Question

If the points having the position vectors \( \mathbf{r}_1 = -i + 4j - 4k\), \( \mathbf{r}_2 = 3i + 2j - 5k\), \( \mathbf{r}_3 = -3i + 8j - 5k\) and \( -3i + 2j + \lambda k\) are coplanar, then \(\lambda = \)

• A. \( 1 \)
• B. \( 2 \)
• C. \( -2 \)
• D. \( -3 \)

Hint:

When testing for coplanarity of vectors, the scalar triple product is a powerful tool. If the scalar triple product equals zero, the vectors are coplanar.

Answer 1

The Correct Option is C

Step 1: Recognize that the vectors are coplanar if the volume of the parallelepiped they form is zero. This volume is calculated using the scalar triple product, which is defined as: \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \] This equation ensures the three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) lie in the same plane. 
Step 2: Define the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) based on the given position vectors: \[ \mathbf{a} = \mathbf{r}_2 - \mathbf{r}_1 = (3i + 2j - 5k) - (-i + 4j - 4k) = 4i - 2j - k \] \[ \mathbf{b} = \mathbf{r}_3 - \mathbf{r}_1 = (-3i + 8j - 5k) - (-i + 4j - 4k) = -2i + 4j - k \] \[ \mathbf{c} = \mathbf{r}_4 - \mathbf{r}_1 = (-3i + 2j + \lambda k) - (-i + 4j - 4k) = -2i - 2j + (\lambda + 4)k \] Step 3: Calculate the cross product \(\mathbf{b} \times \mathbf{c}\). This is done by computing the determinant of the matrix formed by the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of \( \mathbf{b} \) and \( \mathbf{c} \): \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 4 & -1 \\ -2 & -2 & \lambda + 4 \end{vmatrix} \] This determinant expands to: \[ \mathbf{b} \times \mathbf{c} = \mathbf{i} \left( 4(\lambda + 4) - (-1)(-2) \right) - \mathbf{j} \left( -2(\lambda + 4) - (-1)(-2) \right) + \mathbf{k} \left( -2(-2) - 4(-2) \right) \] \[ \mathbf{b} \times \mathbf{c} = \mathbf{i} \left( 4\lambda + 16 - 2 \right) - \mathbf{j} \left( -2\lambda - 8 - 2 \right) + \mathbf{k} \left( 4 + 8 \right) \] \[ \mathbf{b} \times \mathbf{c} = \mathbf{i} \left( 4\lambda + 14 \right) - \mathbf{j} \left( -2\lambda - 10 \right) + \mathbf{k} \left( 12 \right) \] \[ \mathbf{b} \times \mathbf{c} = (4\lambda + 14) \mathbf{i} + (2\lambda + 10) \mathbf{j} + 12 \mathbf{k} \] Step 4: Compute the dot product of \(\mathbf{a}\) with \(\mathbf{b} \times \mathbf{c}\): \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (4i - 2j - k) \cdot \left( (4\lambda + 14) \mathbf{i} + (2\lambda + 10) \mathbf{j} + 12 \mathbf{k} \right) \] \[ = 4(4\lambda + 14) - 2(2\lambda + 10) - 1(12) \] \[ = 16\lambda + 56 - 4\lambda - 20 - 12 \] \[ = 12\lambda + 24 \] Set this equal to 0 for the vectors to be coplanar: \[ 12\lambda + 24 = 0 \] \[ 12\lambda = -24 \] \[ \lambda = -2 \]