Understanding Coplanarity of Points

Four points A, B, C, and D are said to be coplanar if they all lie in the same plane. In vector algebra, a common way to check for the coplanarity of four points is to consider three vectors formed by connecting these points. If these three vectors are coplanar, then the four points are also coplanar.

Let the position vectors of the four points A, B, C, and D be \( \vec{a} \), \( \vec{b} \), \( \vec{c} \), and \( \vec{d} \) respectively. We are given:

• Position vector of A, \( \vec{a} = -\hat{i} + \hat{j} + \hat{k} \)
• Position vector of B, \( \vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k} \)
• Position vector of C, \( \vec{c} = 4\hat{i} - \lambda\hat{j} - \hat{k} \)
• Position vector of D, \( \vec{d} = \hat{i} + \hat{j} + \hat{k} \)

Condition for Coplanarity Using Vectors

To determine the value of \( \lambda \) for which points A, B, C, D are coplanar, we can form three vectors using these points, for example, \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \). The points A, B, C, and D are coplanar if and only if the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) are coplanar.

The condition for three vectors \( \vec{u}, \vec{v}, \vec{w} \) to be coplanar is that their scalar triple product is zero, i.e., \( [\vec{u}, \vec{v}, \vec{w}] = 0 \).

Calculating the Vectors

First, let's calculate the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \):

• \( \vec{AB} = \vec{b} - \vec{a} \)
• \( \vec{AB} = (3\hat{i} - 2\hat{j} + 2\hat{k}) - (-\hat{i} + \hat{j} + \hat{k}) \)
• \( \vec{AB} = (3 - (-1))\hat{i} + (-2 - 1)\hat{j} + (2 - 1)\hat{k} \)
• \( \vec{AB} = 4\hat{i} - 3\hat{j} + \hat{k} \)

• \( \vec{AC} = \vec{c} - \vec{a} \)
• \( \vec{AC} = (4\hat{i} - \lambda\hat{j} - \hat{k}) - (-\hat{i} + \hat{j} + \hat{k}) \)
• \( \vec{AC} = (4 - (-1))\hat{i} + (-\lambda - 1)\hat{j} + (-1 - 1)\hat{k} \)
• \( \vec{AC} = 5\hat{i} - (\lambda + 1)\hat{j} - 2\hat{k} \)

• \( \vec{AD} = \vec{d} - \vec{a} \)
• \( \vec{AD} = (\hat{i} + \hat{j} + \hat{k}) - (-\hat{i} + \hat{j} + \hat{k}) \)
• \( \vec{AD} = (1 - (-1))\hat{i} + (1 - 1)\hat{j} + (1 - 1)\hat{k} \)
• \( \vec{AD} = 2\hat{i} + 0\hat{j} + 0\hat{k} \)

Setting up the Scalar Triple Product

For the points A, B, C, D to be coplanar, the scalar triple product of \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) must be zero:

\( [\vec{AB}, \vec{AC}, \vec{AD}] = \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0 \)

This can be calculated using a determinant formed by the components of the vectors:

\( [\vec{AB}, \vec{AC}, \vec{AD}] = \begin{vmatrix} 4 & -3 & 1 \\ 5 & -(\lambda + 1) & -2 \\ 2 & 0 & 0 \end{vmatrix} \)

Solving for \( \lambda \)

To find the value of \( \lambda \), we set the determinant equal to zero and solve. Expanding the determinant along the third row (because it has two zeros) is the easiest method:

\( 2 \times \begin{vmatrix} -3 & 1 \\ -(\lambda + 1) & -2 \end{vmatrix} - 0 \times \begin{vmatrix} 4 & 1 \\ 5 & -2 \end{vmatrix} + 0 \times \begin{vmatrix} 4 & -3 \\ 5 & -(\lambda + 1) \end{vmatrix} = 0 \)

\( 2 \times ((-3) \times (-2) - (1) \times (-(\lambda + 1))) = 0 \)

\( 2 \times (6 - (-\lambda - 1)) = 0 \)

\( 2 \times (6 + \lambda + 1) = 0 \)

\( 2 \times (7 + \lambda) = 0 \)

Since \( 2 \neq 0 \), we must have:

\( 7 + \lambda = 0 \)

\( \lambda = -7 \)

Thus, the value of \( \lambda \) for which the points A, B, C, D are coplanar is -7.

Revision Table: Key Concepts

Additional Information: Scalar Triple Product Properties

The scalar triple product \( [\vec{u}, \vec{v}, \vec{w}] \) has several useful properties:

• It is cyclic: \( [\vec{u}, \vec{v}, \vec{w}] = [\vec{v}, \vec{w}, \vec{u}] = [\vec{w}, \vec{u}, \vec{v}] \).
• Interchanging any two vectors changes the sign: \( [\vec{u}, \vec{v}, \vec{w}] = -[\vec{v}, \vec{u}, \vec{w}] \).
• If any two vectors are parallel or identical, the scalar triple product is zero.
• If the scalar triple product is zero, the vectors are coplanar. Conversely, if the vectors are coplanar, the scalar triple product is zero.
• It can be computed as the determinant of the matrix formed by the vector components.

Understanding the scalar triple product is crucial for solving problems involving the volume of parallelepipeds and testing for the coplanarity of vectors and points in three-dimensional space.