CUET Mathematics 2024 - If (a-b) *(a+b)=27 and |a|=2|b|, then |b| is : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Vector Algebra

Easy

If $(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})=27$ and $|\vec{a}|=2|\vec{b}|$ , then $|\vec{b}|$ is :

3

2

\frac56

6

✅ Correct Option: 1

Let's find

|\vec{b}|

using the given conditions.

We start by expanding the dot product

(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})

(\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 (

\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

(\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

|\vec{a}| = 2|\vec{b}|

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

Substituting into our equation:

4|\vec{b}|^2 - |\vec{b}|^2 = 27

3|\vec{b}|^2 = 27

|\vec{b}|^2 = 9

|\vec{b}| = 3

Therefore,

|\vec{b}| = 3