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)

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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