Let the position vectors of the vertices A,B and C of a tetrahedron ABCD be i+2j+k,i+3j-2k…

Question

Let the position vectors of the vertices $A,B$ and C of a tetrahedron ABCD be $\hat{i}+2\hat{j}+\hat{k},\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line segment through $A$ of the triangle $ABC$ at the point $E$. If the length of $AD$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of E is

Accepted Answer

Area of $	riangleABC=\frac{1}{2}|\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}|$

$=\frac{1}{2}|5\hat{i}+3\hat{j}+\hat{k}|=\frac{1}{2}\sqrt{35}$

volume of tetrahedron

$\begin{gathered} =\frac{1}{3}	imes 	imes h=\frac{\sqrt{805}}{6\sqrt{2}} \ \frac{1}{3}	imes \frac{1}{2}\sqrt{35}	imes h=\frac{\sqrt{805}}{6\sqrt{2}} \ h=\sqrt{\frac{23}{2}} \end{gathered}$

${AE}^{2}={AD}^{2}-{DE}^{2}=\frac{13}{18}	herefore AE=\sqrt{\frac{13}{18}}$

$\begin{gathered} \overset{	o \leftarrow }{AE}=|AE|\cdot (\frac{\hat{i}-5\hat{k}}{\sqrt{26}}) \ =\sqrt{\frac{13}{18}}\cdot (\frac{\hat{i}-5\hat{k}}{\sqrt{26}}) \ =\sqrt{\frac{13}{18}}\cdot (\frac{\hat{i}-5\hat{k}}{\sqrt{26}})=\frac{\hat{i}-5\hat{k}}{6} \ E=\frac{\hat{i}-5\hat{k}}{6}+\hat{i}+2\hat{j}+\hat{k}=\frac{1}{6}(7\hat{i}+12\hat{j}+\hat{k}) \end{gathered}$

Solutions · 1