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
- A \(\frac{23}{2}\)
- B 14
- C \(\frac{25}{2}\)
- D 10
Answer & Solution
Correct Answer
(B) 14
Step-by-step Solution
Detailed explanation
\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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