Consider two vectors $\overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}$ and…

Consider two vectors $\overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \lambda \gt 0$. The angle between them is given by $\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)$. Let $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{v}}_1+\overrightarrow{\mathrm{v}}_2$, where $\overrightarrow{\mathrm{v}}_1$ is parallel to $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}_2$ is perpendicular to $\overrightarrow{\mathrm{u}}$. Then the value $\left|\overrightarrow{\mathrm{v}}_1\right|^2+\left|\overrightarrow{\mathrm{v}}_2\right|^2$ is equal to
  1. $\frac{23}{2}$
  2. 14
  3. $\frac{25}{2}$
  4. 10

Solution

$\begin{aligned} & \overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}, \overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \\ & \Rightarrow \frac{\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{u}} \| \overrightarrow{\mathrm{v}}|}=\cos \theta \\ & \Rightarrow \frac{5}{\sqrt{10} \sqrt{5+\lambda^2}}=\frac{\sqrt{5}}{2 \sqrt{7}} \\ & \Rightarrow \lambda^2=9 \Rightarrow \lambda=3(\because \lambda \gt 0) \\ & \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{v}}_1+\overrightarrow{\mathrm{v}}_2 \\ & \Rightarrow|\overrightarrow{\mathrm{v}}|^2=\overrightarrow{\mathrm{v}}_1^2+\overrightarrow{\mathrm{v}}_2^2+2 \overrightarrow{\mathrm{v}}_1 \cdot \overrightarrow{\mathrm{v}}_2 \\ & \Rightarrow 14=\overrightarrow{\mathrm{v}}_1^2+\overrightarrow{\mathrm{v}}_2^2+0 \quad\left(\because \overrightarrow{\mathrm{v}}_1 \perp \overrightarrow{\mathrm{v}}_2\right) \\ & \Rightarrow\left|\overrightarrow{\mathrm{v}}_1^2\right|+\left|\overrightarrow{\mathrm{v}}_2^2\right|=14\end{aligned}$ ,

Asked in: JEE Main 2025 (04 Apr Shift 1)

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