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If (a⃗−b⃗)⋅(a⃗+b⃗)=27(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})=27 and ∣a⃗∣=2∣b⃗∣|\vec{a}|=2|\vec{b}|, then ∣b⃗∣|\vec{b}| is :

Solution

✅ Correct Option: 1

Let's find ∣b⃗∣|\vec{b}| using the given conditions.


We start by expanding the dot product (a⃗−b⃗)⋅(a⃗+b⃗)(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b}):

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

Since dot product is commutative (a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}):

(a⃗−b⃗)⋅(a⃗+b⃗)=a⃗⋅a⃗−b⃗⋅b⃗=∣a⃗∣2−∣b⃗∣2=27(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b}) = \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} = |\vec{a}|^2 - |\vec{b}|^2 = 27


Using the relationship ∣a⃗∣=2∣b⃗∣|\vec{a}| = 2|\vec{b}|:

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

Substituting into our equation:

4∣b⃗∣2−∣b⃗∣2=274|\vec{b}|^2 - |\vec{b}|^2 = 27

3∣b⃗∣2=273|\vec{b}|^2 = 27

∣b⃗∣2=9|\vec{b}|^2 = 9

∣b⃗∣=3|\vec{b}| = 3

Therefore, ∣b⃗∣=3|\vec{b}| = 3.

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