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If a⃗\vec{a} and b⃗\vec{b} are two unit vectors and a⃗+b⃗\vec{a} + \vec{b} is also unit vector, the magnitude of a⃗−b⃗\vec{a} - \vec{b} is

Solution

✅ Correct Option: 4

Given:

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

For the magnitude of a vector sum:

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

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

1=1+1+2a⃗⋅b⃗1 = 1 + 1 + 2\vec{a} \cdot \vec{b}

1=2+2a⃗⋅b⃗1 = 2 + 2\vec{a} \cdot \vec{b}

2a⃗⋅b⃗=−12\vec{a} \cdot \vec{b} = -1

a⃗⋅b⃗=−12\vec{a} \cdot \vec{b} = -\frac{1}{2}


For the magnitude of a vector difference:

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

∣a⃗−b⃗∣2=1+1−2(−12)|\vec{a} - \vec{b}|^2 = 1 + 1 - 2\left(-\frac{1}{2}\right)

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

∣a⃗−b⃗∣2=3|\vec{a} - \vec{b}|^2 = 3

∣a⃗−b⃗∣=3|\vec{a} - \vec{b}| = \sqrt{3}

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

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