AP EAMCET · Maths · Vector Algebra
If \(\mathbf{P Q}+\mathbf{Q R}=\left(2 \lambda^2-5\right) \mathbf{R P}\) then, \(\lambda\) is equal to
- A \(\pm 1\)
- B \(\pm \sqrt{2}\)
- C \(\pm \sqrt{3}\)
- D 0
Answer & Solution
Correct Answer
(B) \(\pm \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given, } \mathbf{P Q}+\mathbf{Q R}=\left(2 \lambda^2-5\right) \mathbf{R} \mathbf{P} \\ & \Rightarrow \quad \mathbf{P R}=\left(2 \lambda^2-5\right) \mathbf{R P} \\ & \Rightarrow \quad-\mathbf{R P}=\left(2 \lambda^2-5\right) \mathbf{R P} \\ & \therefore…
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