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

Answer 1

Answer: (A)

$4 \sqrt{6}$ s units

Answer 2

$a = i + j + k, b = i + 3j + 5k$
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)
$\begin{vmatrix}i&j;&k;\\\ 2&4&6\\\ 8&12&16\end{vmatrix}$
$= \frac{1}{2} \left|i\left(64-72\right) - j \left(32 - 48\right) + k \left(24 - 32\right)\right|$
$= \frac{1}{2} \left| - 8i + 6j - 8 k\right|$
$= \sqrt{\left(-4\right)^{2} + \left(8\right)^{2}+\left(-4\right)^{2}}$
$= \sqrt{96}$
$= 4\sqrt{6}$ sq units