NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

Let $\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{w} = \hat{i} + 3\hat{k}$ . If $\vec{u}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \ \vec{v} \ \vec{w}]$ is:

Choose the correct answer:

Explanation

1. Identify the given vectors:

\vec{v} = 2\hat{i} + \hat{j} - \hat{k}

\vec{w} = \hat{i} + 0\hat{j} + 3\hat{k}

Since $\vec{u}$ is a unit vector, $|\vec{u}| = 1$ .

2. Express the Scalar Triple Product:

The scalar triple product $[\vec{u} \ \vec{v} \ \vec{w}]$ is defined as:

[\vec{u} \ \vec{v} \ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})

3. Calculate the cross product $\vec{v} \times \vec{w}$ :

\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix}

\vec{v} \times \vec{w} = \hat{i}(3 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 1)

\vec{v} \times \vec{w} = 3\hat{i} - 7\hat{j} - \hat{k}

4. Find the magnitude of $\vec{v} \times \vec{w}$ :

|\vec{v} \times \vec{w}| = \sqrt{3^2 + (-7)^2 + (-1)^2}

|\vec{v} \times \vec{w}| = \sqrt{9 + 49 + 1} = \sqrt{59}

5. Determine the maximum value:

We know that $\vec{u} \cdot (\vec{v} \times \vec{w}) = |\vec{u}| |\vec{v} \times \vec{w}| \cos \theta$ .

To maximize this value, $\cos \theta$ must be $1$ (i.e., $\vec{u}$ must be in the same direction as $\vec{v} \times \vec{w}$ ).

\text{Max Value} = (1)(\sqrt{59})(1) = \sqrt{59}

Correct Option: (c) $\sqrt{59}$

Choose the correct answer:

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