JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+5\, \hat{\mathrm{j}}+\alpha\, \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+3 \,\hat{\mathrm{j}}+\beta\, \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+2\, \hat{\mathrm{j}}-3 \,\hat{\mathrm{k}}\) be three vectors such that, \(|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|=5 \sqrt{3}\) and \(\overrightarrow{\mathrm{a}}\) is perpendicular to \(\overrightarrow{\mathrm{b}}\). Then the greatest amongst the values of \(|\vec{a}|^{2}\) is .... .
- A \(60\)
- B \(70\)
- C \(80\)
- D \(90\)
Answer & Solution
Correct Answer
(D) \(90\)
Step-by-step Solution
Detailed explanation
since, \(\vec{a} \cdot \vec{b}=0\) \(1+15+\alpha \beta=0 \Rightarrow \alpha \beta=-16\) Also, \(|\vec{b} \times \vec{c}|^{2}=75 \Rightarrow\left(10+\beta^{2}\right) 14-(5-3 \beta)^{2}=75\) \(\Rightarrow 5 \beta^{2}+30 \beta+40=0\) \(\Rightarrow \beta=-4,-2\)…
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