The position vectors of the points A, B, C and D are 3i - 2j - k, 2i + 3j - 4k, -i + j + 2k and 4i +

Q: Why do we use the scalar triple product? Can't we just find the equation of the plane?

You absolutely could find the equation of the plane using the normal vector $\vec{n} = \vec{AB} \times \vec{AC}$ and then checking if point D satisfies it. However, the scalar triple product is a more direct and elegant tool for coplanarity. It essentially checks if the volume of the parallelepiped formed by the three vectors is zero, which is the geometric definition of coplanarity. It saves you from calculating the full plane equation.

Q: Does the order of the vectors in the determinant matter?

Mathematically, swapping rows in a determinant changes its sign, but since we are setting the determinant equal to zero, the sign change doesn't affect the final value of $\lambda$. However, it is good practice to keep the order consistent, such as $\vec{AB}, \vec{AC},$ and $\vec{AD}$.

Q: What if the points were collinear? Would the scalar triple product still work?

Yes, it would! If the points are collinear, they are also coplanar. In that case, the vectors would be linearly dependent, and the scalar triple product would naturally result in zero. The method is robust enough to handle that case as well.

Question

The position vectors of the points $A, B, C$ and $D$ are $3\hat{i} - 2\hat{j} - \hat{k}, 2\hat{i} + 3\hat{j} - 4\hat{k}, -\hat{i} + \hat{j} + 2\hat{k}$ and $4\hat{i} + 5\hat{j} + \lambda\hat{k}$, respectively. If the points $A, B, C$ and $D$ lie on a plane, find the value of $\lambda$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to explore the elegant world of 3D geometry. Imagine you are standing in a vast, three-dimensional space with four points: $A, B, C,$ and $D$.

These points are floating in space, and we are given their position vectors. However, point $D$ contains an unknown coordinate, $\lambda$. Our mission is to find the exact value of $\lambda$ that forces all four points to lie on the same flat, two-dimensional plane, a condition known as **coplanarity**.

### The Vector Toolkit

To solve this, we translate our geometric intuition into the language of vectors. If we fix point $A$ as our anchor, we define three vectors: $\vec{AB}, \vec{AC},$ and $\vec{AD}$.

These vectors represent the displacement from $A$ to the other three points. If these three vectors lie in the same plane, then the four points must also be coplanar.

We check this using the **scalar triple product**, also known as the box product. If the volume of the parallelepiped formed by these three vectors is zero, they are coplanar:

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

### Constructing the Vectors

Let us calculate these vectors step-by-step by subtracting the position vector of $A$ from the others:

$$\vec{AB} = \vec{B} - \vec{A} = (2-3)\hat{i} + (3-(-2))\hat{j} + (-4-(-1))\hat{k} = -\hat{i} + 5\hat{j} - 3\hat{k}$$

$$\vec{AC} = \vec{C} - \vec{A} = (-1-3)\hat{i} + (1-(-2))\hat{j} + (2-(-1))\hat{k} = -4\hat{i} + 3\hat{j} + 3\hat{k}$$

$$\vec{AD} = \vec{D} - \vec{A} = (4-3)\hat{i} + (5-(-2))\hat{j} + (\lambda-(-1))\hat{k} = \hat{i} + 7\hat{j} + (\lambda+1)\hat{k}$$

### The Determinant as a Gatekeeper

Now, we set up the determinant. We place the components of our vectors into a $3 	imes 3$ matrix and set the result to zero:

$$\begin{vmatrix} -1 & 5 & -3 \ -4 & 3 & 3 \ 1 & 7 & \lambda+1 \end{vmatrix} = 0$$

Expanding this determinant along the first row, we obtain the following equation:

$$-1[3(\lambda+1) - 21] - 5[-4(\lambda+1) - 3] - 3[-28 - 3] = 0$$

### The Algebraic Journey

Let us simplify the expression inside the brackets carefully:

$$-1[3\lambda + 3 - 21] - 5[-4\lambda - 4 - 3] - 3[-31] = 0$$

$$-1[3\lambda - 18] - 5[-4\lambda - 7] + 93 = 0$$

Distributing the coefficients yields:

$$-3\lambda + 18 + 20\lambda + 35 + 93 = 0$$

Combining the like terms results in the linear equation:

$$17\lambda + 146 = 0$$

### Final Calculation

We have arrived at a simple linear equation. Solving for $\lambda$ is straightforward:

$$17\lambda = -146$$

$$\lambda = -\frac{146}{17}$$

By enforcing the condition that the volume of the parallelepiped must be zero, we have found the exact coordinate that keeps our points perfectly flat on the plane. The final value is **$\lambda = -\frac{146}{17}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Points

The Coplanarity Condition

Calculating Vector $A B$

Calculating Vector $A C$

Calculating Vector $A D$

Setting up the Determinant

Expanding the Determinant

Simplifying the Expression

Distributing and Combining Terms

Solving for $λ$

Final Takeaway

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Vector Toolkit

Constructing the Vectors

The Determinant as a Gatekeeper

The Algebraic Journey

Final Calculation

Similar Questions

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