The value of `lambda` for which the points `L(1,0,3), M(-1,3,4),N(1,2,1) and P(lambda,2,5)` are coplanar is

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find the value of \( \lambda \) for which the points \( L(1,0,3) \), \( M(-1,3,4) \), \( N(1,2,1) \), and \( P(\lambda,2,5) \) are coplanar, we can use the concept of vectors and the scalar triple product. Here’s a step-by-step solution:

### Step 1: Define the Vectors
We first need to find the vectors \( \vec{LM} \), \( \vec{LN} \), and \( \vec{LP} \).

- The vector \( \vec{LM} \) is calculated as:
\[ \vec{LM} = M - L = (-1 - 1, 3 - 0, 4 - 3) = (-2, 3, 1) \]

- The vector \( \vec{LN} \) is calculated as:
\[ \vec{LN} = N - L = (1 - 1, 2 - 0, 1 - 3) = (0, 2, -2) \]

- The vector \( \vec{LP} \) is calculated as:
\[ \vec{LP} = P - L = (\lambda - 1, 2 - 0, 5 - 3) = (\lambda - 1, 2, 2) \]

### Step 2: Set Up the Scalar Triple Product
The points are coplanar if the scalar triple product of the vectors \( \vec{LM} \), \( \vec{LN} \), and \( \vec{LP} \) is zero:
\[ \vec{LM} \cdot (\vec{LN} \times \vec{LP}) = 0 \]

### Step 3: Calculate the Cross Product \( \vec{LN} \times \vec{LP} \)
We can find the cross product using the determinant:
\[ \vec{LN} \times \vec{LP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -2 \\ \lambda - 1 & 2 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & -2 \\ \lambda - 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 2 \\ \lambda - 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ 2 \cdot 2 - (-2) \cdot 2 = 4 + 4 = 8 \]

2. For \( \hat{j} \):
\[ 0 \cdot 2 - (-2)(\lambda - 1) = 0 + 2(\lambda - 1) = 2\lambda - 2 \]

3. For \( \hat{k} \):
\[ 0 \cdot 2 - 2(\lambda - 1) = -2(\lambda - 1) = -2\lambda + 2 \]

Thus, we have:
\[ \vec{LN} \times \vec{LP} = (8, -(2\lambda - 2), -2\lambda + 2) = (8, -2\lambda + 2, -2\lambda + 2) \]

### Step 4: Calculate the Dot Product
Now we take the dot product with \( \vec{LM} \):
\[ \vec{LM} \cdot (\vec{LN} \times \vec{LP}) = (-2, 3, 1) \cdot (8, -2\lambda + 2, -2\lambda + 2) \]
Calculating this:
\[ = -2 \cdot 8 + 3 \cdot (-2\lambda + 2) + 1 \cdot (-2\lambda + 2) \]
\[ = -16 + 3(-2\lambda + 2) - 2\lambda + 2 \]
\[ = -16 - 6\lambda + 6 - 2\lambda + 2 \]
\[ = -16 + 8 - 8\lambda = -8 - 8\lambda \]

### Step 5: Set the Dot Product to Zero
For coplanarity, we set the dot product to zero:
\[ -8 - 8\lambda = 0 \]
Solving for \( \lambda \):
\[ -8\lambda = 8 \implies \lambda = -1 \]

### Final Answer
The value of \( \lambda \) for which the points are coplanar is:
\[ \lambda = -1 \]