Let a = i + j + k , b =3 i +2 j - k , c =λ j +μ k and d be a unit vector such that a × d = b × d and c · d =1 , If c is perpendicular to a , then |3 λ d +μ c |^2 is equal to _______ .

Correct answer: 0 aligned & a × d - b × d =0 & ( a - b ) × d =0 & ~d =t( a - b ) & d =t(-2 i - j +2 k ) & | d |=1 & |t|=(1)/(3) aligned aligned & c · a =0 & λ+μ=0 & μ=-λ & c =λ( j - k ),| c |^2=2 λ^2 & c · d =1 & t(-2,-1,2) · λ(0,1,-1)=1 & λ t=(-1)/(3) λ^2=1 aligned aligned & |3 λ ~d +μ c|^2=9 λ^2| ~d |^2+μ^2| c |^2+6 λ μ( ~d · c ) & =3 λ^2+2 λ^4 & =5 aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Vector Algebra

Let a = i + j + k , b =3 i +2 j - k , c =λ j +μ k and d be a unit...

Let

\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}

and

\hat{d}

be a unit vector such that

\overrightarrow{\mathrm{a}} \times \hat{\mathrm{d}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{d}}

and

\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1

, If

\vec{c}

is perpendicular to

\vec{a}

, then

|3 \lambda \hat{d}+\mu \overrightarrow{\mathrm{c}}|^2

is equal to _______ .

Correct answer

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=0 \\ & (\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{d}}=0 \\ & \overrightarrow{\mathrm{~d}}=\mathrm{t}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \\ & \overrightarrow{\mathrm{d}}=\mathrm{t}(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\ & |\overrightarrow{\mathrm{d}}|=1 \\ & |\mathrm{t}|=\frac{1}{3}\end{aligned}

\begin{aligned} & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \lambda+\mu=0 \\ & \mu=-\lambda \\ & \overrightarrow{\mathrm{c}}=\lambda(\hat{\mathrm{j}}-\hat{\mathrm{k}}),|\overrightarrow{\mathrm{c}}|^2=2 \lambda^2 \\ & \overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1 \\ & \mathrm{t}(-2,-1,2) \cdot \lambda(0,1,-1)=1 \\ & \lambda \mathrm{t}=\frac{-1}{3} \Rightarrow \lambda^2=1\end{aligned}

\begin{aligned} & |3 \lambda \hat{\mathrm{~d}}+\mu \mathrm{c}|^2=9 \lambda^2|\hat{\mathrm{~d}}|^2+\mu^2|\overrightarrow{\mathrm{c}}|^2+6 \lambda \mu(\hat{\mathrm{~d}} \cdot \overrightarrow{\mathrm{c}}) \\ & =3 \lambda^2+2 \lambda^4 \\ & =5\end{aligned}

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