Question

If diagonals of a parallelogram are given by a=3i-4j-k and b=2i+3j-6k so what will be the area

Answer 1

here you go , I hope this helps you

Answer 2

$\sqrt {42}$ is the Area

Step-by-step explanation:

• The diagonals of a parallelogram bisect each other, forming 4 triangles with equal area.  

Each of the 4 triangles that are formed by diagonals

$\overrightarrow{u}\ and\ \overrightarrow{v}$

have area

= $\frac {1}{2}\left \| \frac {u}{2} \times \frac {v}{2} \right \|$

= $\frac {1}{8}\left \| \overrightarrow{u}\ \times \overrightarrow{v}\ \right \|$

Area of parallelogram  =$\frac {1}{2}\left \| (3i + j-2k)\times(i+3j-4k) \right \|$

= $\frac {1}{2}\left \| 2i-10j+8k \right \|$

= $\frac {1}{2} \sqrt {2^2+10^2+8^2}$

= $\frac {1}{2}\sqrt {168}$

= $\sqrt {42}$