TS EAMCET · Maths · Vector Algebra
\(P\) is the point of intersection of the diagonals of the parallelogram \(A B C D\). If \(S\) is any point in space and \(\mathbf{S A}+\mathbf{S B}+\mathbf{S C}+\mathbf{S D}=\lambda \mathbf{S P}\), then \(\lambda\) equals
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
Taking \(S\) as origin, let the position vectors of \(A, B\), \(C\) and \(D\) be \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{d}\) respectively. In \(\triangle S A C, P\) is the mid-point of \(A C\). \(\therefore \quad {SA}+{SC}=2 {SP}\) In \(\triangle S B D, P\) is the…
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