Let → v = α ^ i + 2 ^ j - 3 ^ k , → w = 2α ^ i + ^ j - ^ k , and → u be a vector such that | → u | = α > 0 . If the minimum value of the scalar triple product [ → u → v → w ] is - α 3401 , and | → u . ^ i | ^ 2 =

Correct answer: 0 We have, [ → u → v → w ] = → u · ( → v × → w ) Now, min([ → u → v → w ] ) = - α 3401 ⇒ - | → u || → v × → w | = - α 3401 ⇒ α| → v × → w | = α 3401 ⇒ | → v × → w | = 3401 ⇒ || cases ^ i & ^ j & ^ k α & 2 & - 3 2α & 1 & - 1 cases | | = 3401 ⇒ | ^ i - 5α ^ j - 3α ^ k | = 3401 ⇒ 1 + 25 α ^ 2 + 9 α ^ 2 = 3401 ⇒ 34 α ^ 2 = 3400 ⇒ α ^ 2 = 100 ⇒ α = 10 (as α > 0 ) So, → u = λ( ^ i - 5α ^ j - 3α ^ k ) ⇒ α ^ 2 = λ ^ 2 (1 + 25 α ^ 2 + 9 α ^ 2 ) ⇒ 100 = λ ^ 2 (1 + 34 × 100) ⇒ λ ^ 2 = (100)/(3401) And, | → u · ^ i | ^ 2 = λ ^ 2 = (100)/(3401) = (m)/(n) Hence, m + n = 3501

JEE Main 20231 FebruaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 1 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2023 · 1 February · Morning Shift · Mathematics · Vector Algebra

Let → v = α ^ i + 2 ^ j - 3 ^ k , → w = 2α ^ i + ^ j - ^ k , and →...

Let

\overset{\to}{v} = \alpha\overset{^}{i} + 2\overset{^}{j} - 3\overset{^}{k}, \overset{\to}{w} = 2\alpha\overset{^}{i} + \overset{^}{j} - \overset{^}{k}

, and

\overset{\to}{u}

be a vector such that \left|\overset{\to}{u}\right| = \alpha > 0

. If the minimum value of the scalar triple product \left[\overset{\to}{u}\overset{\to}{v}\overset{\to}{w}\right]

- \alpha\sqrt{3401}

, and

{\left|\overset{\to}{u} . \overset{^}{i}\right|}^{2} = \dfrac{m}{n}

where

m and n

are coprime natural numbers, then

m + n

is equal to _____ .

Correct answer

Step-by-step solution

We have,

$\left[\overset{\to}{u}\overset{\to}{v}\overset{\to}{w}\right] = \overset{\to}{u} \cdot \left(\overset{\to}{v} \times \overset{\to}{w}\right)Now,$ min\left(\left[\overset{\to}{u}\overset{\to}{v}\overset{\to}{w}\right] $\right) = - \alpha\sqrt{3401}$

$⇒ - \left|\overset{\to}{u}\right|\left|\overset{\to}{v} \times \overset{\to}{w}\right|$ = - \alpha\sqrt{3401} $$ ⇒ \alpha\left|\overset{\to}{v} \times \overset{\to}{w}\right| = \alpha\sqrt{3401} $$ ⇒ \left|\overset{\to}{v} \times \overset{\to}{w}\right| $= \sqrt{3401}$

$⇒ \left|\left|$ \begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ \alpha & 2 & - 3 \\ 2\alpha & 1 & - 1\end{cases} $\right|$ \right| = \sqrt{3401} $$ ⇒ \left|\overset{^}{i} - 5\alpha\overset{^}{j} - 3\alpha\overset{^}{k}\right| = \sqrt{3401} $$ ⇒ \sqrt{1 + 25{\alpha}^{2} + 9{\alpha}^{2}} = \sqrt{3401} $$ ⇒ 34{\alpha}^{2} = 3400 $$ ⇒ {\alpha}^{2} = 100 $$ ⇒ \alpha = 10 $(as$ \alpha > 0 $)So,$ \overset{\to}{u} = \lambda\left(\overset{^}{i} - 5\alpha\overset{^}{j} - 3\alpha\overset{^}{k}\right)

$⇒ {\alpha}^{2} = {\lambda}^{2}\left(1 + 25{\alpha}^{2} + 9{\alpha}^{2}\right)$

$⇒ 100 = {\lambda}^{2}\left(1 + 34 \times 100\right)$ ⇒ {\lambda}^{2} = $\dfrac{100}{3401}And,$ {\left|\overset{\to}{u} \cdot \overset{^}{i}\right| $}^{2} = {\lambda}^{2} =$ \dfrac{100}{3401} = \dfrac{m}{n}

Hence, $m + n = 3501$

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