# Question 1 [JEE adv, 2021]

Let $$\vec{u}, \vec{v}$$ and $$\vec{w}$$ be vectors in three-dimensional space, where $$\vec{u}$$ and $$\vec{v}$$ are unit vectors which are not perpendicular to each other and $\vec{u}\cdot \vec{w} = 1,\quad \vec{v}\cdot \vec{w} = 1\quad \vec{w}\cdot \vec{w} = 4$ If the volume of the parallelopiped, whose adjacent sides are represented by the vectors $$\vec{u}, \vec{v}$$ and $$\vec{w}$$, is $$\sqrt{2}$$, then the value of $$|3\vec{u} + 5\vec{v}|$$ is ______.

## Concepts

The volume of the parallelopiped is given by the scalar triple product of the side vectors.

## Solution

$$\vec{u}$$ and $$\vec{v}$$ are unit vectors, to solve the problem we need the value of $$\vec{u}\cdot\vec{v}$$.

#### Step 1

Use triple product of the vectors which is equal to the given volume. $\vec{u}\cdot(\vec{v}\times\vec{w}) = \sqrt{2}$

#### Step 2

Squaring the above equation gives $(\vec{u}\cdot(\vec{v}\times\vec{w}))^2 = \left |\begin{matrix} \vec{u}\cdot\vec{u} & \vec{u}\cdot\vec{v} & \vec{u}\cdot\vec{w}\\ \vec{v}\cdot\vec{u} & \vec{v}\cdot\vec{v} & \vec{v}\cdot\vec{w}\\ \vec{w}\cdot\vec{u} & \vec{w}\cdot\vec{v} & \vec{w}\cdot\vec{w} \end{matrix}\right | = 2$

This implies $\left |\begin{matrix} 1 & \vec{u}\cdot\vec{v} & 1\\ \vec{v}\cdot\vec{u} & 1 & 1\\ 1 & 1 & 4 \end{matrix}\right | = 2$

Hence, $\begin{eqnarray} 4(\vec{u}\cdot\vec{v})^2 - 2(\vec{u}\cdot\vec{v}) = 0\\ \vec{u}\cdot\vec{v} = 0 \text{~or~} \vec{u}\cdot\vec{v} = \frac{1}{2} \end{eqnarray}$ As $$\vec{u}\cdot\vec{v}\neq 0$$, the only solution is $$\vec{u}\cdot\vec{v} = \dfrac{1}{2}$$

#### Step 3

$$(|3\vec{u} + 5\vec{v}|)^2 = 9 + 25 + 30\times\dfrac{1}{2} = 49$$ This implies $$|3\vec{u} + 5\vec{v}| = 7$$