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 \(a-b+c\), \(λa-3b+4c\), \(-a+2b-3c\) and \(2a-4b+6c\) respectively. If AB AC , and AD are coplanar, then λ is equal to ____.
Correct Answer: 2
Solution and Explanation
Step 1: Condition for coplanarity. For three vectors \( \mathbf{AB}, \mathbf{AC}, \mathbf{AD} \) to be coplanar, the scalar triple product must be zero: \[ \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = 0. \]
Step 2: Expressing \( \mathbf{AB}, \mathbf{AC}, \mathbf{AD} \) in terms of \( \mathbf{a}, \mathbf{b}, \mathbf{c} \). The position vector of point \( A \) is: \[ \mathbf{A} = \mathbf{a} - \mathbf{b} + \mathbf{c}. \] The position vectors of points \( B, C, D \) are given as: \[ \mathbf{B} = \lambda \mathbf{a} - 3 \mathbf{b} + 4 \mathbf{c}, \quad \mathbf{C} = -\mathbf{a} + 2 \mathbf{b} - 3 \mathbf{c}, \quad \mathbf{D} = 2 \mathbf{a} - 4 \mathbf{b} + 6 \mathbf{c}. \] Now, the vectors \( \mathbf{AB}, \mathbf{AC}, \mathbf{AD} \) are given by: \[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = (\lambda \mathbf{a} - 3 \mathbf{b} + 4 \mathbf{c}) - (\mathbf{a} - \mathbf{b} + \mathbf{c}) = (\lambda - 1) \mathbf{a} - 2 \mathbf{b} + 3 \mathbf{c}, \] \[ \mathbf{AC} = \mathbf{C} - \mathbf{A} = (-\mathbf{a} + 2 \mathbf{b} - 3 \mathbf{c}) - (\mathbf{a} - \mathbf{b} + \mathbf{c}) = -2 \mathbf{a} + 3 \mathbf{b} - 4 \mathbf{c}, \] \[ \mathbf{AD} = \mathbf{D} - \mathbf{A} = (2 \mathbf{a} - 4 \mathbf{b} + 6 \mathbf{c}) - (\mathbf{a} - \mathbf{b} + \mathbf{c}) = \mathbf{a} - 3 \mathbf{b} + 5 \mathbf{c}. \]
Step 3: Computing the scalar triple product. Now, we compute the scalar triple product: \[ \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = 0. \] Substitute the expressions for \( \mathbf{AB}, \mathbf{AC}, \mathbf{AD} \): \[ \left( (\lambda - 1) \mathbf{a} - 2 \mathbf{b} + 3 \mathbf{c} \right) \cdot \left( (-2) \mathbf{a} + 3 \mathbf{b} - 4 \mathbf{c} \right) \times \left( \mathbf{a} - 3 \mathbf{b} + 5 \mathbf{c} \right) = 0. \]
Step 4: Expanding and solving for \( \lambda \). After performing the necessary vector cross and dot products, the equation simplifies to: \[ \lambda = 2. \]
Top Questions on Vectors
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LIST-I (Expressions) LIST-II (Values) A. \( i^{49} \) I. 1 B. \( i^{38} \) II. \(-i\) C. \( i^{103} \) III. \(i\) D. \( i^{92} \) IV. \(-1\)
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