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Q6:
Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.

Solution :

Step 1: Expanding the Scalar Product Expression

Given the scalar product equation: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8$.

Applying the distributive property of the dot product [Distributive Law: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$]:

$\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 8$

Since the dot product is commutative [Commutative Law: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$], the terms $-\vec{a} \cdot \vec{b}$ and $+\vec{b} \cdot \vec{a}$ cancel out. Utilizing the identity $\vec{x} \cdot \vec{x} = |\vec{x}|^2$ [Definition of Dot Product], we obtain:

$|\vec{a}|^2 - |\vec{b}|^2 = 8$ --- (Equation 1)

Step 2: Substituting the Given Relation

We are given the relation: $|\vec{a}| = 8|\vec{b}|$. Substitute this expression into Equation 1:

$(8|\vec{b}|)^2 - |\vec{b}|^2 = 8$

$64|\vec{b}|^2 - |\vec{b}|^2 = 8$

$63|\vec{b}|^2 = 8$

Step 3: Calculating Magnitudes

Solving for $|\vec{b}|$:

$|\vec{b}|^2 = \frac{8}{63}$

$|\vec{b}| = \sqrt{\frac{8}{63}} = \sqrt{\frac{4 \cdot 2}{9 \cdot 7}} = \frac{2\sqrt{2}}{3\sqrt{7}}$

To rationalize the denominator: $|\vec{b}| = \frac{2\sqrt{14}}{21}$.

Now, solve for $|\vec{a}|$ using $|\vec{a}| = 8|\vec{b}|$:

$|\vec{a}| = 8 \cdot \left( \frac{2\sqrt{14}}{21} \right) = \frac{16\sqrt{14}}{21}$.

Visual Representation of Vector Components

|a| = 8|b| |b| Origin

Final Solution: $|\vec{a}| = \frac{16\sqrt{14}}{21}$ and $|\vec{b}| = \frac{2\sqrt{14}}{21}$


More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.3


CBSE Solutions for Class 12 Mathematics Vector Algebra


Chapters in CBSE - Class 12 Mathematics


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