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Let a⃗,b⃗\vec{a}, \vec{b} be two vectors such that ∣a⃗∣=2,∣b⃗∣=3,a⃗⋅b⃗=4|\vec{a}| = 2, |\vec{b}| = 3, \vec{a} \cdot \vec{b} = 4, then ∣a⃗−b⃗∣|\vec{a} - \vec{b}| is equal to

Solution

✅ Correct Option: 2

Given ∣a⃗∣=2|\vec{a}| = 2, ∣b⃗∣=3|\vec{b}| = 3, and a⃗⋅b⃗=4\vec{a} \cdot \vec{b} = 4


For any two vectors, the magnitude of their difference is:

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


Substituting the given values:

∣a⃗−b⃗∣2=(2)2+(3)2−2(4)|\vec{a} - \vec{b}|^2 = (2)^2 + (3)^2 - 2(4)

∣a⃗−b⃗∣2=4+9−8|\vec{a} - \vec{b}|^2 = 4 + 9 - 8

∣a⃗−b⃗∣2=5|\vec{a} - \vec{b}|^2 = 5


Taking the square root:

∣a⃗−b⃗∣=5|\vec{a} - \vec{b}| = \sqrt{5}

Therefore, ∣a⃗−b⃗∣=5|\vec{a} - \vec{b}| = \sqrt{5}

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