NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

The sum of two vectors $\vec{a}$ and $\vec{b}$ is a vector $\vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = 2$ . Then, the magnitude of $\vec{a} - \vec{b}$ is equal to:

Choose the correct answer:

Explanation

Solution

1. Given Data:

$\vec{a} + \vec{b} = \vec{c}$
$|\vec{a}| = 2, |\vec{b}| = 2, |\vec{c}| = 2$

2. Finding the Dot Product:

We know that for any vector, the square of the magnitude is equal to the dot product of the vector with itself: $\vec{A} \cdot \vec{A} = |\vec{A}|^2$ .

Taking the dot product of both sides of the equation $\vec{a} + \vec{b} = \vec{c}$ with themselves:

(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{c} \cdot \vec{c}

|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{c}|^2

Substitute the given magnitudes (all are equal to $2$ ):

2^2 + 2^2 + 2\vec{a} \cdot \vec{b} = 2^2

4 + 4 + 2\vec{a} \cdot \vec{b} = 4

2\vec{a} \cdot \vec{b} = 4 - 8

2\vec{a} \cdot \vec{b} = -4

3. Finding the Magnitude of $(\vec{a} - \vec{b})$ :

Now, we calculate the square of the magnitude of the difference:

|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})

|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}

Substitute the values we know ( $|\vec{a}|^2 = 4, |\vec{b}|^2 = 4$ and $2\vec{a} \cdot \vec{b} = -4$ ):

|\vec{a} - \vec{b}|^2 = 4 + 4 - (-4)

|\vec{a} - \vec{b}|^2 = 4 + 4 + 4

|\vec{a} - \vec{b}|^2 = 12

Taking the square root:

|\vec{a} - \vec{b}| = \sqrt{12}

|\vec{a} - \vec{b}| = 2\sqrt{3}

Correct Option: 1 ( $2\sqrt{3}$ )