CUET UG 2025 — Mathematics Arithmetic

Given $|\vec{a}| = 1$∣a∣=1, $|\vec{b}| = 2$∣b∣=2, and $|2\vec{a}+\vec{b}| = 2\sqrt{3}$∣2a+b∣=23​.

Starting with $|2\vec{a}+\vec{b}| = 2\sqrt{3}$∣2a+b∣=23​:

$$|2\vec{a}+\vec{b}|^2 = (2\sqrt{3})^2$$∣2a+b∣2=(23​)2

$$|2\vec{a}+\vec{b}|^2 = 12$$∣2a+b∣2=12

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Expanding the left side using the dot product:

$$(2\vec{a}+\vec{b}) \cdot (2\vec{a}+\vec{b}) = 12$$(2a+b)⋅(2a+b)=12

$$4\vec{a} \cdot \vec{a} + 4\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 12$$4a⋅a+4a⋅b+b⋅b=12

Substituting $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$a⋅a=∣a∣2=1 and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 4$b⋅b=∣b∣2=4:

$$4(1) + 4\vec{a} \cdot \vec{b} + 4 = 12$$4(1)+4a⋅b+4=12

$$8 + 4\vec{a} \cdot \vec{b} = 12$$8+4a⋅b=12

$$4\vec{a} \cdot \vec{b} = 4$$4a⋅b=4

$$\vec{a} \cdot \vec{b} = 1$$a⋅b=1

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To find $|\vec{a}-\vec{b}|$∣a−b∣:

$$|\vec{a}-\vec{b}|^2 = (\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b})$$∣a−b∣2=(a−b)⋅(a−b)

$$|\vec{a}-\vec{b}|^2 = \vec{a} \cdot \vec{a} - 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$$∣a−b∣2=a⋅a−2a⋅b+b⋅b

$$|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$$∣a−b∣2=∣a∣2−2a⋅b+∣b∣2

$$|\vec{a}-\vec{b}|^2 = 1 - 2(1) + 4$$∣a−b∣2=1−2(1)+4

$$|\vec{a}-\vec{b}|^2 = 3$$∣a−b∣2=3

$$|\vec{a}-\vec{b}| = \sqrt{3}$$∣a−b∣=3​