Let → a = ^ i + 5 ^ j + α ^ k , → b = ^ i + 3 ^ j + β ^ k and → c = - ^ i + 2 ^ j - 3 ^ k be three vectors such that, | → b × → c | = 5 3 and → a is perpendicular to → b . Then the greatest amongst the values of | → a |

Correct answer: 0 → a · → b = 0 1 + 15 + αβ = 0 αβ = - 16 . . . ( 1 ) Also, | → b × → c | ^ 2 = 75 (10 + β ^ 2 ) · 14 - (5 - 3β) ^ 2 = 75 5 β ^ 2 + 30β + 40 = 0 β = - 4, - 2 α = 4, 8 | → a | _ ^ 2 = (26 + α ^ 2 ) _ = 90

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Vector Algebra

Let → a = ^ i + 5 ^ j + α ^ k , → b = ^ i + 3 ^ j + β ^ k and → c =...

Let

\overset{\to}{a} = \overset{^}{i} + 5\overset{^}{j} + \alpha\overset{^}{k}, \overset{\to}{b} = \overset{^}{i} + 3\overset{^}{j} + \beta\overset{^}{k}

and

\overset{\to}{c} = - \overset{^}{i} + 2\overset{^}{j} - 3\overset{^}{k}

be three vectors such that,

| \overset{\to}{b} \times \overset{\to}{c} | = 5\sqrt{3}

and

\overset{\to}{a}

is perpendicular to

\overset{\to}{b} .

Then the greatest amongst the values of

| \overset{\to}{a}{|}^{2}

is ________.

Correct answer

Step-by-step solution

\overset{\to}{a} \cdot \overset{\to}{b} = 0 $$ 1 + 15 + \alpha\beta = 0 $$ \alpha\beta = - 16 . . . ( 1 ) $Also, $ {\left|\overset{\to}{b} \times \overset{\to}{c}\right|}^{2} = 75 $\left(10 + {\beta}^{2}\right) \cdot 14 - {\left(5 - 3\beta\right)}^{2} = 75$

$5{\beta}^{2} + 30\beta + 40 = 0$

$\beta = - 4, - 2$

$\alpha = 4, 8$

${\left|\overset{\to}{a}\right|}_\max^{2} = {\left(26 + {\alpha}^{2}\right)}_\max = 90$

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