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JEE Mains · Maths · STD 12 - 10. vector algebra

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. A \(\frac{23}{2}\)
  2. B 14
  3. C \(\frac{25}{2}\)
  4. D 10
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(B) 14

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\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…

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