NIMCET 2025 — Mathematics PYQ

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Question

NIMCET 2025 — Mathematics PYQ

NIMCET | Mathematics | 2025

Let $\vec{a}$ and $\vec{c}$ be unit vectors such that the angle between them is $\cos^{-1}\left(\frac{1}{4}\right)$ . If $\vec{b} = 2\vec{c} + \lambda\vec{a}$ , where $\lambda > 0$ and $|\vec{b}| = 4$ , then $\lambda$ is equal to:

Choose the correct answer:

Explanation

Solution

Given:

$|\vec{a}| = 1, |\vec{c}| = 1$
$\cos \theta = \frac{1}{4}$ (where $\theta$ is the angle between $\vec{a}$ and $\vec{c}$ )
$\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos \theta = 1 \cdot 1 \cdot \frac{1}{4} = \frac{1}{4}$
$\vec{b} = 2\vec{c} + \lambda\vec{a}$
$|\vec{b}| = 4 \implies |\vec{b}|^2 = 16$

Step 1: Square the vector equation

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

16 = (2\vec{c} + \lambda\vec{a}) \cdot (2\vec{c} + \lambda\vec{a})

Step 2: Expand the dot product

16 = 4|\vec{c}|^2 + \lambda^2|\vec{a}|^2 + 4\lambda(\vec{a} \cdot \vec{c})

Step 3: Substitute the known values

16 = 4(1)^2 + \lambda^2(1)^2 + 4\lambda\left( \frac{1}{4} \right)

16 = 4 + \lambda^2 + \lambda

Step 4: Solve the quadratic equation

\lambda^2 + \lambda - 12 = 0

(\lambda + 4)(\lambda - 3) = 0

\lambda = -4 \text{ or } \lambda = 3

Step 5: Apply the condition

Since it is given that $\lambda > 0$ , we discard $\lambda = -4$ .

\lambda = 3

Final Answer:

The value of $\lambda$ is 3. The correct option is (B).

Choose the correct answer:

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