CUET UG 2025 — Mathematics Arithmetic

Given:

• $|\vec{a}| = 1$∣a∣=1 (unit vector)
• $|\vec{b}| = 1$∣b∣=1 (unit vector)
• $|\vec{a} + \vec{b}| = 1$∣a+b∣=1 (unit vector)

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For the magnitude of a vector sum:

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

Since $|\vec{a} + \vec{b}| = 1$∣a+b∣=1, $|\vec{a}| = 1$∣a∣=1, and $|\vec{b}| = 1$∣b∣=1:

$$1 = 1 + 1 + 2\vec{a} \cdot \vec{b}$$1=1+1+2a⋅b

$$1 = 2 + 2\vec{a} \cdot \vec{b}$$1=2+2a⋅b

$$2\vec{a} \cdot \vec{b} = -1$$2a⋅b=−1

$$\vec{a} \cdot \vec{b} = -\frac{1}{2}$$a⋅b=−21​

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For the magnitude of a vector difference:

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

$$|\vec{a} - \vec{b}|^2 = 1 + 1 - 2\left(-\frac{1}{2}\right)$$∣a−b∣2=1+1−2(−21​)

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

$$|\vec{a} - \vec{b}|^2 = 3$$∣a−b∣2=3

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

Therefore, the magnitude of $\vec{a} - \vec{b}$a−b is $\sqrt{3}$3​.