Q&A - Ask Doubts and Get Answers
Q

Give answer! - Vector Algebra - JEE Main

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)
115 Views

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.

Exams
Articles
Questions