Q76 / 160 · AP EAMCET 2009 · Mathematics

The volume of the tetrahedron having the edges $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ as coterminous, is $\frac{2}{3}$ cubic unit. Then $\lambda$ equals

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    4

Solution

$\begin{aligned} & \text { Let } \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} \end{aligned}$ Since, volume of tetrahedron $=\frac{1}{6}[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$ $\begin{array}{ll} \Rightarrow & \frac{2}{3}=\frac{1}{6}\left|\begin{array}{rrr} 1 & 2 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & \lambda \end{array}\right| \\ \Rightarrow & \frac{2}{3}=\frac{1}{6}[1(\lambda+1)-2(\lambda-1)-1(-1-1)] \\ \Rightarrow & 4=[-\lambda+5] \Rightarrow \lambda=1 \end{array}$

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

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