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The length of the edge of the regular tetrahedron D-ABC is $k$.Point E and F are taken on the edges AD and BD respectively such that $E$ divides $\vec{DA}$ and $F$ divides $\vec{BD}$ in the ratio 2:1 each.Then find the area of triangle $CEF$


Let the position vectors of $D,A,B,C$ are $\vec{0},\vec{a},\vec{b},\vec{c}$
Then by section formula,position vector of E is $\frac{2\vec{a}}{3}$ and position vector of F is $\frac{\vec{b}}{3}$
So the area of triangle $CEF=\frac{1}{2}|\vec{CE}\times\vec{CF}|=\frac{1}{2}|(\frac{2\vec{a}}{3}-\vec{c})\times (\frac{\vec{b}}{3}-\vec{c})|$
$=\frac{1}{2}|\frac{2}{9}\vec{a}\times\vec{b}-\frac{2}{3}\vec{a}\times\vec{c}-\frac{1}{3}\vec{c}\times\vec{b}|=\frac{7\sqrt3k^2}{36}$

Using the property that angle between two faces of regular tetrahedron is $\arccos{\frac{1}{3}}$


But the answer given in my book is $\frac{5\sqrt3k^2}{36}$.I dont know where i am wrong.

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I obtain the same result, indeed let consider the origin in C, then

  • $\vec{CE}=\vec d + \frac13(\vec a-\vec d)$

  • $\vec{CF}=\vec d + \frac13(\vec b-\vec d)$

then

$$CEF=\frac{1}{2}\left|\vec{CE}\times\vec{CF}\right|=\frac{1}{2}\left|\frac13\vec d \times \vec a+\frac13\vec d \times \vec b+\frac19\vec a \times \vec b \right|=\frac12\frac{7}{9}\frac{\sqrt 3}{2}k^2=\frac{7\sqrt 3}{36}k^2$$

otherwise for

  • $\vec{CE}=\vec d + \frac23(\vec a-\vec d)$

  • $\vec{CF}=\vec d + \frac23(\vec b-\vec d)$

then

$$CEF=\frac{1}{2}\left|\vec{CE}\times\vec{CF}\right|=\frac{1}{2}\left|\frac23\vec d \times \vec a+\frac23\vec d \times \vec b+\frac49\vec a \times \vec b \right|=\frac12\frac{16}{9}\frac{\sqrt 3}{2}k^2=\frac{4\sqrt 3}{9}k^2$$

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This same question was written in my maths module and I couldn't solve it. So, I searched it on this site and discovered that the answer is still mystery. However, after trying it myself I got the correct answer by using simple geometry as $\displaystyle \frac{5k^2}{12\sqrt 3}$ as given in OP's book. Here it is:

enter image description here

This is the sketch of the given situation. Now, break the three sides of the tetrahedron having common vertex $C$. It will look something like this

enter image description here

It is clear from the second figure that the triangles $\triangle CEA \cong \triangle CFD$. Hence, $CE=CF$.

Now, $CE$ can be calculated by applying Cosine Rule in $\triangle CEA$ knowing that $\displaystyle AE=FD=\frac{k}{3}$. We get, $$CE=CF=\frac{k\sqrt 7}{3}.$$ Similarly, in $\triangle DEF$, $$EF=\frac{k}{\sqrt 3}$$

Finally, you get $\triangle CEF$ as an isosceles one.

enter image description here

Now, I don't think it's a tough task to find the area of this triangle, which comes out to be $\displaystyle \frac{5k^2}{12\sqrt 3}$.

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