NIMCET 2023 — Mathematics PYQ

1. Identify Given Properties

Since $\vec{a}$ and $\vec{b}$ are unit vectors, their magnitudes are equal to $1$:

$$|\vec{a}|=1$$

$$|\vec{b}|=1$$

We are given the magnitude condition:

$$|2\vec{a}+\vec{b}|=3$$

2. Square Both Sides

To eliminate the magnitude bars, square both sides of the equation:

$$|2\vec{a}+\vec{b}{|}^2=3^2$$

Using the vector property $|\vec{v}{|}^2=\vec{v}\cdot \vec{v}$:

$$(2\vec{a}+\vec{b})\cdot (2\vec{a}+\vec{b})=9$$

3. Expand the Dot Product

$$4(\vec{a}\cdot \vec{a})+2(\vec{a}\cdot \vec{b})+2(\vec{b}\cdot \vec{a})+(\vec{b}\cdot \vec{b})=9$$

Since dot product is commutative ($\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$), and $\vec{v}\cdot \vec{v}=|\vec{v}{|}^2$:

$$4|\vec{a}{|}^2+4(\vec{a}\cdot \vec{b})+|\vec{b}{|}^2=9$$

4. Substitute Unit Values

Substitute $|\vec{a}|=1$ and $|\vec{b}|=1$ into the equation:

$$4(1{)}^2+4(\vec{a}\cdot \vec{b})+(1{)}^2=9$$

$$4+4(\vec{a}\cdot \vec{b})+1=9$$

$$5+4(\vec{a}\cdot \vec{b})=9$$

Subtract $5$ from both sides:

$$4(\vec{a}\cdot \vec{b})=4$$

$$\vec{a}\cdot \vec{b}=1$$

5. Evaluate the Angle Between Vectors

The formula for the dot product is:

$$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$$

Substitute the known values ($1=1\cdot 1\cdot \cos \theta$):

$$\cos \theta =1$$

The angle where cosine is $1$ is:

$$\theta =0^{\circ }$$

An angle of $\theta =0^{\circ }$ means that the directions match perfectly. Therefore, $\vec{a}$ is parallel to $\vec{b}$.