The area (in sq. units) of the parallelogram whose diagonals are along the vectors Option 1) 26 Option 2) 65 Option 3) 20 Option 4) 52

As we learnt in

Vector Product of two vectors(cross product) -

If $\vec{a}$ and $\vec{b}$ are two vectors and $\Theta$ is the angle between them , then $\vec{a}\times \vec{b}=\left |\vec{a} \left | \right |\vec{b} \right |Sin\Theta \hat{n}$

- wherein

$\hat{n}$ is unit vector perpendicular to both $\vec{a} \: and \: \vec{b}$

and

Position vector of a point -

$\mid \overrightarrow{OP}\mid = \sqrt{x^{2}+y^{2}+z^{2}} = r$

- wherein

Area of parallelogram = $\frac{1}{2}\left | \vec{a}\times \vec{b} \right |$

where $\vec{a} \:and \:\vec{b}$ are diagonals

$= \frac{1}{2}\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 8 &-6 & 0\\ 3 & 4 & -12 \end{vmatrix}$

$= \left | \frac{1}{2}\left ( 72\hat{i} +96\hat{j}+50\hat{k}\right ) \right |$

$=\left | 36\hat{i}+48\hat{j}+25\hat{k} \right |$

magnitude$=\sqrt{36^{2}+48^{2}+25^{2}}$

$=65$

Option 1)

26

This option is incorrect.

Option 2)

65

This option is correct.

Option 3)

20

This option is incorrect.

Option 4)

52

This option is incorrect.

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