Question

If the volume of the tetrahedron whose verticles are with position vectors

$\hat{i}-6\hat{j}+10\hat{k},-\hat{i}-3\hat{j}+7\hat{k},5\hat{i}-\hat{j}+\lambda \hat{k}$ and $7\hat{i}-4\hat{j}+7\hat{k}$ is 11 cubic units, then find the value of $\lambda$.

Answer 1

$Volume=\frac{1}{6}\left[\begin{array}{ccc}a& b& c\end{array}\right]=\frac{1}{6}\overline{a}.(\overline{b}\times \overline{c})$

$Positionvectors:$

$\overline{A}=\stackrel{\wedge }{i}-6\stackrel{\wedge }{j}+10\stackrel{\wedge }{k}$

$\overline{B}=-\stackrel{\wedge }{i}-3\stackrel{\wedge }{j}+7\stackrel{\wedge }{k}$

$\overline{C}=5\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+\lambda \stackrel{\wedge }{k}$

$\overline{D}=7\stackrel{\wedge }{i}-4\stackrel{\wedge }{j}+7\stackrel{\wedge }{k}$

$\overline{AB}=\stackrel{\wedge }{i}(-1-1)+\stackrel{\wedge }{j}(-3+6)+\stackrel{\wedge }{k}(7-10)$

$=-2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}-3\stackrel{\wedge }{k}$

$\overline{AC}=\stackrel{\wedge }{i}(5-1)+\stackrel{\wedge }{j}(-1+6)+\stackrel{\wedge }{k}(\lambda -10)$

$=\stackrel{\wedge }{i}\left(4\right)+\stackrel{\wedge }{j}\left(5\right)+\stackrel{\wedge }{k}(\lambda -10)$

$\overline{AD}=\stackrel{\wedge }{i}(7-1)+\stackrel{\wedge }{j}(-4+6)+\stackrel{\wedge }{k}(7-10)=\stackrel{\wedge }{i}\left(6\right)+\stackrel{\wedge }{j}\left(2\right)+\stackrel{\wedge }{k}(-3)$

$Volume=\frac{1}{6}\left|\begin{array}{ccc}-2& 3& -3\\ 4& 5\lambda & -10\\ 6& 2& -3\end{array}\right|$

$=\frac{1}{6}[-2(-15-2(\lambda -10))-3(-12-6(\lambda -10))-3(8-30\left)\right]$

$=\frac{1}{6}[30+4\lambda -40+36+18\lambda -180-24+90]$

$=\frac{1}{6}[66+22\lambda -154]$

$\because V=11$

$⟹11=\frac{1}{6}[66+22\lambda -154]$

$⟹22\lambda -154$

$⟹22\lambda =154$

$⟹\lambda =7$