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 $\vec{ V }_{4}=\frac{1}{2}\{(\vec{ c }-\vec{ a }) \times(\vec{ b }-\vec{ a })\}$
$\therefore \vec{ V }_{1}+\vec{ V }_{2}+\vec{ V }_{3}+\vec{ V }_{4}=\frac{1}{2}(\vec{ a } \times \vec{ b }+\vec{ b } \times \vec{ c }+\vec{ c } \times \vec{ a }+$
$+\vec{ c } \times \vec{ b }-\vec{ c } \times \vec{ a }-\vec{ a } \times \vec{ b })=\vec{0}$
$\left|\vec{ V }_{1}+\vec{ V }_{2}+\vec{ V }_{3}+\vec{ V }_{4}\right|=\vec{0}$