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The area (in sq. units) of the parallelogram whose diagonals are along the vectors

$\small 8 \hat{i}- 6 \hat{j} \; \; and\: \: 3 \hat{i} + 4\hat{j} -12\hat{k},$

Option 1)
26

Option 2)
65

Option 3)
20

Option 4)
52

Answers (1)

best_answer

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

fig 5

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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prateek

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