Exams Joint Entrance Examination Mathematics Vector Algebra Question 106 Question 106 of 379 Medium NUMERICAL
Let a → = i ^ + j ^ + k ^ , b → = − i ^ − 8 j ^ + 2 k ^ \overset{\to }{a}=\hat{i}+\hat{j}+\hat{k},\overset{\to }{b}=-\hat{i}-8\hat{j}+2\hat{k} a → = i ^ + j ^ + k ^ , b → = − i ^ − 8 j ^ + 2 k ^ and c → = 4 i ^ + c 2 j ^ + c 3 k ^ \overset{\to }{c}=4\hat{i}+{c}_{2}\hat{j}+{c}_{3}\hat{k} c → = 4 i ^ + c 2 j ^ + c 3 k ^ be three vectors such that b → × a → = c → × a → \overset{\to }{b}\times \overset{\to }{a}=\overset{\to }{c}\times \overset{\to }{a} b → × a → = c → × a → . If the angle between the vector c → \overset{\to }{c} c → and the vector 3 i ^ + 4 j ^ + k ^ 3\hat{i}+4\hat{j}+\hat{k} 3 i ^ + 4 j ^ + k ^ is θ \theta θ , then the greatest integer less than or equal to tan 2 θ {\tan }^{2}\theta tan 2 θ is \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 38
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a ⃗ = i ^ + j ^ + k b ⃗ = i ^ + 8 j ^ + 2 k c ⃗ = 4 i ^ + c 2 j ^ + c 3 k b ⃗ × a ⃗ = c ⃗ × a ⃗ ( b ⃗ − c ⃗ ) × a ⃗ = 0 b ⃗ − c ⃗ = λ α ⃗ b ⃗ = c ⃗ + λ α ⃗ \begin{gathered} \vec{a}=\hat{i}+\hat{j}+k \\ \vec{b}=\hat{i}+8\hat{j}+2k \\ \vec{c}=4\hat{i}+{c}_{2}\hat{j}+{c}_{3}k \\ \vec{b}\times \vec{a}=\vec{c}\times \vec{a} \\ (\vec{b}-\vec{c})\times \vec{a}=0 \\ \vec{b}-\vec{c}=\lambda \vec{\alpha } \\ \vec{b}=\vec{c}+\lambda \vec{\alpha } \end{gathered} a = i ^ + j ^ + k b = i ^ + 8 j ^ + 2 k c = 4 i ^ + c 2 j ^ + c 3 k b × a = c × a ( b − c ) × a = 0 b − c = λ α b = c + λ α
− i ^ − 8 j ^ + 2 k = ( 4 i ^ + c 2 j ^ + c 3 k ) + λ ( i ^ + j ^ + k ) λ + 4 = − 1 ⇒ λ = − 5 λ + c 2 = − 8 ⇒ c 2 = − 3 λ + c 3 = 2 ⇒ c 3 = 7 c → = 4 i ^ − 3 j ^ + 7 k \begin{gathered} -\hat{i}-8\hat{j}+2k=(4\hat{i}+{c}_{2}\hat{j}+{c}_{3}k)+\lambda (\hat{i}+\hat{j}+k) \\ \lambda +4=-1\Rightarrow \lambda =-5 \\ \lambda +{c}_{2}=-8\Rightarrow {c}_{2}=-3 \\ \lambda +{c}_{3}=2\Rightarrow {c}_{3}=7 \\ \overset{\to }{c}=4\hat{i}-3\hat{j}+7k \end{gathered} − i ^ − 8 j ^ + 2 k = ( 4 i ^ + c 2 j ^ + c 3 k ) + λ ( i ^ + j ^ + k ) λ + 4 = − 1 ⇒ λ = − 5 λ + c 2 = − 8 ⇒ c 2 = − 3 λ + c 3 = 2 ⇒ c 3 = 7 c → = 4 i ^ − 3 j ^ + 7 k
cos θ = 12 − 12 + 7 26 ⋅ 74 = 7 26 ⋅ 74 = 7 2 481 tan 2 θ = 625 × 3 49 [ tan 2 θ ] = 38 \begin{gathered} \cos \theta =\frac{12-12+7}{\sqrt{26}\cdot \sqrt{74}}=\frac{7}{\sqrt{26}\cdot \sqrt{74}}=\frac{7}{2\sqrt{481}} \\ {\tan }^{2}\theta =\frac{625\times 3}{49} \\ [{\tan }^{2}\theta ]=38 \end{gathered} cos θ = 26 ⋅ 74 12 − 12 + 7 = 26 ⋅ 74 7 = 2 481 7 tan 2 θ = 49 625 × 3 [ tan 2 θ ] = 38 Answer the question to unlock