Question

Let $ABCD$ is a regular tetrahedron, and $E$ and $F$ are the mid-points of the edges $AC$ and $AB$ respectively and $G$ is the centroid of the face $ABC$. Let $\theta$ is the angle between the vectors $\overrightarrow{EG}$ and $\overrightarrow{DF}$, then

• A. The angle between $\overrightarrow{DA}$ and $\overrightarrow{BC}$ is $\frac{\pi}{3}$
• B. The angle between $\overrightarrow{AB}$ and $\overrightarrow{DC}$ is $\frac{\pi}{2}$
• C. $\theta = \cos^{-1}(-\frac{1}{2\sqrt{3}})$
• D. $\theta = \tan^{-1}(\sqrt{11})$

Answer 1

Answer: (B)

The angle between $\overrightarrow{AB}$ and $\overrightarrow{DC}$ is $\frac{\pi}{2}$

Answer 2

Let the side length of the regular tetrahedron be $a$. We place vertex $A$ at the origin, so $\overrightarrow{A} = \mathbf{0}$. Let the position vectors of the other vertices be $\overrightarrow{B}$, $\overrightarrow{C}$, and $\overrightarrow{D}$. For a regular tetrahedron with side length $a$: $|\overrightarrow{B}| = |\overrightarrow{C}| = |\overrightarrow{D}| = a$. The angle between any two edges originating from the same vertex is $60^\circ$ ($\pi/3$). Thus, the dot products are: $\overrightarrow{B} \cdot \overrightarrow{C} = |\overrightarrow{B}| |\overrightarrow{C}| \cos(\pi/3) = a \cdot a \cdot \frac{1}{2} = \frac{a^2}{2}$. Similarly, $\overrightarrow{B} \cdot \overrightarrow{D} = \frac{a^2}{2}$ and $\overrightarrow{C} \cdot \overrightarrow{D} = \frac{a^2}{2}$.

$E$ is the midpoint of $AC$, so $\overrightarrow{AE} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}\overrightarrow{C}$. $F$ is the midpoint of $AB$, so $\overrightarrow{AF} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}\overrightarrow{B}$. $G$ is the centroid of the face $ABC$. The position vector of the centroid of a triangle with vertices $A, B, C$ is $\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$. Since $A$ is the origin, $\overrightarrow{AG} = \frac{\overrightarrow{0}+\overrightarrow{B}+\overrightarrow{C}}{3} = \frac{\overrightarrow{B}+\overrightarrow{C}}{3}$.

Now, we find the vectors $\overrightarrow{EG}$ and $\overrightarrow{DF}$: $\overrightarrow{EG} = \overrightarrow{AG} - \overrightarrow{AE} = \frac{\overrightarrow{B}+\overrightarrow{C}}{3} - \frac{\overrightarrow{C}}{2} = \frac{2(\overrightarrow{B}+\overrightarrow{C}) - 3\overrightarrow{C}}{6} = \frac{2\overrightarrow{B} - \overrightarrow{C}}{6}$. $\overrightarrow{DF} = \overrightarrow{AF} - \overrightarrow{AD} = \frac{\overrightarrow{B}}{2} - \overrightarrow{D}$.

To find the angle $\theta$ between $\overrightarrow{EG}$ and $\overrightarrow{DF}$, we use the formula $\cos \theta = \frac{\overrightarrow{EG} \cdot \overrightarrow{DF}}{|\overrightarrow{EG}| |\overrightarrow{DF}|}$.

Calculate the dot product $\overrightarrow{EG} \cdot \overrightarrow{DF}$: $\overrightarrow{EG} \cdot \overrightarrow{DF} = \left(\frac{2\overrightarrow{B} - \overrightarrow{C}}{6}\right) \cdot \left(\frac{\overrightarrow{B}}{2} - \overrightarrow{D}\right)$ $= \frac{1}{12} (2\overrightarrow{B} - \overrightarrow{C}) \cdot (\overrightarrow{B} - 2\overrightarrow{D})$ $= \frac{1}{12} (2\overrightarrow{B}\cdot\overrightarrow{B} - 4\overrightarrow{B}\cdot\overrightarrow{D} - \overrightarrow{C}\cdot\overrightarrow{B} + 2\overrightarrow{C}\cdot\overrightarrow{D})$ Substitute the dot product values: $= \frac{1}{12} \left(2a^2 - 4\left(\frac{a^2}{2}\right) - \frac{a^2}{2} + 2\left(\frac{a^2}{2}\right)\right)$ $= \frac{1}{12} \left(2a^2 - 2a^2 - \frac{a^2}{2} + a^2\right) = \frac{1}{12} \left(\frac{a^2}{2}\right) = \frac{a^2}{24}$.

Calculate the magnitudes $|\overrightarrow{EG}|$ and $|\overrightarrow{DF}|$: $|\overrightarrow{EG}|^2 = \left|\frac{2\overrightarrow{B} - \overrightarrow{C}}{6}\right|^2 = \frac{1}{36} (2\overrightarrow{B} - \overrightarrow{C}) \cdot (2\overrightarrow{B} - \overrightarrow{C})$ $= \frac{1}{36} (4|\overrightarrow{B}|^2 - 4\overrightarrow{B}\cdot\overrightarrow{C} + |\overrightarrow{C}|^2)$ $= \frac{1}{36} \left(4a^2 - 4\left(\frac{a^2}{2}\right) + a^2\right) = \frac{1}{36} (4a^2 - 2a^2 + a^2) = \frac{3a^2}{36} = \frac{a^2}{12}$. $|\overrightarrow{EG}| = \sqrt{\frac{a^2}{12}} = \frac{a}{2\sqrt{3}}$.

$|\overrightarrow{DF}|^2 = \left|\frac{\overrightarrow{B}}{2} - \overrightarrow{D}\right|^2 = \left(\frac{\overrightarrow{B}}{2} - \overrightarrow{D}\right) \cdot \left(\frac{\overrightarrow{B}}{2} - \overrightarrow{D}\right)$ $= \frac{1}{4}|\overrightarrow{B}|^2 - \overrightarrow{B}\cdot\overrightarrow{D} + |\overrightarrow{D}|^2$ $= \frac{1}{4}a^2 - \frac{a^2}{2} + a^2 = a^2 \left(\frac{1}{4} - \frac{1}{2} + 1\right) = \frac{3a^2}{4}$. $|\overrightarrow{DF}| = \sqrt{\frac{3a^2}{4}} = \frac{a\sqrt{3}}{2}$.

Now, calculate $\cos \theta$: $\cos \theta = \frac{a^2/24}{\left(\frac{a}{2\sqrt{3}}\right) \left(\frac{a\sqrt{3}}{2}\right)} = \frac{a^2/24}{\frac{a^2 (\sqrt{3})^2}{4}} = \frac{a^2/24}{3a^2/4}$ $\cos \theta = \frac{a^2}{24} \cdot \frac{4}{3a^2} = \frac{4}{72} = \frac{1}{18}$.

Let's check the given options: A. The angle between $\overrightarrow{DA}$ and $\overrightarrow{BC}$ is $\frac{\pi}{3}$. The angle between $\overrightarrow{AD}$ and $\overrightarrow{BC}$ is $\pi/2$ (as their dot product is 0). The angle between $\overrightarrow{DA}$ and $\overrightarrow{BC}$ is also $\pi/2$. So, A is false.

B. The angle between $\overrightarrow{AB}$ and $\overrightarrow{DC}$ is $\frac{\pi}{2}$. The angle between $\overrightarrow{AB}$ and $\overrightarrow{DC}$ is $\pi/2$ (as their dot product is 0). So, B is true.

C. $\theta = \cos^{-1}(-\frac{1}{2\sqrt{3}})$. This implies $\cos \theta = -\frac{1}{2\sqrt{3}}$. This is not $1/18$. So, C is false.

D. $\theta = \tan^{-1}(\sqrt{11})$. If $\tan \theta = \sqrt{11}$, then $\sec^2 \theta = 1 + \tan^2 \theta = 1 + (\sqrt{11})^2 = 1 + 11 = 12$. So, $\cos^2 \theta = \frac{1}{\sec^2 \theta} = \frac{1}{12}$. This implies $\cos \theta = \pm \frac{1}{\sqrt{12}} = \pm \frac{1}{2\sqrt{3}}$. This is not $1/18$. So, D is false.

Based on the calculations, only statement B is correct. The calculated value for $\theta$ does not match options C or D.