JEE Main 20254 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 4 April
- Shift
- Morning Shift
- Subject
- Mathematics
- Chapter
- Vector Algebra
JEE Main 2025 · 4 April · Morning Shift · Mathematics · Vector Algebra
Consider two vectors u =3 i - j and v =2 i + j -λ k , λ 0 . The...
Consider two vectors \overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \lambda \gt 0\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{v}}_1+\overrightarrow{\mathrm{v}}_2\overrightarrow{\mathrm{v}}_1\overrightarrow{\mathrm{u}}\overrightarrow{\mathrm{v}}_2\overrightarrow{\mathrm{u}} is equal to
Options
- A
- B14
- C
- D10
Correct answer
B. 14
Step-by-step solution
\begin{aligned} & \overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}, \overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \\ & \Rightarrow \frac{\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{u}} \| \overrightarrow{\mathrm{v}}|}=\cos \theta \\ & \Rightarrow \frac{5}{\sqrt{10} \sqrt{5+\lambda^2}}=\frac{\sqrt{5}}{2 \sqrt{7}} \\ & \Rightarrow \lambda^2=9 \Rightarrow \lambda=3(\because \lambda \gt 0) \\ & \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{v}_1+\overrightarrow{\mathrm{v}_2 \\ & \Rightarrow|\overrightarrow{\mathrm{v}}|^2=\overrightarrow{\mathrm{v}_1^2+\overrightarrow{\mathrm{v}_2^2+2 \overrightarrow{\mathrm{v}_1 \cdot \overrightarrow{\mathrm{v}_2 \\ & \Rightarrow 14=\overrightarrow{\mathrm{v}_1^2+\overrightarrow{\mathrm{v}_2^2+0 \quad\left(\because \overrightarrow{\mathrm{v}_1 \perp \overrightarrow{\mathrm{v}_2\right) \\ & \Rightarrow\left|\overrightarrow{\mathrm{v}_1^2\right|+\left|\overrightarrow{\mathrm{v}_2^2\right|=14\end{aligned}