Question

Abcd is a parallelogram in which ab=3i-6j+3k and ad = 6i+6j+3k. P is a point of ab such that ap

Answer 1

Solution:

Concept used:

Area of a parallelogram having adjacent sides $\vec{a}\:and\:\vec{b}$ is

$|\vec{a}X\vec{b}|$

$\vec{AB}=3\vec{i}-6\vec{j}+3\vec{k}\\\vec{AD}=6\vec{i}+6\vec{j}+3\vec{k}$

Area of paralleloram ABCD

$=|\vec{AB}X\vec{AD}|$

$\vec{AB}X\vec{AD}$

$=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\3&-6&3\\6&6&3\end{array}\right|$

$=\vec{i}(-18-18)-\vec{j}(9-18)+\vec{k}(18+36)\\\\=-36\vec{i}+9\vec{j}+54\vec{k}\\\\=9(-4\vec{i}+\vec{j}+6\vec{k})$

$|\vec{AB}X\vec{AD}|=9\sqrt{{(-4)}^2+{1}^2+{6}^2}=9\sqrt{16+1+36}=9\sqrt{53}$

Area of the parallelogram ABCD is

$9\sqrt{53}$square units