Let ( a = 2 i - j +3 k , ~b = 3 i -5 j + k ) and ( c ) be a vector such that ( a × c = c × b ) and (( a + c ) ·( b + c )= 168). Then the maximum value of (| c |^2) is :

Correct answer: D. 308 aligned & a =2 i - j +3 k & ~b =3 i -5 j +3 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 aligned 14+ c ·( a + b )+77 λ^2=168 using equation (1) aligned & 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 aligned Maximum value of | c |^2 occurs when λ=-2 aligned & | c |^2=77 λ^2 & =77 × 4 & =308 aligned

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Vector Algebra

Let ( a = 2 i - j +3 k , ~b = 3 i -5 j + k ) and ( c ) be a vector...

Let (

\overrightarrow{\mathrm{a}}=

2 \hat{i}-\hat{j}+3 \hat{k},

\overrightarrow{\mathrm{~b}}=

3 \hat{i}-5 \hat{j}+\hat{k}) and (

\overrightarrow{\mathrm{c}}

) be a vector such that (

\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}

) and ((\vec{a}+\vec{c})

\cdot(\vec{b}+\vec{c})=

168). Then the maximum value of (|\vec{c}|^2) is :

Options

A462
B77
C154
D308

Correct answer

D. 308

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ & \overrightarrow{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \ldots . .(1) \\ & |\overrightarrow{\mathrm{c}}|^2=\lambda^2(25+36+16) \\ & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168 \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~b}}+|\overrightarrow{\mathrm{c}}|^2=168 \end{aligned}

14+\vec{c} \cdot(\vec{a}+\vec{b})+77 \lambda^2=168

using equation (1)

\begin{aligned} & \lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^2+77 \lambda^2=154 \\ & 77 \lambda+77 \lambda^2-154=0 \\ & \lambda^2+\lambda-2=0 \\ & \lambda=-2,1 \end{aligned}

\therefore

Maximum value of

|\overrightarrow{\mathrm{c}}|^2

occurs when

\lambda=-2

\begin{aligned} & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & =77 \times 4 \\ & =308 \end{aligned}

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