Question

If.${v_1},{v_2},{v_3}$. are the volumes of parallelepiped, triangular prism and tetrahedron respectively and three conterminous edges of all figures are $\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $ then ${v_1},{v_2},{v_3}$ equals to
$
  A)\,1:3:6 \\
  B)\,6:1:3 \\
  C)\,6:3:1 \\
  D)None\,\,of\,these \\
$

Answer 1

Hint: Use the volume formula of volume when conterminous edges are given for the given figures and then put them in ratio form.
We are going to use the formula for the volume of parallelepiped, the triangular prism and tetrahedron. Since they have vector edges which are conterminous. We have a volume formula for them. We use them and put them in ratio form with the help of like terms and find the ratio.

Complete step by step solution:
 We are given volume of parallelepiped, triangular prism and tetrahedron as ${v_1},{v_2},{v_3}$ and also we are given the three conterminous edges of these figures are $\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $.
We know that the formula for the volume of parallelepiped whose edges are $\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $ is
${v_1} = \left[ {\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to } \right]$
We know that the formula for the volume of triangular prism whose edges are $\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $is
${v_2} = \dfrac{1}{2}\left[ {\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to } \right]$

We know that the formula for the volume of tetrahedron whose edges are $\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $is
${v_3} = \dfrac{1}{6}\left[ {\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to } \right]$
Now that we have volumes of the parallelepiped, triangular prism, tetrahedron, now we can put them in ratio form as they have similar terms.
So, we get

${v_1}:{v_2}:{v_3} = \left[ {\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to } \right]:\dfrac{1}{2}\left[ {\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to } \right]:\dfrac{1}{6}\left[ {\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to } \right]$
Now, cancel the like terms and then we get
${v_1}:{v_2}:{v_3} = 1:\dfrac{1}{2}:\dfrac{1}{6}$
We, have to simplify this such that we find an answer from the given options
Then, we get
${v_1}:{v_2}:{v_3} = 6:3:1$

Hence, option C is the right answer.

Note: Here, we have to be careful while taking ratio, as we must not assign the volume formula of one figure to another formula for volume of another figure, due to which will give us the wrong ratio.