Let ^ i , ^ j and ^ k be the unit vectors along the three positive coordinate axes. Let → a = 3 ^ i + ^ j - ^ k → b = ^ i + b _ 2 ^ j + b _ 3 ^ k , b _ 2 , b _ 3 ∈ ℝ → c = c _ 1 ^ i + c _ 2 ^ j + c _ 3 ^ k , c _ 1 , c _

Correct answer: B. → b · → c - = 0 Given, → a = 3 ^ i + ^ j - ^ k → b = ^ i + b _ 2 ^ j + b _ 3 ^ k → c = c _ 1 ^ i + c _ 2 ^ j + c _ 3 ^ k [ cases 0 & - c _ 3 & c _ 2 c _ 3 & 0 & - c _ 1 - c _ 2 & c _ 1 & 0 cases ][ array ll 1 b _ 2 b _ 3 array ] = [ array ll 3 - c _ 1 1 - c _ 2 - 1 - c _ 3 array ] Now multiply and compare we get, b _ 2 c _ 3 - b _ 3 c _ 2 = c _ 1 - 3 . . . . . . . (1) c _ 3 - b _ 3 c _ 1 = 1 - c _ 2 . . . . . . . . . . . (2) c _ 2 - b _ 2 c _ 1 = 1 + c _ 3 . . . . . . . . . . . (3) Now applying operation equation(1) ^ i - (2) ^ j + (3) ^ k we get, ^ i ( b _ 2 c _ 3 - c _ 2 b _ 3 ) - ^ j ( c _ 3 - b _ 3 c _ 1 ) + ^ k ( c _ 2 - b _ 2 c _ 1 ) = c _ 1 ^ i + c _ 2 ^ j + c _ 3 ^ k - 3 ^ i - ^ j + ^ k ⇒ → b × → c = → c - → a . . . . (i) ⇒ ( → b × → c ) · → b = → c · → b - → a · → b ⇒ → b · → c = 0 \ given → a . → b =0\ Again from (i) ( → c × → b ) · → c = → a · → c - | → c | ^ 2 = 0 ⇒ | → c | ^ 2 = | → a || → c |cosθ , where θ = → a ∧ → c ⇒ | → c | ≥ | → a | ⇒ | → c | ≥ 11 Given that → a · → b = 0 ⇒ b _ 2 - b _ 3 + 3 = 0 ⇒ b _ 3 - b _ 2 = 3 Also b _ 2 - b _ 3 > 0 Now | → b | = 1 + b _ 2 ^ 2 + b _ 3 ^ 2 = 1 + ( b _ 3 - b _ 2 ) ^ 2 + 2 b _ 2 b _ 3 = 10 + 2 b _ 2 b _ 3 ⇒ | → b | ^ 2 > 10 ⇒ | → b | > 10

JEE Advanced 2022Paper 2

Exam: JEE Advanced
Year: 2022
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2022 · Paper 2 · Mathematics · Vector Algebra

Let ^ i , ^ j and ^ k be the unit vectors along the three positive...

Let $\overset{^}{i}, \overset{^}{j} and$ \overset{^}{k} $be the unit vectors along the three positive coordinate axes. Let$ \overset{\to}{a} = 3\overset{^}{i} + \overset{^}{j} - \overset{^}{k} $$ \overset{\to}{b} = \overset{^}{i} + {b}_{2}\overset{^}{j} + {b}_{3}\overset{^}{k}, {b}_{2}, {b}_{3} \in \mathbb{R} $$ \overset{\to}{c} = {c}_{1}\overset{^}{i} + {c}_{2}\overset{^}{j} + {c}_{3}\overset{^}{k}, {c}_{1}, {c}_{2}, {c}_{3} \in \mathbb{R} $be three vectors such that$ {b}_{2}{b}_{3} > 0, \overset{\to}{a} \cdot \overset{\to}{b} = 0 $and \left($ \begin{cases}0 & - {c}_{3} & {c}_{2} \\ {c}_{3} & 0 & - {c}_{1} \\ - {c}_{2} & {c}_{1} & 0\end{cases} $\right)\left($ $\begin{array}{ll}1 \\ {b}_{2} \\ {b}_{3}\end{array}$ $\right) = \left($ $\begin{array}{ll}3 - {c}_{1} \\ 1 - {c}_{2} \\ - 1 - {c}_{3}\end{array}$ $ $\right) .$ Then, which of the following is/are TRUE?

Options

A $\overset{\to}{a} \cdot \overset{\to}{c} = 0$
B $\overset{\to}{b} \cdot \overset{\to}{c} - = 0$
C $\left|\overset{\to}{b}\right| > \sqrt{10}$
D $\left|\overset{\to}{c}\right| \le \sqrt{11}$

Correct answer

B. $\overset{\to}{b} \cdot \overset{\to}{c} - = 0$

Step-by-step solution

Given,

$\overset{\to}{a} = 3\overset{^}{i} + \overset{^}{j} - \overset{^}{k}$

$\overset{\to}{b} = \overset{^}{i} + {b}_{2}\overset{^}{j} + {b}_{3}\overset{^}{k}$

$\overset{\to}{c} = {c}_{1}\overset{^}{i} + {c}_{2}\overset{^}{j} + {c}_{3}\overset{^}{k}$

\left[ $\begin{cases}0 & - {c}_{3} & {c}_{2} \\ {c}_{3} & 0 & - {c}_{1} \\ - {c}_{2} & {c}_{1} & 0\end{cases}$ \right]\left[ $\begin{array}{ll}1 \\ {b}_{2} \\ {b}_{3}\end{array}$ \right] = \left[ $\begin{array}{ll}3 - {c}_{1} \\ 1 - {c}_{2} \\ - 1 - {c}_{3}\end{array}$ \right] $Now multiply and compare we get,$ {b}_{2}{c}_{3} - {b}_{3}{c}_{2} = {c}_{1} - 3 . . . . . . . \left(1\right) $$ {c}_{3} - {b}_{3}{c}_{1} = 1 - {c}_{2} . . . . . . . . . . . \left(2\right) $$ {c}_{2} - {b}_{2}{c}_{1} = 1 + {c}_{3} . . . . . . . . . . . \left(3\right) $Now applying operation$ \text{equation}\left(1\right)\overset{^}{i} - \left(2\right)\overset{^}{j} + \left(3\right)\overset{^}{k} $we get,$ \overset{^}{i}\left({b}_{2}{c}_{3} - {c}_{2}{b}_{3}\right) - \overset{^}{j}\left({c}_{3} - {b}_{3}{c}_{1}\right) + \overset{^}{k}\left({c}_{2} - {b}_{2}{c}_{1}\right) = {c}_{1}\overset{^}{i} + {c}_{2}\overset{^}{j} + {c}_{3}\overset{^}{k} - 3\overset{^}{i} - \overset{^}{j} + \overset{^}{k} $$ ⇒ \overset{\to}{b} \times \overset{\to}{c} = \overset{\to}{c} - \overset{\to}{a} . . . . \left(i\right) $$ ⇒ \left(\overset{\to}{b} \times \overset{\to}{c}\right) \cdot \overset{\to}{b} = \overset{\to}{c} \cdot \overset{\to}{b} - \overset{\to}{a} \cdot \overset{\to}{b} $$ ⇒ \overset{\to}{b} \cdot \overset{\to}{c} = 0 $\left\{\text{given}\overset{\to}{a}\text{.}\overset{\to}{b}\text{=0}\right\}$

Again from (i)

\left(\overset{\to}{c} \times \overset{\to}{b}\right) \cdot \overset{\to}{c} = \overset{\to}{a} \cdot \overset{\to}{c} - {\left|\overset{\to}{c}\right|}^{2} = 0 $$ ⇒ {\left|\overset{\to}{c}\right|}^{2} = \left|\overset{\to}{a}\right|\left|\overset{\to}{c}\right|cos\theta $, where$ \theta = \overset{\to}{a} ∧\overset{\to}{c} $$ ⇒ \left|\overset{\to}{c}\right| \ge \left|\overset{\to}{a}\right| ⇒ \left|\overset{\to}{c}\right| \ge \sqrt{11} $Given that$ \overset{\to}{a} \cdot \overset{\to}{b} = 0 ⇒ {b}_{2} - {b}_{3} + 3 = 0 $$ ⇒ {b}_{3} - {b}_{2} = 3 $Also$ {b}_{2} - {b}_{3} > 0 $Now \left|\overset{\to}{b}\right| = 1 + {b}_{2}^{2} + {b}_{3}^{2}$

$= 1 + {\left({b}_{3} - {b}_{2}\right)}^{2} + 2{b}_{2}{b}_{3}$

$= 10 + 2{b}_{2}{b}_{3}$

$⇒ {\left|\overset{\to}{b}\right|}^{2} > 10 ⇒ \left|\overset{\to}{b}\right| > \sqrt{10}$

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