Let a = i + j + k , b =2 i +2 j + k and d = a × b . If c is a vector such that a · c =| c | , | c -2 a |^2=8 and the angle between d and c is (π)/(4) , then |10-3 ~b · c |+| d × c |^2 is equal to

Correct answer: 0 aligned & a = i + j + k & b =2 i +2 j + k & d = a × b & =- i + j & | c -2 a |^2=8 & | c |^2+4| a |^2-4 a · c =8 & | c |^2+12-4| c |=8 & | c |^2-4| c |+4=0 & | c |^2=2 & d = a × b & d × c =( a × b ) × c & (| d | ×| c | (π)/(4))^2=(( a · c ) b -( b · c ) a )^2 & 4=4| b |^2+( b · c ) 2(| a |^2)-2( b · c )( a · b ) aligned Let b · c =x aligned & 4=36+3 x^2-20 x & 3 x^2-20 x+32=0 & x=(8)/(3), 4 & b · c =(8)/(3), 4 & b · c =(8)/(3) aligned Now, |10-3 b · c |+| d × c |^2 =|10-8|+(2)^2=6

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Vector Algebra

Let a = i + j + k , b =2 i +2 j + k and d = a × b . If c is a...

Let

\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}

and

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

. If

\overrightarrow{\mathrm{c}}

is a vector such that

\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|

|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8

and the angle between

\overrightarrow{\mathrm{d}}

and

\overrightarrow{\mathrm{c}}

\frac{\pi}{4}

, then

|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2

is equal to

Correct answer

Step-by-step solution

\begin{aligned} & \vec{a}=\hat{i}+\hat{j}+\hat{k} \\ & \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k} \\ & \vec{d}=\vec{a} \times \vec{b} \\ & =-\hat{i}+\hat{j} \\ & |\vec{c}-2 \vec{a}|^2=8 \\ & |\vec{c}|^2+4|\vec{a}|^2-4 \vec{a} \cdot \vec{c}=8 \\ & |\vec{c}|^2+12-4|\vec{c}|=8 \\ & |\vec{c}|^2-4|\vec{c}|+4=0 \\ & |\vec{c}|^2=2 \\ & \vec{d}=\vec{a} \times \vec{b} \\ & \vec{d} \times \vec{c}=(\vec{a} \times \vec{b}) \times \vec{c} \\ & \left(|\vec{d}| \times|\vec{c}| \sin \frac{\pi}{4}\right)^2=((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a})^2 \\ & 4=4|\vec{b}|^2+(\vec{b} \cdot \vec{c}) 2\left(|\vec{a}|^2\right)-2(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{b}) \end{aligned}

Let

\vec{b} \cdot \vec{c}=x

\begin{aligned} & 4=36+3 x^2-20 x \\ & 3 x^2-20 x+32=0 \\ & x=\frac{8}{3}, 4 \\ & \Rightarrow \vec{b} \cdot \vec{c}=\frac{8}{3}, 4 \\ & \Rightarrow \vec{b} \cdot \vec{c}=\frac{8}{3} \end{aligned}

Now,

|10-3 \vec{b} \cdot \vec{c}|+|\vec{d} \times \vec{c}|^2 =|10-8|+(2)^2=6$

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