Exams Joint Entrance Examination Mathematics Vector Algebra Question 109 Question 109 of 379 Medium NUMERICAL
The least positive integral value of α \alpha α , for which the angle between the vectors α i ^ − 2 j ^ + 2 k ^ \alpha \hat{i}-2\hat{j}+2\hat{k} α i ^ − 2 j ^ + 2 k ^ and α i ^ + 2 α j ^ − 2 k ^ \alpha \hat{i}+2\alpha \hat{j}-2\hat{k} α i ^ + 2 α j ^ − 2 k ^ is acute, is \_\_\_\_\_\_\_\_\_\_\_.
Answer: 5
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& cos θ = ( α i ^ − 2 j ^ + 2 k ^ ) ⋅ ( α i ^ + 2 α j ^ − 2 k ^ ) α 2 + 4 + 4 α 2 + 4 α 2 + 4 & cos θ = α 2 − 4 α − 4 α 2 + 8 5 α 2 + 4 & ⇒ α 2 − 4 α − 4 > 0 ⇒ ( α − 2 ) 2 > 8 & ⇒ α 2 − 4 α + 4 > 8 α − 22 2 α − 22 + 2 2 α L e a s t p o s i t i v e i n t e g r a l v a l u e o f \& \cos \theta =\frac{(\alpha \hat{i}-2\hat{j}+2\hat{k})\cdot (\alpha \hat{i}+2\alpha \hat{j}-2\hat{k})}{\sqrt{{\alpha }^{2}+4+4}\sqrt{{\alpha }^{2}+4{\alpha }^{2}+4}} \ \& \cos \theta =\frac{{\alpha }^{2}-4\alpha -4}{\sqrt{{\alpha }^{2}+8}\sqrt{5{\alpha }^{2}+4}} \ \& \Rightarrow {\alpha }^{2}-4\alpha -4>0\, \Rightarrow (\alpha -2{)}^{2}>8 \ \& \Rightarrow {\alpha }^{2}-4\alpha +4>8\, \alpha -22\sqrt{2}\alpha -22+2\sqrt{2}\alpha Least positive integral value of & cos θ = α 2 + 4 + 4 α 2 + 4 α 2 + 4 ( α i ^ − 2 j ^ + 2 k ^ ) ⋅ ( α i ^ + 2 α j ^ − 2 k ^ ) & cos θ = α 2 + 8 5 α 2 + 4 α 2 − 4 α − 4 & ⇒ α 2 − 4 α − 4 > 0 ⇒ ( α − 2 ) 2 > 8 & ⇒ α 2 − 4 α + 4 > 8 α − 22 2 α − 22 + 2 2 α L e a s tp os i t i v e in t e g r a l v a l u eo f Answer the question to unlock