Question

The volume of the tetrahedron with edges $\mathrm{i}+2\mathrm{j}-\mathrm{k},\mathrm{i}+\mathrm{j}+\mathrm{k}$ and $\mathrm{i}-\mathrm{j}+\lambda \mathrm{k}$ as coterminous, is $\frac{2}{3}$ cubic unit Then $\lambda$ equals?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$1$

Explanation of the correct option:

Given : The edges of of the tetrahedron are $\mathrm{i}+2\mathrm{j}-\mathrm{k},\mathrm{i}+\mathrm{j}+\mathrm{k}$ and $\mathrm{i}-\mathrm{j}+\lambda \mathrm{k}$ and volume is $\frac{2}{3}$cubic unit

SInce, Volume of tetrahedron$=\frac{1}{6}\left[\begin{array}{ccc}{u}_1& u_2& u_3\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right]$

$\Rightarrow$ $\frac{2}{3}=\frac{1}{6}\left[\begin{array}{ccc}1& 2& -1\\ 1& 1& 1\\ 1& -1& \lambda \end{array}\right]$

$\Rightarrow$ $4=1\left(\lambda +1\right)-1\left(2\lambda -1\right)+1\left(2+1\right)$

$\Rightarrow$ $4=\lambda +1-2\lambda +1+2+1$

$\Rightarrow$ $4=-\lambda +5$

$\Rightarrow$ $\lambda =1$

Therefore, the value of $\lambda$ is $1$.

Hence option A is the correct option.