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

Q: Why is the altitude DE perpendicular to the base ABC?

By definition, the altitude of a tetrahedron from a vertex to the opposite face is the perpendicular line segment from that vertex to the plane containing the opposite face. Thus, DE is perpendicular to every line in the plane ABC passing through E, including the median AM.

Q: Can we use the scalar triple product for the volume?

Yes, you absolutely can! The volume of a tetrahedron with vertices A, B, C, and D is given by (1/6) * |(AB x AC) . AD|. However, since the problem already provides the volume and the base area, using the formula V = (1/3) * Area * h is much faster and less prone to calculation errors.

Q: How do I know the ratio AE/AM is 1/3?

We calculated the length of AE using the Pythagorean theorem in triangle ADE, and we calculated the length of AM using the distance formula. By taking the ratio of these two lengths, we found that AE/AM = 1/3. This tells us that E divides the median AM in the ratio 1:2, which is a beautiful geometric property!

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

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Base Triangle

To begin our analysis of the tetrahedron, we first define the base triangle $ABC$ with vertices $A(1, 2, 1)$, $B(1, 3, -2)$, and $C(2, 1, -1)$. We calculate the vectors $\vec{AB}$ and $\vec{AC}$ as follows:

$$\vec{AB} = (1-1)\hat{i} + (3-2)\hat{j} + (-2-1)\hat{k} = \hat{j} - 3\hat{k}$$

$$\vec{AC} = (2-1)\hat{i} + (1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - \hat{j} - 2\hat{k}$$

To find the area of the base, we determine the normal vector $\vec{n} = \vec{AB} 	imes \vec{AC}$ using the determinant method:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -3 \ 1 & -1 & -2 \end{vmatrix} = -5\hat{i} - 3\hat{j} - \hat{k}$$

The area of $	riangle ABC$ is given by $\frac{1}{2} |\vec{n}|$:

$$	ext{Area} = \frac{1}{2} \sqrt{(-5)^2 + (-3)^2 + (-1)^2} = \frac{\sqrt{35}}{2}$$

### Determining the Altitude

The volume of a tetrahedron is defined by the formula $V = \frac{1}{3} 	imes 	ext{Base Area} 	imes h$. Given the volume $V = \frac{\sqrt{805}}{6\sqrt{2}}$, we solve for the altitude $h$:

$$\frac{\sqrt{805}}{6\sqrt{2}} = \frac{1}{3} 	imes \frac{\sqrt{35}}{2} 	imes h$$

Solving this equation yields the length of the altitude $DE$:

$$h = \sqrt{\frac{23}{2}}$$

### Geometric Projection

We now consider the right-angled triangle $	riangle ADE$. Given the hypotenuse $AD = \frac{\sqrt{110}}{3}$, we apply the Pythagorean theorem to find $AE^2$:

$$AE^2 = AD^2 - DE^2 = \frac{110}{9} - \frac{23}{2} = \frac{220 - 207}{18} = \frac{13}{18}$$

Next, we find the median $AM$ of the base. The midpoint $M$ of $BC$ is $M(\frac{3}{2}, 2, -\frac{3}{2})$, and the vector $\vec{AM}$ is:

$$\vec{AM} = \left(\frac{3}{2}-1ight)\hat{i} + (2-2)\hat{j} + \left(-\frac{3}{2}-1ight)\hat{k} = \frac{1}{2}\hat{i} - \frac{5}{2}\hat{k}$$

The squared length $|\vec{AM}|^2$ is:

$$|\vec{AM}|^2 = \left(\frac{1}{2}ight)^2 + \left(-\frac{5}{2}ight)^2 = \frac{26}{4} = \frac{13}{2}$$

### Final Calculation of Point E

Comparing the squared lengths, we find the ratio $\frac{AE^2}{AM^2} = \frac{13/18}{13/2} = \frac{1}{9}$, which implies $\frac{AE}{AM} = \frac{1}{3}$. Consequently, $\vec{AE} = \frac{1}{3} \vec{AM}$.

The position vector of $E$ is calculated as $\vec{E} = \vec{A} + \vec{AE}$:

$$\vec{E} = (\hat{i} + 2\hat{j} + \hat{k}) + \frac{1}{3} \left(\frac{1}{2}\hat{i} - \frac{5}{2}\hat{k}ight) = \frac{7}{6}\hat{i} + 2\hat{j} + \frac{1}{6}\hat{k}$$

The final coordinates of point $E$ are **$\frac{1}{6}(7, 12, 1)$**.

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Visualized Solution

Define Base Vertices

Calculate Base Vectors

Find Normal Vector $n$

Calculate Area of $△ A B C$

Volume Formula for Altitude $h$

Calculate Altitude $h$

Right Triangle $△ A D E$

Apply Pythagoras Theorem

Locate Median Point $M$

Calculate Vector $A M$

Ratio of $A E$ to $A M$

Vector Equation for $E$

Final Calculation for $E$

Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Base Triangle

Determining the Altitude

Geometric Projection

Final Calculation of Point E

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Select Answer:

Visualized Solution

Define Base Vertices

Calculate Base Vectors

Find Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Calculate Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

Volume Formula for Altitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Calculate Altitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Right Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ADE