Question

Four vertices O, A, B, C of a tetrahedron satisfy

$\left[\begin{array}{c}\overrightarrow{OA}\times \overrightarrow{OB}=\hat{i}-\hat{j}+\hat{k}\\ \overrightarrow{OB}\times \overrightarrow{OC}=\hat{i}\\ \overrightarrow{OC}\times \overrightarrow{OA}=-\hat{i}+\hat{j}\end{array}\right]$ where 'O' is the the origin.

Then $|\overrightarrow{CA}\times \overrightarrow{CB}|$ has the value equal to

• A. $1/2$
• B. $1/\sqrt{2}$
• C. $\sqrt{2}$
• D. $2$

Answer 1

The correct option is C $\sqrt{2}$

The concept here is "the sum of all normal vectors in a tetrahedron is zero"
$\overrightarrow{OA}\times \overrightarrow{OB}+\overrightarrow{OB}\times \overrightarrow{OC}+\overrightarrow{OC}\times \overrightarrow{OA}+\overrightarrow{CB}\times \overrightarrow{CA}=0$
$\overrightarrow{CB}\times \overrightarrow{CA}=-\left(\right(\hat{i}-\hat{j}+\hat{k})+(\hat{i})+(-\hat{i}+\hat{j}\left)\right)$
$|\overrightarrow{CB}\times \overrightarrow{CA}|=$ Magnitude is $\sqrt{2}$