Exams Joint Entrance Examination Mathematics Vector Algebra Question 31 Question 31 of 379 Medium MCQ
Let a ⃗ = 2 i ^ − j ^ + 3 k ^ , b ⃗ = 3 i ^ − 5 j ^ + k ^ \vec{a}=2\hat{i}-\hat{j}+3\hat{k},\vec{b}=3\hat{i}-5\hat{j}+\hat{k} a = 2 i ^ − j ^ + 3 k ^ , b = 3 i ^ − 5 j ^ + k ^ and c ⃗ \vec{c} c be a vector such that a ⃗ × c ⃗ = a ⃗ × b ⃗ = c ⃗ × b ⃗ \vec{a}\times \vec{c}=\vec{a}\times \vec{b}=\vec{c}\times \vec{b} a × c = a × b = c × b and ( a ⃗ + c ⃗ ) ⋅ ( b ⃗ + c ⃗ ) = 168 (\vec{a}+\vec{c})\cdot (\vec{b}+\vec{c})=168 ( a + c ) ⋅ ( b + c ) = 168 . Then the maximum value of ∣ c ⃗ ∣ 2 |\vec{c}{|}^{2} ∣ c ∣ 2 is :
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a → = 2 i ^ − j ^ + 3 k ^ b → ← = 3 i ^ − 5 j ^ + k ^ a → × c → = c → × b → a → × c → + b → × c → = 0 ( a → + b → ) × c → = 0 ⇒ c → = λ ( a → + b → ) c → = λ ( 5 i ^ − 6 j ^ + 4 k ^ ) … . . ( 1 ) ∣ c → ∣ 2 = λ 2 ( 25 + 36 + 16 ) ∣ c → ∣ 2 = 77 λ 2 ( a → + c → ) ⋅ ( b → + c → ) = 168 a → ⋅ b → ← + a → ⋅ c → + c → ⋅ b → ← + ∣ c → ∣ 2 = 168 14 + c → ⋅ ( a → + b → ) + 77 λ 2 = 168 \begin{gathered} \overset{\to }{a}=2\hat{i}-\hat{j}+3\hat{k} \\ \overset{\to \leftarrow }{b}=3\hat{i}-5\hat{j}+\hat{k} \\ \overset{\to }{a}\times \overset{\to }{c}=\overset{\to }{c}\times \overset{\to }{b} \\ \overset{\to }{a}\times \overset{\to }{c}+\overset{\to }{b}\times \overset{\to }{c}=0 \\ (\overset{\to }{a}+\overset{\to }{b})\times \overset{\to }{c}=0 \\ \Rightarrow \overset{\to }{c}=\lambda (\overset{\to }{a}+\overset{\to }{b}) \\ \overset{\to }{c}=\lambda (5\hat{i}-6\hat{j}+4\hat{k})\ldots ..(1) \\ |\overset{\to }{c}{|}^{2}={\lambda }^{2}(25+36+16) \\ |\overset{\to }{c}{|}^{2}=77{\lambda }^{2} \\ (\overset{\to }{a}+\overset{\to }{c})\cdot (\overset{\to }{b}+\overset{\to }{c})=168 \\ \overset{\to }{a}\cdot \overset{\to \leftarrow }{b}+\overset{\to }{a}\cdot \overset{\to }{c}+\overset{\to }{c}\cdot \overset{\to \leftarrow }{b}+|\overset{\to }{c}{|}^{2}=168 \\ 14+\overset{\to }{c}\cdot (\overset{\to }{a}+\overset{\to }{b})+77{\lambda }^{2}=168 \end{gathered} a → = 2 i ^ − j ^ + 3 k ^ b →← = 3 i ^ − 5 j ^ + k ^ a → × c → = c → × b → a → × c → + b → × c → = 0 ( a → + b → ) × c → = 0 ⇒ c → = λ ( a → + b → ) c → = λ ( 5 i ^ − 6 j ^ + 4 k ^ ) … .. ( 1 ) ∣ c → ∣ 2 = λ 2 ( 25 + 36 + 16 ) ∣ c → ∣ 2 = 77 λ 2 ( a → + c → ) ⋅ ( b → + c → ) = 168 a → ⋅ b →← + a → ⋅ c → + c → ⋅ b →← + ∣ c → ∣ 2 = 168 14 + c → ⋅ ( a → + b → ) + 77 λ 2 = 168
using equation (1)
λ ∣ 5 i ^ − 6 j ^ + 4 k ^ ∣ 2 + 77 λ 2 = 154 77 λ + 77 λ 2 − 154 = 0 λ 2 + λ − 2 = 0 λ = − 2 , 1 \begin{gathered} \lambda |5\hat{i}-6\hat{j}+4\hat{k}{|}^{2}+77{\lambda }^{2}=154 \\ 77\lambda +77{\lambda }^{2}-154=0 \\ {\lambda }^{2}+\lambda -2=0 \\ \lambda =-2,1 \end{gathered} λ ∣5 i ^ − 6 j ^ + 4 k ^ ∣ 2 + 77 λ 2 = 154 77 λ + 77 λ 2 − 154 = 0 λ 2 + λ − 2 = 0 λ = − 2 , 1
∴ \therefore ∴ Maximum value of ∣ c → ∣ 2 |\overset{\to }{c}{|}^{2} ∣ c → ∣ 2 occurs when λ = − 2 \lambda =-2 λ = − 2
∣ c → ∣ 2 = 77 λ 2 = 77 × 4 = 308 \begin{gathered} |\overset{\to }{c}{|}^{2}=77{\lambda }^{2} \\ =77\times 4 \\ =308 \end{gathered} ∣ c → ∣ 2 = 77 λ 2 = 77 × 4 = 308 Answer the question to unlock