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 $\vec{AB}, \vec{AC}, \vec{AD}$ instead of just the position vectors $\vec{A}, \vec{B}, \vec{C}, \vec{D}$?

That is a brilliant question! The position vectors $\vec{A}, \vec{B}, \vec{C}, \vec{D}$ are defined relative to an arbitrary origin. However, coplanarity is a property of the *relative* positions of the points. By constructing vectors like $\vec{AB} = \vec{B} - \vec{A}$, we are essentially shifting our origin to point $A$. If $A, B, C, D$ are coplanar, then the vectors $\vec{AB}, \vec{AC}, \vec{AD}$ must all lie in that same plane. This is a much more powerful and direct way to use the scalar triple product.

Q: Why does the scalar triple product being zero imply the determinant of coefficients is zero?

The scalar triple product $[\vec{u}, \vec{v}, \vec{w}]$ is defined as $\vec{u} \cdot (\vec{v} \times \vec{w})$. If we express $\vec{u}, \vec{v}, \vec{w}$ as linear combinations of non-coplanar basis vectors $\vec{a}, \vec{b}, \vec{c}$, the scalar triple product becomes the determinant of the coefficients multiplied by the scalar triple product of the basis vectors $[\vec{a}, \vec{b}, \vec{c}]$. Since the problem states $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, their scalar triple product is non-zero. Therefore, for the whole expression to be zero, the determinant of the coefficients *must* be zero.

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

If $\vec{a}, \vec{b}, \vec{c}$ were coplanar, their scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$ would be zero. In that case, the determinant of the coefficients would be zero regardless of the vectors $\vec{AB}, \vec{AC}, \vec{AD}$, because the entire space would be flattened. The problem specifically tells us they are non-coplanar to ensure that the determinant condition is both necessary and sufficient.

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:

Sigma Percentile · Accepted Answer

### The Geometry of Flatness: A Journey into Coplanarity

Welcome, future engineer. Today, we are not just solving a vector algebra problem; we are exploring the very nature of space.

Imagine you are standing in a vast, three-dimensional room. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, which act as your compass—they are non-coplanar, meaning they span the entire 3D space.

Now, we are given four points, $A, B, C,$ and $D$, floating in this space. We are told they are coplanar.

This means if you were to take a giant, flat sheet of glass and pass it through the room, you could align it so that all four points touch the glass simultaneously. This is the geometric reality we are about to translate into the language of algebra.

### Step 1: Shifting the Perspective

To understand the relationship between these points, we need to look at them relative to one another. We fix point $A$ as our anchor.

By constructing the vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$, we are essentially creating a local coordinate system originating from $A$. If the points $A, B, C,$ and $D$ lie on a single plane, then the vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ must also lie on that same plane.

We calculate these vectors by subtracting the position vector of $A$ from the others. For $\vec{AB}$, we have:

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

We repeat this for $\vec{AC}$ and $\vec{AD}$, carefully tracking our coefficients. This is where most students stumble—a simple sign error here can derail the entire calculation.

### Step 2: The Scalar Triple Product

Now, we enter the heart of the problem. How do we mathematically enforce the condition that three vectors lie in the same plane?

We use the scalar triple product, denoted as $[\vec{AB}, \vec{AC}, \vec{AD}]$. Geometrically, this product represents the volume of the parallelepiped formed by these three vectors.

If the vectors are coplanar, they cannot span any volume. The parallelepiped collapses into a flat shape, and its volume becomes zero.

Thus, the condition for coplanarity is simply:

$$[\vec{AB}, \vec{AC}, \vec{AD}] = 0$$

Because our basis vectors $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, this condition simplifies to the determinant of the coefficients being zero. This is the elegance of linear algebra: we have reduced a complex 3D geometric constraint into a simple determinant equation.

### Step 3: The Determinant Engine

We set up our determinant using the coefficients we derived:

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

Now, we expand along the first row. We have:

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

Let us simplify this step-by-step. The first term becomes $(\lambda - 1)(15 - 12) = 3(\lambda - 1)$.

The second term becomes $+2(-10 + 4) = 2(-6) = -12$. The third term becomes $3(6 - 3) = 3(3) = 9$.

Combining these, we get $3(\lambda - 1) - 12 + 9 = 0$. This simplifies to $3\lambda - 3 - 3 = 0$, or $3\lambda - 6 = 0$.

Solving for $\lambda$, we find **$\lambda = 2$**.

### Conclusion: The Beauty of the Result

We have arrived at our destination. By understanding the geometric meaning of coplanarity and translating it into the language of determinants, we have unlocked the value of $\lambda$.

It is not just a number; it is the specific value that forces these four points into perfect alignment. Remember this process: visualize the geometry, identify the constraint, and let the algebra guide you to the solution.

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $a - b + c, λ a - 3 b + 4 c, - a + 2 b - 3 c$ and $2 a - 4 b + 6 c$ respectively. If AB, AC and AD are coplanar, then $λ$ is:

Enter Numerical Value:

Visualized Solution

Visualizing the Points in Space

The Coplanarity Condition

Calculating Vector $A B$

Calculating Vector $A C$

Calculating Vector $A D$

Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant

Simplifying the Equation

Final Value of $λ$

The Sigma Insight: Scalar Triple Product

The Geometry of Flatness

Step 1

Step 2

Step 3

Conclusion

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