Question

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

• A. $4\sqrt{6}$ sq. units
• B. $\frac{1}{2}\sqrt{21}$ sq. units
• C. $48$
• D. $\sqrt{6}$ sq. units

Answer 1

Answer: (A)

$a=i+j+k,b=i+3j+5k$
and $c=7i+9j+11k$
Let $A=a+b$
$=(i+j+k)+(i+3j+5k)$
$=2i+4j+6k$
and $B=b+c$
$=(i+3j+5k)+(7i+9j+11k)$
$=8i+12j+16k$
$\therefore$ Area of parallelogram
$=\frac{1}{2}\left|A\times B\right|$
($\because A$ and $B$ are diagonals)
$\left|\begin{array}{ccc}i& j& k\\ 2& 4& 6\\ 8& 12& 16\end{array}\right|$
$=\frac{1}{2}\left|i\left(64-72\right)-j\left(32-48\right)+k\left(24-32\right)\right|$
$=\frac{1}{2}\left|-8i+6j-8k\right|$
$=\sqrt{{\left(-4\right)}^2+{\left(8\right)}^2+{\left(-4\right)}^2}$
$=\sqrt{96}$
$=4\sqrt{6}$ sq units