Q94 / 200 · Manipal (MET) 2010 · Mathematics

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
  • Correct answer
    C
    $\overrightarrow{0}$
  • D
    None of these

Solution

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

Hence, the correct answer is option: 'C'.

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