AP EAMCET · Maths · Vector Algebra
It \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are non-coplanar vectors and the four points with position vectors \(2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}\), \(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}, 3 \mathbf{a}+4 \mathbf{b}+2 \mathbf{c}\) and \(k \mathbf{a}-6 \mathbf{b}+6 \mathbf{c}\) are coplanar, then \(k=\)
- A 0
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Let A(2a + 3b − c), B(a − 2b + 3c) C(3a + 4b − 2c) and D(ka − 6b + 6c) ∴ AB = − a − 5b + 4c \(\begin{aligned} & \mathbf{A C}=\mathbf{a}+\mathbf{b}-\mathbf{c} \\ & \mathbf{A D}=(k-2) \mathbf{a}-9 \mathbf{b}+7 \mathbf{c} \end{aligned}\) Since, \(A, B, C, D\) are coplanar…
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