Let → u , → v and → w be vectors in three-dimensional space, where → u and → v are unit vectors which are not perpendicular to each other and → u · → w = 1, → v · → w = 1, → w · → w = 4 . If the volume of the

Correct answer: 0 Volume of the parallelopiped is [ → u → v → w ] = 2 ⇒ [ → u → v → w ] ^ 2 = 2 ⇒ | cases → u · → u & → u · v & → u · → w → v · → u & → v · → v & → v · → w → w · → u & → w · → v & → w · → w cases | = 2 Using → u · → w = 1, → v · → w = 1, → w · → w = 4 , we get | cases → u · → u & → u · v & → u · → w → v · → u & → v · → v & → v · → w → w · → u & → w · → v & → w · → w cases | = | cases 1 & → u · → v & 1 → v · → u & 1 & 1 1 & 1 & 4 cases | = 2 ⇒ 3 - ( → u · → v )(4 → u · → v - 1) + → v · → u - 1 = 2 ⇒ ( → u · → v ) ( 1 - 4 → u · → v + 1 ) = 0 ⇒ → u · → v = (1)/(2) ( ∵ → u · → v ≠ 0 ) Also, | → u | = 1, | → v | = 1 So, | 3 → u + 5 → v | = (| 3 → u + 5 → v |) ^ 2 = 9 | → u | ^ 2 + 25 | → v | ^ 2 + 30 → u · → v = 7

JEE Advanced 2021Paper 1

Exam: JEE Advanced
Year: 2021
Paper: Paper 1
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2021 · Paper 1 · Mathematics · Vector Algebra

Let → u , → v and → w be vectors in three-dimensional space, where...

Let

\overset{\to}{u}, \overset{\to}{v}

and

\overset{\to}{w}

be vectors in three-dimensional space, where

\overset{\to}{u}

and

\overset{\to}{v}

are unit vectors which are not perpendicular to each other and

\overset{\to}{u} \cdot \overset{\to}{w} = 1, \overset{\to}{v} \cdot \overset{\to}{w} = 1, \overset{\to}{w} \cdot \overset{\to}{w} = 4

.
If the volume of the parallelopiped, whose adjacent sides are represented by the vectors

\overset{\to}{u}, \overset{\to}{v}

and

\overset{\to}{w}

, is

\sqrt{2}

, then the value of

| 3\overset{\to}{u} + 5\overset{\to}{v} |

Correct answer

Step-by-step solution

Volume of the parallelopiped is $\left[\overset{\to}{u}\overset{\to}{v}\overset{\to}{w}\right] = \sqrt{2}$

$⇒ {\left[\overset{\to}{u}\overset{\to}{v}\overset{\to}{w}\right]}^{2} = 2$

$⇒ \left|$ \begin{cases}\overset{\to}{u} \cdot \overset{\to}{u} & \overset{\to}{u} \cdot v & \overset{\to}{u} \cdot \overset{\to}{w} \\ \overset{\to}{v} \cdot \overset{\to}{u} & \overset{\to}{v} \cdot \overset{\to}{v} & \overset{\to}{v} \cdot \overset{\to}{w} \\ \overset{\to}{w} \cdot \overset{\to}{u} & \overset{\to}{w} \cdot \overset{\to}{v} & \overset{\to}{w} \cdot \overset{\to}{w}\end{cases} $\right| = 2$

Using $\overset{\to}{u} \cdot \overset{\to}{w} = 1, \overset{\to}{v} \cdot \overset{\to}{w} = 1, \overset{\to}{w} \cdot \overset{\to}{w} = 4$ , we get

\left| $\begin{cases}\overset{\to}{u} \cdot \overset{\to}{u} & \overset{\to}{u} \cdot v & \overset{\to}{u} \cdot \overset{\to}{w} \\ \overset{\to}{v} \cdot \overset{\to}{u} & \overset{\to}{v} \cdot \overset{\to}{v} & \overset{\to}{v} \cdot \overset{\to}{w} \\ \overset{\to}{w} \cdot \overset{\to}{u} & \overset{\to}{w} \cdot \overset{\to}{v} & \overset{\to}{w} \cdot \overset{\to}{w}\end{cases}$ \right| = \left| $\begin{cases}1 & \overset{\to}{u} \cdot \overset{\to}{v} & 1 \\ \overset{\to}{v} \cdot \overset{\to}{u} & 1 & 1 \\ 1 & 1 & 4\end{cases}$ \right| = 2 $$ ⇒ 3 - \left(\overset{\to}{u} \cdot \overset{\to}{v}\right)\left(4\overset{\to}{u} \cdot \overset{\to}{v} - 1\right) + \overset{\to}{v} \cdot \overset{\to}{u} - 1 = 2 $$ ⇒ \left(\overset{\to}{u} \cdot \overset{\to}{v}\right) ( 1 - 4\overset{\to}{u} \cdot \overset{\to}{v} + 1 ) = 0 $$ ⇒ \overset{\to}{u} \cdot \overset{\to}{v} = \dfrac{1}{2} ( ∵ \overset{\to}{u} \cdot \overset{\to}{v} \ne 0 ) $Also, \left|\overset{\to}{u}\right| = 1, \left|\overset{\to}{v}\right| = 1$

So, $| 3\overset{\to}{u} + 5\overset{\to}{v} | = \sqrt{{\left(| 3\overset{\to}{u} + 5\overset{\to}{v} |\right)}^{2}}$

$= \sqrt{9{\left|\overset{\to}{u}\right|}^{2} + 25{\left|\overset{\to}{v}\right|}^{2} + 30\overset{\to}{u} \cdot \overset{\to}{v}}$

$= 7$

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