Examples On Vectors Applied To Tetrahedrons Set-2

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Example – 42

Find the angle between any two faces of a regular tetrahedral.

Solution: Refer to Fig - 51 Example 40

The angle between any two faces will obviously equal the angle between the normals to the two faces. Let us find the (acute) angle \(\theta \) between the normals to the faces OAC and OAB.

\[\begin{align}&{{\hat n}_{OAC}} = \frac{{\vec c \times \vec a}}{{\left| {\vec c \times \vec a} \right|}} = \frac{{\vec c \times \vec a}}{{\left| {\vec c} \right|\left| {\vec a} \right|\sin \pi /3}} \hfill \\\\&
  {{\hat n}_{OAB}} = \frac{{\vec a \times \vec b}}{{\left| {\vec a \times \vec b} \right|}} = \frac{{\vec a \times \vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|\sin \pi /3}} \hfill \\\\&
   \Rightarrow  \quad {{\hat n}_{OAC}} \cdot {{\hat n}_{OAB}} = \frac{{\left( {\vec c \times \vec a} \right) \cdot \left( {\vec a \times \vec b} \right)}}{{{{\left| {\vec a} \right|}^2}{{\left| {\vec b} \right|}^2}{{\sin }^2}\pi /3}}\left( {\because \,\,\,\,\,\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec c} \right|} \right) \hfill \\\\&\qquad\qquad\qquad\qquad= \frac{{\left( {\vec c \cdot \vec a} \right)\left( {\vec a \cdot \vec b} \right) - \left( {\vec c \cdot \vec b} \right)\left( {\vec a \cdot \vec a} \right)}}{{{{\left| {\vec a} \right|}^4}{{\sin }^2}\pi /3}} \hfill \\\\&\qquad\qquad\qquad\qquad= \frac{{\frac{{{{\left| {\vec a} \right|}^2}\left| {\vec b} \right|\left| {\vec c} \right|}}{4} - \frac{{{{\left| {\vec a} \right|}^2}\left| {\vec b} \right|\left| {\vec c} \right|}}{2}}}{{{{\left| {\vec a} \right|}^4}{{\sin }^2}\pi /3}} \hfill \\\\&\qquad\qquad\qquad\qquad = \frac{{ - \frac{1}{4}{{\left| {\vec a} \right|}^4}}}{{{{\left| {\vec a} \right|}^4} \cdot \frac{3}{4}}} = \frac{{ - 1}}{3} \hfill \\\\&\Rightarrow  \quad \cos \theta  = \frac{1}{3} \hfill \\\\&\Rightarrow\quad \theta ={{\cos }^{-1}}\left( {}^{1}\!\!\diagup\!\!{}_{3}\; \right) \hfill \\ \end{align} \]

Thus, the angle between any two faces equals \(\begin{align}{{\cos }^{-1}}\left( \frac{1}{3} \right).\end{align}\)

Example – 43

The position vectors of the vertices A, B and C of a tetrahedron ABCD are  \(\hat{i}+\hat{j}+\hat{k},\,\,\hat{i}\,and\,\,3\hat{i}\) respectively. The altitude from the vertex D to the opposite face ABC meets the median line through A of the triangle ABC at E. If AD is 4 units and the volume of the tetrahedron is  \(\begin{align}\frac{2\sqrt{2}}{3}\end{align}\) cubic units, find all the possible position vectors of the point E.

Solution:

\[\begin{align} & V=\frac{1}{3}\times \left( \frac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{AC} \right| \right)\times \left( DE \right) \\\\ & \Rightarrow \quad\frac{2\sqrt{2}}{3}=\frac{\left( DE \right)}{6}\times \left\{ \left| \left( -\hat{j}-\hat{k} \right)\times \left( 2\hat{i}-\hat{j}-\hat{k} \right) \right| \right\} \\\\ & \qquad\qquad\quad=\frac{DE}{6}\times \left\{ \left| -2\hat{j}+2\hat{k} \right| \right\}\\\\&\qquad\qquad\quad =\frac{2\sqrt{2}}{6}\times DE \\\\ & \Rightarrow\; \quad DE\;=\;2 \\ \end{align}\]

Since \(\Delta ADE\) is right - angled at E, we have

\[\begin{align}& AE=\sqrt{A{{D}^{2}}-D{{E}^{2}}} \\\\  &\quad\;\; =\sqrt{16-4} \\\\  & \quad\;\;=2\sqrt{3} \\ \end{align}\]

It is given that E lies on the median AF. Since F is the mid-point of BC, the position vector of F is \(2\hat{i}.\)

Now, our problem has been reduced to this; we know the position vectors of A and F and we know that \(AE=2\sqrt{3}.\) We need to find the position vector(s) of E.

Let us suppose E divides AF in the ratio \(\lambda :1.\) Then the position vector of E is

\[\begin{align} &\; E\equiv \frac{\lambda (2i)+1\cdot (\hat{i}+\hat{j}+\hat{k})}{1+\lambda } \\\\ &\quad=\left( \frac{1+2\lambda }{1+\lambda } \right)\hat{i}+\left( \frac{1}{1+\lambda } \right)\hat{j}+\left( \frac{1}{1+\lambda } \right)\hat{k}...\left( 1 \right) \\\\  & \Rightarrow \quad \left| \overrightarrow{AE} \right|=\left| \frac{\lambda }{\lambda +1}\left( \hat{i}-\hat{j}-\hat{k} \right) \right| \\\\ &\qquad\qquad\;\; =\left| \frac{\sqrt{3}\lambda }{\lambda +1} \right| \\ \end{align}\]

Since \(AE=2\sqrt{3},\) we have

\[\begin{align}& \quad\;\pm \frac{\sqrt{3}\lambda }{\lambda +1}=2\sqrt{3} \\ & \Rightarrow\quad \frac{\lambda }{\lambda +1}=\pm 2 \\  & \Rightarrow \quad\lambda =-2,\,\,-\frac{2}{3} \\ \end{align}\]

Using these values of  \(\lambda \) in (1), the possible position vectors of E are  \((3\hat{i}-\hat{j}-\hat{k})\,\,and\,\,(-\hat{i}+3\hat{j}+3\hat{k}).\)

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