Question

If $a=i+j+k,b=i+3j+5k$ and $c=7i+9j+11k$, then the area of Parallelogram having diagonals a+b and b+c is.

• A. $4\sqrt{6}$ sq. units
• B. $\frac{1}{2}\sqrt{21}$ sq. units
• C. $\frac{\sqrt{6}}{2}$ sq.units
• D. $\sqrt{6}$ sq. units

Answer 1

The correct option is A $4\sqrt{6}$ sq. units

$a=i+j+k,b=i+3j+5k$ and $c=7i+9j+11k$

Let $A=a+b$
$=(i+k)+(i+3j+5k)$
$=2i+4j+6k$
and $B=b+c$
$=(i+3j+5k)+(7i+9j+11k)$
$=8i+12j+16k$
$\therefore$ Area of parallelogram
$\frac{1}{2}|A\times B|$
$=\frac{1}{2}\left|\begin{array}{ccc}i& j& k\\ 2& 4& 6\\ 8& 12& 16\end{array}\right|$ ($\because$ A and B are diagram)

$=\frac{1}{2}\left|i\right(64-72)-j(32-48)+k(24-32\left)\right|$

$=\frac{1}{2}|-8i+16j-8k|$

$=\sqrt{(-4{)}^2+(8{)}^2+(-4{)}^2}$

$=\sqrt{96}=4\sqrt{6}$ sq units

Hence, area of parallelogram $=4\sqrt{6}$ sq units.