Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be , and respectively. The altitude from the vertes 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

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Solution

Q.
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 [2025]

1 $\frac{1}{2} (\hat{i} + 4 \hat{j} + 7 \hat{k})$

2 $\frac{1}{6} (7 \hat{i} + 12 \hat{j} + \hat{k})$

3 $\frac{1}{6} (12 \hat{i} + 12 \hat{j} + \hat{k})$

4 $\frac{1}{12} (7 \hat{i} + 4 \hat{j} + 3 \hat{k})$

Ans.
(2)

Coordinates of F are $(\frac{3}{2}, 2, \frac{- 3}{2})$ .

Area of $△$ ABC $= \frac{1}{2}$ $| \vec{A B} \times \vec{A C} |$

$= \frac{1}{2}$ $| - 5 \hat{i} - 3 \hat{j} - \hat{k} |$ $= \frac{\sqrt{35}}{2} sq . units$

Volume of Tetrahedron $= \frac{1}{3} \times Base Area \times Height$

$= \frac{1}{3} \times \frac{\sqrt{35}}{2} \times D E$

$∴ \frac{\sqrt{35}}{6} \times D E = \frac{\sqrt{805}}{6 \sqrt{2}} \Rightarrow D E = \sqrt{\frac{23}{2}}$

$∴ In △ A D E, A E = \sqrt{A D^{2} - D E^{2}} = \sqrt{\frac{13}{18}}$ $[∵ A D = \frac{\sqrt{110}}{3}]$

$∴ \vec{A E} =$ $| \vec{A E} |$ $(\frac{\hat{i} - 5 \hat{k}}{\sqrt{26}}) = \sqrt{\frac{13}{18}} (\frac{\hat{i} - 5 \hat{k}}{\sqrt{26}}) = \frac{\hat{i} - 5 \hat{k}}{6}$

$∴$ Position vector of 'E' $= (\frac{i - 5 k}{6}) + (\hat{i} + 2 \hat{j} + \hat{k})$

$= \frac{1}{6} (7 \hat{i} + 12 \hat{j} + \hat{k})$ .

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