The volume of a tetrahedron with coterminus e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V_{tetra} = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.Gi

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(C) The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V_{tetra} = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.Given $V_{tetra} = \frac{64}{3}$,we have $\frac{1}{6} |[\bar{a} \bar{b} \bar{c}]| = \frac{64}{3}$,which implies $|[\bar{a} \bar{b} \bar{c}]| = 128$.The volume of a parallelepiped with coterminus edges $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is given by the scalar triple product $|[(\bar{a}+\bar{b}) (\bar{b}+\bar{c}) (\bar{c}+\bar{a})]|$.Using the property of scalar triple products,$[(\bar{a}+\bar{b}) (\bar{b}+\bar{c}) (\bar{c}+\bar{a})] = 2 [\bar{a} \bar{b} \bar{c}]$.Thus,the volume is $|2 [\bar{a} \bar{b} \bar{c}]| = 2 	imes 128 = 256$ cubic units.

The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is $\frac{64}{3}$ cubic units. Then,the volume of a parallelepiped with coterminus edges given by the vectors $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is ... cubic units.

Similar Questions

The volume of the tetrahedron with $\hat{i}-\lambda \hat{j}+\hat{k}$, $\lambda \hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ as coterminous edges is $2$. If $\lambda$ is an integer, then $|\lambda \hat{i}-3 \lambda \hat{j}+3 \hat{k}|=$

If $|\vec{a}|=5, |\vec{b}|=3, |\vec{c}|=4$ and $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$ such that the angle between $\vec{b}$ and $\vec{c}$ is $\frac{5 \pi}{6}$,then $[\vec{a} \vec{b} \vec{c}]=$

The value of $(\vec{a} + 2\vec{b} - \vec{c}) \cdot \{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})\}$ is equal to

$i \cdot(j \times k)+j \cdot(k \times i)+k \cdot(j \times i)$ is equal to

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$. If $\vec{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$,then the projection of the vector $\vec{a}$ on the vector $2 \hat{i} + 2 \hat{j} + \hat{k}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module