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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 u→=3i^−j^and\overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}} and\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \lambda \gt 0.Theanglebetweenthemisgivenby. The angle between them is given by\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right).Let. Let\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{v}}_1+\overrightarrow{\mathrm{v}}_2,where, where\overrightarrow{\mathrm{v}}_1isparalleltois parallel to\overrightarrow{\mathrm{u}}andand\overrightarrow{\mathrm{v}}_2isperpendiculartois perpendicular to\overrightarrow{\mathrm{u}}.Thenthevalue∣v→1∣2+∣v→2∣2. Then the value \left|\overrightarrow{\mathrm{v}}_1\right|^2+\left|\overrightarrow{\mathrm{v}}_2\right|^2 is equal to

Options

  1. A232\frac{23}{2}
  2. B14
  3. C252\frac{25}{2}
  4. 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}
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