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 \(|3\vec{u} + 5\vec{v}| = 7\)