Question

The area of the parallelogram having diagonals $\vec a = 3\hat i + \hat j - 2\hat k$ and $\vec b = \hat i - 3\hat j + 4\hat k$ is

1. $10\sqrt 3$
2. $5\sqrt 3$
3. $8$
4. $4$

Answer 1

Hint : To find the area of the parallelogram we have to use the formula of area of parallelogram whose diagonals are given, which is,
Area of parallelogram $= \dfrac{1}{2}\left| {\vec a \times \vec b} \right|$
Where, $\vec a$ and $\vec b$ are the diagonals of the parallelogram.
So, we are to find the cross product between the diagonals first by using the formula,
$\vec a \times \vec b = \left( {{x_1}\hat i + {y_1}\hat j + {z_1}\hat k} \right) \times \left( {{x_2}\hat i + {y_2}\hat j + {z_2}\hat k} \right)$
$\Rightarrow \vec a \times \vec b = \left( {{y_1}{z_2} - {z_1}{y_2}} \right)\hat i - \left( {{x_1}{z_2} - {x_2}{z_1}} \right)\hat j + \left( {{x_1}{y_2} - {x_2}{y_1}} \right)\hat k$
Then, we have to find the modulus value of the cross product of the diagonals, as to find the area we have to find $\left| {\vec a \times \vec b} \right|$ .
The formula of the modulus of a vector is,
$\left| {x\hat i + y\hat j + z\hat k} \right| = \sqrt {{x^2} + {y^2} + {z^2}}$ .

Complete step-by-step answer :
The two given diagonals of the parallelogram are,
$\vec a = 3\hat i + \hat j - 2\hat k$
$\vec b = \hat i - 3\hat j + 4\hat k$
Therefore, we know, the area of a parallelogram, if it’s two diagonals are given, is,
Area of parallelogram $= \dfrac{1}{2}\left| {\vec a \times \vec b} \right|$
Therefore, the cross product of the diagonals is,
$\vec a \times \vec b = \left( {{x_1}\hat i + {y_1}\hat j + {z_1}\hat k} \right) \times \left( {{x_2}\hat i + {y_2}\hat j + {z_2}\hat k} \right)$
$\Rightarrow \vec a \times \vec b = \left( {{y_1}{z_2} - {z_1}{y_2}} \right)\hat i - \left( {{x_1}{z_2} - {x_2}{z_1}} \right)\hat j + \left( {{x_1}{y_2} - {x_2}{y_1}} \right)\hat k$
Now, substituting the given values, we get,
$\Rightarrow \vec a \times \vec b = \left( {\left( 1 \right)\left( 4 \right) - \left( { - 3} \right)\left( { - 2} \right)} \right)\hat i - \left( {\left( 3 \right)\left( 4 \right) - \left( 1 \right)\left( { - 2} \right)} \right)\hat j + \left( {\left( 3 \right)\left( { - 3} \right) - \left( 1 \right)\left( 1 \right)} \right)\hat k$
Opening the brackets, we get,
$\Rightarrow \vec a \times \vec b = \left( {4 - 6} \right)\hat i - \left( {12 + 2} \right)\hat j + \left( { - 9 - 1} \right)\hat k$
Now, simplifying, we get,
$\Rightarrow \vec a \times \vec b = \left( { - 2} \right)\hat i - \left( {14} \right)\hat j + \left( { - 10} \right)\hat k$
$\Rightarrow \vec a \times \vec b = - 2\hat i - 14\hat j - 10\hat k$
Now, we know, $\left| {x\hat i + y\hat j + z\hat k} \right| = \sqrt {{x^2} + {y^2} + {z^2}}$
Therefore, to find the area of the parallelogram, we have to find, $\left| {\vec a \times \vec b} \right|$ .
So, $\left| {\vec a \times \vec b} \right| = \left| { - 2\hat i - 14\hat j - 10\hat k} \right|$
$\Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 14} \right)}^2} + {{\left( { - 10} \right)}^2}}$
Simplifying the equation, we get,
$\Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt {4 + 196 + 100}$
$\Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt {300}$
Therefore, we get,
$\Rightarrow \left| {\vec a \times \vec b} \right| = 10\sqrt 3$
Therefore, the area of parallelogram $= \dfrac{1}{2}\left| {\vec a \times \vec b} \right|$
Now, substituting the values, we get,
$= \dfrac{1}{2} \times 10\sqrt 3$
$= 5\sqrt 3$ sq. units.
Therefore, the area of the parallelogram is $5\sqrt 3$ sq. units.
So, the correct answer is “ $5\sqrt 3$ sq. units”.

Note : -The cross product of vectors is derived from the determinant of two vectors. If in case of diagonals the vectors of two sides of the parallelogram were given, then the area of the parallelogram would have been $\left| {\vec a \times \vec b} \right|$ , where $\vec a$ and $\vec b$ are the vectors of the two sides of a parallelogram.