Question

Consider a tetrahedron with faces $F_1,F_2,F_3,F_4$. Let ${\vec{V}}_1,{\vec{V}}_2,{\vec{V}}_3,{\vec{V}}_4$be the vectors whose magnitudes are respectively equal to areas of $F_1,F_2,F_3,F_4$ and whose directions are perpendicular to these faces in outward direction, then $\left|{\vec{V}}_1+{\vec{V}}_2+{\vec{V}}_3+{\vec{V}}_4\right|$ equals

• A. $1$
• B. $4$
• C. $\vec{0}$
• D. None of these

Answer 1

Answer: (C)

We have,
${\vec{V}}_1=\frac{1}{2}(\vec{a}\times \vec{b}),{\vec{V}}_2=\frac{1}{2}(\vec{b}\times \vec{c}),{\vec{V}}_3=\frac{1}{2}(\vec{c}\times \vec{a})$
and ${\vec{V}}_4=\frac{1}{2}\{(\vec{c}-\vec{a})\times (\vec{b}-\vec{a})\}$
$\therefore {\vec{V}}_1+{\vec{V}}_2+{\vec{V}}_3+{\vec{V}}_4=\frac{1}{2}(\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}+$
$+\vec{c}\times \vec{b}-\vec{c}\times \vec{a}-\vec{a}\times \vec{b})=\vec{0}$
$\left|{\vec{V}}_1+{\vec{V}}_2+{\vec{V}}_3+{\vec{V}}_4\right|=\vec{0}$