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  • Volume of ter\trahedron whoes three edges are along  and <img alt="\vec{c}"

Volume of ter\trahedron whoes three edges are along \vec{a},\vec{b} and \vec{c} is 4 cubic units , then the volume of tetrahedron ,whoes three edges are along 2\vec{a}, 3\vec{b} and \frac{\vec{c}}{2} will be (in cubic units)

Option: 1

6


Option: 2

12


Option: 3

18


Option: 4

24


Answers (1)

best_answer

As we learn

Properties of Scalar Triple Product -

\left [K \vec{a} \vec{b} \vec{c} \right ]=K\left [ \vec{a} \vec{b} \vec{c} \right ]

- wherein

\vec{a}, \vec{b} and \vec{c} are the three vectors.

 

 \left | \frac{1}{6}\left [ \vec{a}\ \vec{b}\ \vec{c}\ \right ] \right |=4\Rightarrow \left | \left [\vec{a}\vec{b}\vec{c} \right ] \right |=24

New volume \left | \frac{1}{6}\left [ 2\vec{a}\ 3\vec{b}\ \frac{\ \vec{c}}{2} \right ] \right |

\left [ 2\vec{a}\ 3\vec{b}\ \ \frac{\vec{c}}{2} \right ]= 2\vec{a}\cdot (3\vec{b}*\frac{\vec{c}}{2})=3\vec{a}\cdot (\vec{b}*\vec{c}) =3 \begin{bmatrix} \vec{a} &\vec{b} & \vec{c} \end{bmatrix}

\therefore required volume =\left | \frac{1}{6}*3\begin{bmatrix} \vec{a} &\vec{b} &\vec{c} \end{bmatrix} \right |=\left | \frac{\begin{bmatrix} \vec{a} &\vec{b} &\vec{c} \end{bmatrix}}{2} \right |=12

Posted by

Ritika Harsh

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