Let the position vectors of the points A, B, C and D be 5i + 5j + 2λk, i + 2j + 3k, -2i + λj + 4k an

Q: Wait, why do we use point A as the anchor? Can we use another point like B?

That is a fantastic question! You absolutely can use any point as the anchor. If you choose B, you would construct vectors $\vec{BA}$, $\vec{BC}$, and $\vec{BD}$. The scalar triple product $[\vec{BA} \\ \vec{BC} \\ \vec{BD}]$ would still equal zero because the volume of the parallelepiped formed by these vectors remains zero if the points are coplanar. Point A is just a convenient choice to keep the algebra consistent.

Q: What if the determinant is not zero? What does that mean physically?

If the scalar triple product is non-zero, it means the three vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ are linearly independent and span a 3D volume. Physically, this means the four points form a tetrahedron with a non-zero volume, meaning they are not coplanar. The condition of zero volume is the mathematical 'seal' that forces them onto a single plane.

Q: Is there a faster way than expanding the determinant? It feels prone to calculation errors.

You are right to be cautious! While expansion is the most reliable method, you can sometimes use row or column operations to simplify the determinant before expanding. For example, you could subtract one row from another to create zeros. However, for a $3 \times 3$ determinant with a variable like $\lambda$, direct expansion is often just as fast and less prone to 'clever' mistakes. Just stay focused on the signs!

Question

Let the position vectors of the points $A, B, C$ and $D$ be $5\hat{i} + 5\hat{j} + 2\lambda\hat{k}$, $\hat{i} + 2\hat{j} + 3\hat{k}$, $-2\hat{i} + \lambda\hat{j} + 4\hat{k}$ and $-\hat{i} + 5\hat{j} + 6\hat{k}$. Let the set $S = \{\lambda \in \mathbb{R} : 	ext{the points } A, B, C 	ext{ and } D 	ext{ are coplanar}\}$. The $\sum_{\lambda \in S} (\lambda + 2)^2$ is equal to

Options:

(A) 25

(B) $\frac{37}{2}$

(C) 14

(D) 41

Sigma Percentile · Accepted Answer

### The Geometry of Flatness: A Journey into 3D Space

Welcome, fellow explorer of the mathematical universe! Today, we are going to tackle a problem that sits at the heart of 3D geometry.

We are given four points in space, $A$, $B$, $C$, and $D$, and we are tasked with finding the values of a mysterious parameter $\lambda$ that forces these four points to lie on a single, flat plane. This is not just about crunching numbers; it is about understanding the spatial relationship between vectors.

### Phase 1: Anchoring Our Perspective

Imagine you are standing in a 3D room. You have four markers representing points $A$, $B$, $C$, and $D$. To understand how they relate to each other, we need a reference point.

Let us choose point $A$ as our anchor. By creating vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$, we are essentially drawing arrows from $A$ to the other three points. These vectors define the "reach" of our points from our anchor.

We calculate these by subtracting the coordinates of $A$ from the others:
$$\vec{AB} = (1-5, 2-5, 3-2\lambda) = (-4, -3, 3-2\lambda)$$
$$\vec{AC} = (-2-5, \lambda-5, 4-2\lambda) = (-7, \lambda-5, 4-2\lambda)$$
$$\vec{AD} = (-1-5, 0-5, 6-2\lambda) = (-6, -5, 6-2\lambda)$$

It might look like a mess of variables, but remember: $\lambda$ is just a number waiting to be discovered.

### Phase 2: The Scalar Triple Product

Here is the core physical intuition: if these four points are coplanar, then the three vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ must also lie within that same plane. If they lie in the same plane, they cannot span a 3D volume.

In vector calculus, the volume of the parallelepiped formed by three vectors is given by their scalar triple product, denoted as $[\vec{AB} \ \vec{AC} \ \vec{AD}]$. If the volume is zero, the points are coplanar.

This is our golden key: the determinant of the matrix formed by these vectors must be zero.

### Phase 3: The Determinant Expansion

Now, we set up our determinant:

$$\begin{vmatrix} -4 & -3 & 3 - 2\lambda \ -7 & \lambda - 5 & 4 - 2\lambda \ -6 & -5 & 6 - 2\lambda \end{vmatrix} = 0$$

This is where the magic happens. We expand along the first row. Take a deep breath—this is where most students stumble, not because the math is hard, but because of the negative signs.

We take $-4$ multiplied by the minor, then subtract $-3$ (which becomes $+3$) multiplied by its minor, and finally add $(3-2\lambda)$ multiplied by its minor. By carefully simplifying these terms, we reduce the complexity of the expression step by step.

### Phase 4: The Quadratic Revelation

After the expansion and careful algebraic simplification, we arrive at a beautiful, clean quadratic equation:

$$-4\lambda^2 + 20\lambda - 24 = 0$$

Dividing by $-4$ gives us $\lambda^2 - 5\lambda + 6 = 0$. This is a classic quadratic that factorizes into $(\lambda - 2)(\lambda - 3) = 0$.

We have found our values: $\lambda = 2$ and $\lambda = 3$. These are the two specific configurations where the points align perfectly on a plane.

### Conclusion: The Final Sum

Finally, we calculate the sum $\sum_{\lambda \in S} (\lambda + 2)^2$.

Substituting our values, we get:

$$(2+2)^2 + (3+2)^2 = 4^2 + 5^2 = 16 + 25 = 41$$

We have navigated the 3D space, used the power of the scalar triple product, and arrived at our destination. The final result is **41**. Remember, every complex problem is just a series of simple steps. Keep practicing, and keep visualizing!

Select Answer:

Visualized Solution

Identifying the Points

Constructing Vectors

Simplifying the Vectors

Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant

Simplifying the Terms

Forming the Quadratic Equation

Solving for $λ$

Setting up the Final Sum

Final Calculation

The Sigma Insight: Scalar Triple Product

The Geometry of Flatness

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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