JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{ AB }=2 \hat{ i }+4 \hat{ j }-5 k\) and \(\overrightarrow{ AD }=\hat{ i }+2 \hat{ j }+\lambda k , \lambda \in R\). Let the projection of the vector \(\overrightarrow{ v }=\hat{ i }+\hat{ j }+\hat{ k }\) on the diagonal \(\overrightarrow{ AC }\) of the parallelogram ABCD be of length one unit. If \(\alpha, \beta\), where \(\alpha>\beta\), be the roots of the equation \(\lambda^2 x ^2- 6 \lambda x +5=0\), then \(2 \alpha-\beta\) is equal to
- A 1
- B 4
- C 3
- D 6
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ AC }=3 \hat{ i }+6 \hat{ j }+(\lambda-5) k\) \(\overrightarrow{ V } \cdot AC =1 \Rightarrow 3+6+\lambda-5=\sqrt{9+36+(\lambda-5)^2}\) \(\Rightarrow \lambda^2+8 \lambda+16=\lambda^2-10 \lambda+70\) \(\Rightarrow \lambda=\frac{54}{18}=3\) ∴ Quadratic :…
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