Exams Joint Entrance Examination Mathematics Vector Algebra Question 29 Question 29 of 379 Medium MCQ
Let a ^ \hat{a} a ^ be a unit vector perpendicular to the vectors b ⃗ = i ^ − 2 j ^ + 3 k ^ \vec{b}=\hat{i}-2\hat{j}+3\hat{k} b = i ^ − 2 j ^ + 3 k ^ and c ⃗ = 2 i ^ + 3 j ^ − k ^ \vec{c}=2\hat{i}+3\hat{j}-\hat{k} c = 2 i ^ + 3 j ^ − k ^ , and a ^ \hat{a} a ^ makes an angle of cos − 1 ( − 1 3 ) {\cos }^{-1}(-\frac{1}{3}) cos − 1 ( − 3 1 ) with the vector i ^ + j ^ + k ^ \hat{i}+\hat{j}+\hat{k} i ^ + j ^ + k ^ . If a ^ \hat{a} a ^ makes an angle of π 3 \frac{\pi }{3} 3 π with the vector i ^ + α j ^ + k ^ \hat{i}+\alpha \hat{j}+\hat{k} i ^ + α j ^ + k ^ , then the value of a a a is:
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\begin{gathered} \overset{\to }{v}=\hat{i}+\hat{j}+\hat{k} \\ \begin{gathered} \overset{\to }{b} \quad \times \overset{\to }{c}=|\begin{array}{|c|}\hat{i} \quad \hat{j} \quad \hat{k} \\ 1 \quad -2 \quad 3 \\ 2 \quad 3 \quad -1\end{array}| \\ =-7\hat{i}+7\hat{j}+7\hat{k} \\ =-7(\hat{i}-\hat{j}-\hat{k}) \end{gathered}
Now a ^ = i ^ − j ^ − k ^ 3 \hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} a ^ = 3 i ^ − j ^ − k ^ or − i ^ + j ^ + k ^ 3 \frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} 3 − i ^ + j ^ + k ^
↓ ↓ \, \, \downarrow \, \, \downarrow ↓ ↓
\cos \theta = \frac{\hat{a}\cdot \overset{\to }{v}}{|\overset{\to }{v}|}=\frac{1-1-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3}\, \cos \theta =\frac{\hat{a}\cdot \overset{\to }{v}}{|\overset{\to }{v}|}=\frac{-1+1+1}{3}=\frac{1}{3}\, \\ \Rightarrow \hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} \end{gathered}
cos π 3 = a ^ ⋅ ( i ^ + α j ^ + k ^ ) 1 + α 2 + 1 ⇒ 1 2 = 1 − α − 1 3 α 2 + 2 ⇒ 3 2 α 2 + 2 = − α ( ∴ α < 0 ) 3 α 2 + 6 = 4 α 2 ⇒ α = − 6 \begin{gathered} \cos \frac{\pi }{3}=\frac{\hat{a}\cdot (\hat{i}+\alpha \hat{j}+\hat{k})}{\sqrt{1+{\alpha }^{2}+1}} \\ \Rightarrow \frac{1}{2}=\frac{1-\alpha -1}{\sqrt{3}\sqrt{{\alpha }^{2}+2}} \\ \Rightarrow \frac{\sqrt{3}}{2}\sqrt{{\alpha }^{2}+2}=-\alpha \, (\therefore \alpha <0) \\ 3{\alpha }^{2}+6=4{\alpha }^{2} \\ \Rightarrow \alpha =-\sqrt{6} \end{gathered} cos 3 π = 1 + α 2 + 1 a ^ ⋅ ( i ^ + α j ^ + k ^ ) ⇒ 2 1 = 3 α 2 + 2 1 − α − 1 ⇒ 2 3 α 2 + 2 = − α ( ∴ α < 0 ) 3 α 2 + 6 = 4 α 2 ⇒ α = − 6 Answer the question to unlock