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If a⃗\vec{a} is a unit vector and (x⃗−a⃗)⋅(x⃗+a⃗)=15(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15, then the value of ∣x⃗∣|\vec{x}| is:

Solution

✅ Correct Option: 3

Given a⃗\vec{a} is a unit vector, so ∣a⃗∣=1|\vec{a}| = 1

Also given: (x⃗−a⃗)⋅(x⃗+a⃗)=15(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15


Expanding the dot product:

(x⃗−a⃗)⋅(x⃗+a⃗)(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a})

=x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗= \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a}


Since the dot product is commutative, x⃗⋅a⃗=a⃗⋅x⃗\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}, the middle terms cancel:

=x⃗⋅x⃗−a⃗⋅a⃗= \vec{x} \cdot \vec{x} - \vec{a} \cdot \vec{a}

=∣x⃗∣2−∣a⃗∣2= |\vec{x}|^2 - |\vec{a}|^2


Since ∣a⃗∣=1|\vec{a}| = 1:

∣x⃗∣2−1=15|\vec{x}|^2 - 1 = 15

∣x⃗∣2=16|\vec{x}|^2 = 16

∣x⃗∣=4|\vec{x}| = 4


Therefore, ∣x⃗∣=4|\vec{x}| = 4

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