The position vectors of the vertices A, B and C of a tetrahedron ABCD are i + j + k, i and 3i, respe

Q: Wait, why is the altitude $DE$ perpendicular to the base? Isn't it just a line?

In geometry, an 'altitude' of a tetrahedron is strictly defined as the perpendicular segment from a vertex to the plane containing the opposite face. It is not just any line; it is the shortest path from $D$ to the plane $ABC$. That perpendicularity is the geometric key that allows us to use the normal vector to define the position of $D$!

Q: Why can't we just assume $E$ is the centroid of $\triangle ABC$?

That is a very tempting trap! The centroid is a specific point on the median, but the problem states $E$ is the foot of the altitude from $D$. Unless the tetrahedron is regular or possesses specific symmetries, the foot of the altitude is generally not the centroid. We must solve for $E$ using the given constraints, not assume its location based on intuition.

Q: Why are there two possible positions for $E$?

Think about the base $ABC$ as a flat sheet of paper. The altitude $h=2$ allows the vertex $D$ to be placed either 'above' the paper or 'below' the paper, while still maintaining the same volume and distance $AD$. Both positions are physically valid, which is why we get two distinct solutions for $E$.

Question

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 vertex $D$ to the opposite face $ABC$ meets the median line through $A$ of the triangle $ABC$ at a point $E$. If the length of the side $AD$ is $4$ and the volume of the tetrahedron is $\frac{2\sqrt{2}}{3}$, find the position vector of the point $E$ for all its possible positions.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system, looking at a tetrahedron $ABCD$. It is a beautiful, rigid structure floating in space. We are given the vertices $A(1, 1, 1)$, $B(1, 0, 0)$, and $C(3, 0, 0)$.

Our mission is to find the position vector of point $E$, which sits on the median $AM$ of the base triangle $ABC$. This is not just a calculation; it is a dance of vectors.

### Mapping the Median

First, we must define the floor of our tetrahedron, the triangle $ABC$. The median $AM$ starts at $A$ and bisects the side $BC$.

The midpoint $M$ of $BC$ is simply the average of $B$ and $C$:
$$M = \left( \frac{1+3}{2}, \frac{0+0}{2}, \frac{0+0}{2} ight) = (2, 0, 0)$$

The direction vector of this median is $\vec{v} = M - A = (1, -1, -1)$. Any point $E$ on this line can be parameterized as:
$$E = A + t\vec{v} = (1+t, 1-t, 1-t)$$

### The Volume and the Altitude

We are given the volume $V = \frac{2\sqrt{2}}{3}$. We know the volume of a tetrahedron is $V = \frac{1}{3} 	imes 	ext{Area}(ABC) 	imes h$, where $h$ is the altitude $DE$.

To find the area, we compute the cross product of $\vec{AB}$ and $\vec{AC}$. With $\vec{AB} = (0, -1, -1)$ and $\vec{AC} = (2, -1, -1)$, the cross product is:
$$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & -1 & -1 \ 2 & -1 & -1 \end{vmatrix} = (0, -2, 2)$$

The area is $\frac{1}{2} |\vec{AB} 	imes \vec{AC}| = \sqrt{0^2 + (-2)^2 + 2^2} / 2 = \sqrt{8}/2 = \sqrt{2}$. Plugging this into our volume formula:
$$\frac{2\sqrt{2}}{3} = \frac{1}{3} 	imes \sqrt{2} 	imes h$$

The $\sqrt{2}$ and the $3$ cancel out beautifully, leaving us with the altitude **$h = 2$**.

### The Normal Vector and Vertex $D$

Now, we need to locate $D$. Since $DE$ is the altitude, it must be parallel to the normal vector of the base plane. The normal vector is $\vec{n} = (0, -2, 2)$, which simplifies to the unit normal $\hat{n} = \frac{(0, -1, 1)}{\sqrt{2}}$.

Vertex $D$ is located at a distance $h=2$ from $E$ along this normal. Thus:
$$D = E \pm 2\hat{n} = (1+t, 1-t \mp \sqrt{2}, 1-t \pm \sqrt{2})$$

### The Final Constraint

We are given $AD = 4$, so $AD^2 = 16$. Using the distance formula between $A(1, 1, 1)$ and $D$:
$$(1+t-1)^2 + (1-t \mp \sqrt{2} - 1)^2 + (1-t \pm \sqrt{2} - 1)^2 = 16$$

This simplifies to:
$$t^2 + (-t \mp \sqrt{2})^2 + (-t \pm \sqrt{2})^2 = 16$$

Expanding this, we get:
$$t^2 + (t^2 + 2 \pm 2\sqrt{2}t) + (t^2 + 2 \mp 2\sqrt{2}t) = 16$$

Notice the magic? The cross terms $\pm 2\sqrt{2}t$ and $\mp 2\sqrt{2}t$ cancel out perfectly. We are left with:
$$3t^2 + 4 = 16 \Rightarrow 3t^2 = 12 \Rightarrow t^2 = 4$$

Thus, $t = \pm 2$. Substituting these values back into our expression for $E = (1+t, 1-t, 1-t)$, we find the two possible positions:
**$E_1 = (3, -1, -1)$** and **$E_2 = (-1, 3, 3)$**.

Visualized Solution

Visualize the Tetrahedron $A B C D$

Find the Median $A M$ and Point $E$

Calculate Area of $△ A B C$

Find Altitude Height $h$

Define Coordinates of Vertex $D$

Set up Equation using $A D = 4$

Solve for Parameter $t$

Final Position Vectors of $E$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Mapping the Median

The Volume and the Altitude

The Normal Vector and Vertex $D$

The Final Constraint

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Visualized Solution

Visualize the Tetrahedron .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

Find the Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

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

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

Define Coordinates of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Set up Equation using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD=4

Solve for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Final Position Vectors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E