JEE Mains · Maths · STD 12 - 10. vector algebra
Let the position vectors of points \(A\) and \(B\) be \(\hat{ i }+\hat{ j }+\hat{ k }\) and \(2 \hat{ i }+\hat{ j }+3 \hat{ k },\) respectively. A point \(P\) divides the line segment \(AB\) internally in the ratio \(\lambda: 1(\lambda>0) .\) If \(O\) is the origin and \(\overline{ OB } \cdot \overrightarrow{ OP }-3|\overrightarrow{ OA } \times \overrightarrow{ OP }|^{2}=6,\) then \(\lambda\) is equal to
- A \(0.6\)
- B \(0.7\)
- C \(0.8\)
- D \(0.5\)
Answer & Solution
Correct Answer
(C) \(0.8\)
Step-by-step Solution
Detailed explanation
Using section formula we get \(\overline{ OP }=\frac{2 \lambda+1}{\lambda+1} \hat{ i }+\frac{\lambda+1}{\lambda+1} \hat{ j }+\frac{3 \lambda+1}{\lambda+1} \hat{ k }\) \(Now \overrightarrow{ OB } \cdot \overrightarrow{ OP }=\frac{4 \lambda+2+\lambda+1+9 \lambda+3}{\lambda+1}\)…
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