Let → a = ^ i - α ^ j + β ^ k , → b = 3 ^ i + β ^ j - α ^ k and → c = - α ^ i - 2 ^ j + ^ k , where α and β are integers. If → a · → b = - 1 and → b · → c = 10, then ( → a × → b ) · → c is equal to ______.

Correct answer: 0 We have, → a = (1, - α, β) , → b = (3, β, - α) and → c = (- α, - 2, 1); α, β ∈ I → a · → b = - 1 ⇒ 3 - αβ - αβ = - 1 ⇒ αβ = 2 . . . (1) Now, → b · → c = 10 ⇒ - 3α - 2β - α = 10 ⇒ 2α + β + 5 = 0 . . . (2) From equations (1) and (2) , ⇒ 2α + (2)/(α) +5 = 0 ⇒ 2 α ^ 2 + 2 + 5α = 0 ⇒ 2 α ^ 2 + 5α + 2 = 0 ⇒ 2 α ^ 2 + 4α + α + 2 = 0 ⇒ 2α(α + 2) + 1(α + 2) = 0 ⇒ (2α + 1)(α + 2) = 0 α = (- 1)/(2) is neglected since α is an integer. ∴ α = - 2 Put α = - 2 in equation (1), we get β = - 1 . Now, [ → a → b → c ] = | cases 1 & 2 & - 1 3 & - 1 & 2 2 & - 2 & 1 cases | = 1(- 1 + 4) - 2 (3 - 4) - 1 (- 6 + 2) = 3 + 2 + 4 = 9 .

JEE Main 202127 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2021 · 27 July · Evening Shift · Mathematics · Vector Algebra

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

Let

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

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

where

\alpha

and

\beta

are integers. If

\overset{\to}{a} \cdot \overset{\to}{b} = - 1

and

\overset{\to}{b} \cdot \overset{\to}{c} = 10,

then \left(\overset{\to}{a} \times \overset{\to}{b}\right) \cdot \overset{\to}{c}

is equal to ______.

Correct answer

Step-by-step solution

We have, $\overset{\to}{a} = \left(1, - \alpha, \beta\right)$ , \overset{\to}{b} = \left(3, \beta, - \alpha\right) $and$ \overset{\to}{c} = \left(- \alpha, - 2, 1\right); \alpha, \beta \in I $$ \overset{\to}{a} \cdot \overset{\to}{b} = - 1 ⇒ 3 - \alpha\beta - \alpha\beta = - 1 $$ ⇒ \alpha\beta = 2 . . . \left(1\right)

Now, $\overset{\to}{b} \cdot \overset{\to}{c} = 10$

$⇒ - 3\alpha - 2\beta - \alpha = 10$

$⇒ 2\alpha + \beta + 5 = 0 . . . \left(2\right)$

From equations \left(1\right) $and \left(2\right)$ ,

$⇒ 2\alpha + \dfrac{2}{\alpha} +5 = 0$

$⇒ 2{\alpha}^{2} + 2 + 5\alpha = 0$

$⇒ 2{\alpha}^{2} + 5\alpha + 2 = 0$

$⇒ 2{\alpha}^{2} + 4\alpha + \alpha + 2 = 0$

$⇒ 2\alpha\left(\alpha + 2\right) + 1\left(\alpha + 2\right) = 0$

$⇒ \left(2\alpha + 1\right)\left(\alpha + 2\right) = 0$

$\alpha = \dfrac{- 1}{2} is neglected since$ \alpha $is an integer.$ ∴ \alpha = - 2 $Put$ \alpha = - 2 in $equation \left(1\right), we get$ \beta = - 1 $.Now, \left[\overset{\to}{a}\overset{\to}{b}\overset{\to}{c}\right] = \left|$ \begin{cases}1 & 2 & - 1 \\ 3 & - 1 & 2 \\ 2 & - 2 & 1\end{cases} $\right|$ = 1\left(- 1 + 4\right) $- 2$ \left(3 - 4\right) $- 1 \left(- 6 + 2\right)$

$= 3 + 2 + 4 = 9 .$

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