides of rectangle ABCD is given as (AB, BC)
So, the position vectors of these sides is given as
$\left( {\vec A\vec B} \right) = \left( {\vec O\vec B} \right) - \left( {\vec O\vec A} \right)$
Now substitute the above values in above equation we have,
$\Rightarrow \left( {\vec A\vec B} \right) = \left( {\hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) - \left( { - \hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) \\ \Rightarrow \left( {\vec A\vec B} \right) = \left( {1 + 1} \right)\hat i + \left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + \left( {4 - 4} \right)\hat k = 2\hat i \\$
And, $\left( {\vec B\vec C} \right) = \left( {\vec O\vec C} \right) - \left( {\vec O\vec B} \right)$
Now substitute the above values in above equation we have,

$\Rightarrow \left( {\vec B\vec C} \right) = \left( {\hat i - \dfrac{1}{2}\hat j + 4\hat k} \right) - \left( {\hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) \\ \Rightarrow \left( {\vec B\vec C} \right) = \left( {1 - 1} \right)\hat i + \left( { - \dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + \left( {4 - 4} \right)\hat k = - \hat j \\$
Now we all know that the area of the rectangle whose adjacent sides are $\vec a$ and $\vec b$ is the modulus of cross product of these vectors which is $\left| {\left( {\vec a \times \vec b} \right)} \right|$.
So the area of the rectangle having adjacent sides $\left( {\vec A\vec B} \right)$ and $\left( {\vec B\vec C} \right)$ is $\left| {\left( {\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right)} \right)} \right|$
Now first calculate cross product of these vectors we have
$\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\vec j}&{\vec k} \\ 2&0&0 \\ 0&{ - 1}&0 \end{array}} \right| = \left( {0 - 0} \right)\hat i - \left( {0 - 0} \right)\vec j + \left( { - 2 - 0} \right)\vec k = - 2\vec k$
Now the modulus of above cross product is
$\left| {\left( {\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right)} \right)} \right| = \left| { - 2\hat k} \right| = \sqrt {{{\left( { - 2} \right)}^2}} = 2$
So, the area of the rectangle is 2 sq. units.
Hence, option (c) is correct.

Note: In such types of questions the key concept we have to remember is that always recall that the area of rectangle whose adjacent sides are $\vec a$ and $\vec b$ is the modulus of cross product of these vectors which is$\left| {\left( {\vec a \times \vec b} \right)} \right|$, so first calculate the adjacent sides of the rectangle as above then apply this concept, we will get the required area of the rectangle.