MHT CET · Maths · Three Dimensional Geometry
If \(|\overline{\mathrm{u}}|=2\) and \(\overline{\mathrm{u}}\) makes angles of \(60^{\circ}\) and \(120^{\circ}\) with axes \(\mathrm{OX}\) and OY in the origin, then \(\bar{u}=\)
- A \(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}\)
- B \(2(\hat{\mathrm{i}}+\hat{\mathrm{j}} \pm \sqrt{2} \hat{\mathrm{k}})\)
- C \(2(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}})\)
- D \(2(\hat{\mathrm{i}}-\hat{\mathrm{j}} \pm \sqrt{2} \hat{\mathrm{k}})\)
Answer & Solution
Correct Answer
(D) \(2(\hat{\mathrm{i}}-\hat{\mathrm{j}} \pm \sqrt{2} \hat{\mathrm{k}})\)
Step-by-step Solution
Detailed explanation
We have \(|\overline{\mathrm{u}}|=2\) and \(\cos \alpha=60^{\circ}=\frac{1}{2}\) and \(\cos \beta=\cos 120^{\circ}=-\frac{1}{2}\) Now \(\cos ^2 \gamma=1-\left(\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{2} \quad \Rightarrow \cos \gamma= \pm \frac{1}{\sqrt{2}}\)
Thus direction of \(\overline{\mathrm{u}}\) are \(1,-1, \pm \sqrt{2}\) \(\therefore \overline{\mathrm{u}}=2(\hat{\mathrm{i}}-\hat{\mathrm{j}} \pm \sqrt{2} \hat{\mathrm{k}})\)
Thus direction of \(\overline{\mathrm{u}}\) are \(1,-1, \pm \sqrt{2}\) \(\therefore \overline{\mathrm{u}}=2(\hat{\mathrm{i}}-\hat{\mathrm{j}} \pm \sqrt{2} \hat{\mathrm{k}})\)
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