Let a, b and c be three non-zero non-coplanar vectors. Let the position vectors of four points A, B,

Q: Wait, why do we use $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$? Can't we use $\vec{BA}$, $\vec{BC}$, and $\vec{BD}$?

That is a fantastic question! Yes, you absolutely can. The condition for coplanarity is that the vectors must lie in the same plane. As long as you choose three vectors that originate from a common point (like A, B, C, or D) and lie on that plane, the Scalar Triple Product will be zero. The choice of the starting point is arbitrary, but using a common point makes the vector construction consistent and less prone to errors.

Q: What if the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ were coplanar? Would the determinant method still work?

No, it would not. The determinant method relies on the fact that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are non-coplanar, meaning they form a basis for 3D space. If they were coplanar, their scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$ would be zero, and the determinant would be zero regardless of the points' positions. The problem specifically states they are non-coplanar to ensure that the determinant condition is both necessary and sufficient.

Q: Is there a faster way to solve this than expanding the determinant?

While you could potentially use row operations to simplify the determinant before expanding, for a $3 \times 3$ matrix, direct expansion is usually the most reliable and efficient method. It minimizes the risk of algebraic slips. Always remember: in JEE, accuracy is just as important as speed!

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $\vec{a}-\vec{b}+\vec{c}$, $\lambda\vec{a}-3\vec{b}+4\vec{c}$, $-\vec{a}+2\vec{b}-3\vec{c}$ and $2\vec{a}-4\vec{b}+6\vec{c}$ respectively. If AB, AC and AD are coplanar, then $\lambda$ is :

Options:

(A) 1

(B) 2

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### The Geometry of Flatness: Understanding Coplanarity

Imagine you are standing in a vast, three-dimensional space. You have three foundational vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, which act as your coordinate axes. They are non-coplanar, meaning they do not lie on a single flat sheet; they span the entire 3D world.

Now, we introduce four points: $A$, $B$, $C$, and $D$. The problem asks us to find a mysterious scalar $\lambda$ such that these four points sit perfectly on a single, flat plane. This is the essence of coplanarity.

### Constructing the Vectors

To determine if these points are coplanar, we need to translate their positions into vectors. We can think of these points as locations in space.

If we fix point $A$ as our origin on this plane, we can create three vectors: $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$. These vectors represent the displacement from $A$ to the other three points. If the points are coplanar, these three vectors must also be coplanar—they must lie flat on that same sheet.

We calculate them by subtracting the position vector of $A$ from the position vectors of $B$, $C$, and $D$ respectively:

$$\vec{AB} = \vec{OB} - \vec{OA} = (\lambda\vec{a} - 3\vec{b} + 4\vec{c}) - (\vec{a} - \vec{b} + \vec{c}) = (\lambda-1)\vec{a} - 2\vec{b} + 3\vec{c}$$

$$\vec{AC} = \vec{OC} - \vec{OA} = (-\vec{a} + 2\vec{b} - 3\vec{c}) - (\vec{a} - \vec{b} + \vec{c}) = -2\vec{a} + 3\vec{b} - 4\vec{c}$$

$$\vec{AD} = \vec{OD} - \vec{OA} = (2\vec{a} - 4\vec{b} + 6\vec{c}) - (\vec{a} - \vec{b} + \vec{c}) = \vec{a} - 3\vec{b} + 5\vec{c}$$

### The Power of the Scalar Triple Product

Now, how do we mathematically enforce the condition that these three vectors are coplanar? We use the Scalar Triple Product, $[\vec{AB}, \vec{AC}, \vec{AD}]$.

Geometrically, this product represents the volume of a parallelepiped formed by these three vectors. If the vectors are coplanar, the parallelepiped is crushed into a flat shape with zero height, and thus, zero volume.

Therefore, the condition for coplanarity is simply $[\vec{AB}, \vec{AC}, \vec{AD}] = 0$. Because $\vec{a}$, $\vec{b}$, and $\vec{c}$ are non-coplanar, this condition is equivalent to the determinant of the coefficients of our vectors being zero:

$$\begin{vmatrix} \lambda-1 & -2 & 3 \ -2 & 3 & -4 \ 1 & -3 & 5 \end{vmatrix} = 0$$

### The Final Calculation

Now, let us expand this determinant with care. Expanding along the first row, we get:

$$(\lambda-1)[(3)(5) - (-4)(-3)] - (-2)[(-2)(5) - (-4)(1)] + 3[(-2)(-3) - (3)(1)] = 0$$

Simplifying the terms inside the brackets:

$$(\lambda-1)[15 - 12] + 2[-10 + 4] + 3[6 - 3] = 0$$

$$3(\lambda-1) + 2(-6) + 3(3) = 0$$

$$3\lambda - 3 - 12 + 9 = 0$$

$$3\lambda - 6 = 0$$

Solving for $\lambda$, we find **$\lambda = 2$**. It is truly elegant how the complex geometry of 3D space collapses into a simple linear equation. By understanding the physical meaning of the scalar triple product, we have navigated the problem with precision.

Select Answer:

Visualized Solution

Defining the Position Vectors

The Concept of Coplanar Vectors

Calculating Vector $A B$

Calculating Vector $A C$

Calculating Vector $A D$

The Scalar Triple Product Condition

Expanding the Determinant

Simplifying the Terms

Solving for $λ$

Conclusion and Key Takeaway

The Sigma Insight: Scalar Triple Product

The Geometry of Flatness

Constructing the Vectors

The Power of the Scalar Triple Product

The Final Calculation

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