Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|2\vec{a} + 3\vec{b}| = |3\vec{a} + \vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$. If $\frac{1}{8}\vec{a}$ is a unit vector, then $|\vec{b}|$ is equal to : Options: (A) 4 (B) 6 (C) 5 (D) 8

### Analyzing the Setup We are given two vectors, $\vec{a}$ and $\vec{b}$, originating from a common point with an angle of $\theta = 60^\circ$ between them. We are provided with the condition that $\frac{1}{8}\vec{a}$ is a unit vector. By definition, the magnitude of a unit vector is $1$. $$|\frac{1}{8}\vec{a}| = 1$$ Since $\frac{1}{8}$ is a positive scalar, we can extract it from the modulus: $$\frac{1}{8}|\vec{a}| = 1 \Rightarrow |\vec{a}| = 8$$ ### The Squaring Strategy We are tasked with solving the equation $|2\vec{a} + 3\vec{b}| = |3\vec{a} + \vec{b}|$. To eliminate the modulus bars, we square both sides of the equation. $$(|2\vec{a} + 3\vec{b}|)^2 = (|3\vec{a} + \vec{b}|)^2$$ Squaring is a powerful technique because it allows us to utilize the property that the square of a magnitude is equivalent to the dot product of the vector with itself, specifically $|\vec{u}|^2 = \vec{u} \cdot \vec{u}$. ### The Algebraic Expansion Using the identity $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2(\vec{u} \cdot \vec{v})$, we expand both sides of the equation: $$4|\vec{a}|^2 + 9|\vec{b}|^2 + 12(\vec{a} \cdot \vec{b}) = 9|\vec{a}|^2 + |\vec{b}|^2 + 6(\vec{a} \cdot \vec{b})$$ By rearranging the terms to one side, we simplify the expression: $$8|\vec{b}|^2 + 6(\vec{a} \cdot \vec{b}) - 5|\vec{a}|^2 = 0$$ ### The Final Bridge: The Dot Product We resolve the dot product using the definition $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$. Given $|\vec{a}| = 8$ and $\theta = 60^\circ$: $$\vec{a} \cdot \vec{b} = 8 \cdot |\vec{b}| \cdot \cos(60^\circ) = 8 \cdot |\vec{b}| \cdot \frac{1}{2} = 4|\vec{b}|$$ Substituting this value back into our simplified equation: $$8|\vec{b}|^2 + 6(4|\vec{b}|) - 5(8^2) = 0$$ $$8|\vec{b}|^2 + 24|\vec{b}| - 320 = 0$$ Dividing the entire equation by $8$, we obtain the quadratic equation: $$|\vec{b}|^2 + 3|\vec{b}| - 40 = 0$$ Factoring the quadratic yields: $$(|\vec{b}| + 8)(|\vec{b}| - 5) = 0$$ Since the magnitude of a vector must be non-negative, we discard the negative root. Thus, the final result is: **$|\vec{b}| = 5$**

I got confused with the cross terms. Why is it $2(\vec{u} \cdot \vec{v})$ and not just $2\vec{u}\vec{v}$?

Ah, the classic trap! Remember that vectors do not multiply like scalar numbers. When you expand $(2\vec{a} + 3\vec{b}) \cdot (2\vec{a} + 3\vec{b})$, you are performing a dot product. The distributive property gives you $4(\vec{a} \cdot \vec{a}) + 6(\vec{a} \cdot \vec{b}) + 6(\vec{b} \cdot \vec{a}) + 9(\vec{b} \cdot \vec{b})$. Since the dot product is commutative, $6(\vec{a} \cdot \vec{b}) + 6(\vec{b} \cdot \vec{a})$ becomes $12(\vec{a} \cdot \vec{b})$. Always treat the dot product as the 'multiplication' operator for vectors.

JEE Main 2021 (31 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :

Q: Why do we reject the negative root when solving the quadratic equation for $|\vec{b}|$?

Remember, $|\vec{b}|$ represents the magnitude of a vector. By definition, magnitude is the length of a vector, and length is a scalar quantity that measures distance. Distance, in our physical world, cannot be negative. Even if the algebra gives you a negative result, it is an 'extraneous solution' that doesn't fit the physical reality of the problem.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. When you see an equation involving the modulus of vector sums, like

∣2 a + 3 b ∣ = ∣3 a + b ∣

, your first instinct should be to remove the modulus. How do we do that? By squaring both sides. This transforms the geometric magnitude problem into an algebraic one involving dot products.

Hint 2

Hint 2: The Arsenal. Remember the fundamental identity for the square of a vector sum:

∣ u + v ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2} + 2 (u \cdot v)

. You will need to apply this to both sides of your equation. Also, recall that the dot product is defined as

a \cdot b = ∣ a ∣∣ b ∣ cos θ

Hint 3

Hint 3: The Execution. Start by squaring both sides:

(∣2 a + 3 b ∣)^{2} = (∣3 a + b ∣)^{2}

. Expand these using the identity mentioned in Hint 2. Substitute the known value

∣ a ∣ = 8

and the angle

θ = 6 0^{\circ}

into the resulting expression. This will leave you with a quadratic equation in terms of

∣ b ∣

. Solve for

∣ b ∣

and remember to reject any negative roots.

Visualized Solution

Geometry of the Vectors

Let's visualize the two vectors $a$ and $b$ .
They originate from a common point.
The angle between them is given as $6 0^{\circ}$ .

Magnitude of Vector $a$

Given: $\frac{1}{8} a$ is a unit vector.
By definition of a unit vector: $∣ \frac{1}{8} a ∣ = 1$
This implies: $\frac{1}{8} ∣ a ∣ = 1$
Therefore, the magnitude of vector $a$ is: $∣ a ∣ = 8$

The Magnitude Equation

Given equation: $∣2 a + 3 b ∣ = ∣3 a + b ∣$
To eliminate the modulus, we square both sides:
$(∣2 a + 3 b ∣)^{2} = (∣3 a + b ∣)^{2}$

Expanding the Left Hand Side

Expanding LHS: $∣2 a + 3 b ∣^{2}$
Using the identity $∣ u + v ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2} + 2 u \cdot v$ :
$= 4∣ a ∣^{2} + 9∣ b ∣^{2} + 2 (2 a \cdot 3 b)$
$= 4∣ a ∣^{2} + 9∣ b ∣^{2} + 12 a \cdot b$

Expanding the Right Hand Side

Expanding RHS: $∣3 a + b ∣^{2}$
$= 9∣ a ∣^{2} + ∣ b ∣^{2} + 2 (3 a \cdot b)$
$= 9∣ a ∣^{2} + ∣ b ∣^{2} + 6 a \cdot b$

Equating and Rearranging

$4∣ a ∣^{2} + 9∣ b ∣^{2} + 12 a \cdot b = 9∣ a ∣^{2} + ∣ b ∣^{2} + 6 a \cdot b$
Grouping like terms:
$(9 - 1) ∣ b ∣^{2} + (12 - 6) a \cdot b + (4 - 9) ∣ a ∣^{2} = 0$
$8∣ b ∣^{2} + 6 a \cdot b - 5∣ a ∣^{2} = 0$

Evaluating the Dot Product

Formula: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$
Substitute $∣ a ∣ = 8$ and $θ = 6 0^{\circ}$ :
$a \cdot b = 8 \cdot ∣ b ∣ \cdot cos 6 0^{\circ}$
Since $cos 6 0^{\circ} = \frac{1}{2}$ :
$a \cdot b = 8 \cdot ∣ b ∣ \cdot \frac{1}{2} = 4∣ b ∣$

Forming the Quadratic Equation

Substitute $a \cdot b = 4∣ b ∣$ and $∣ a ∣ = 8$ :
$8∣ b ∣^{2} + 6 (4∣ b ∣) - 5 (8^{2}) = 0$
$8∣ b ∣^{2} + 24∣ b ∣ - 5 (64) = 0$
$8∣ b ∣^{2} + 24∣ b ∣ - 320 = 0$

Simplifying the Quadratic

Divide the entire equation by $8$ :
$\frac{8∣ b ∣ ^{2}}{8} + \frac{24∣ b ∣}{8} - \frac{320}{8} = 0$
$∣ b ∣^{2} + 3∣ b ∣ - 40 = 0$
Factorizing: $(∣ b ∣ + 8) (∣ b ∣ - 5) = 0$

Final Solution for $∣ b ∣$

Possible values: $∣ b ∣ = - 8$ or $∣ b ∣ = 5$
Since magnitude represents length, it must be positive: $∣ b ∣ > 0$
Therefore, we reject $- 8$ .
Final Answer: $∣ b ∣ = 5$

00:00 / 00:00

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

We are given two vectors,

a

and

b

, originating from a common point with an angle of

θ = 6 0^{\circ}

between them.

We are provided with the condition that

\frac{1}{8} a

is a unit vector. By definition, the magnitude of a unit vector is

1

∣ \frac{1}{8} a ∣ = 1

Since

\frac{1}{8}

is a positive scalar, we can extract it from the modulus:

\frac{1}{8} ∣ a ∣ = 1 \Rightarrow ∣ a ∣ = 8

The Squaring Strategy

We are tasked with solving the equation

∣2 a + 3 b ∣ = ∣3 a + b ∣

. To eliminate the modulus bars, we square both sides of the equation.

(∣2 a + 3 b ∣)^{2} = (∣3 a + b ∣)^{2}

Squaring is a powerful technique because it allows us to utilize the property that the square of a magnitude is equivalent to the dot product of the vector with itself, specifically

∣ u ∣^{2} = u \cdot u

The Algebraic Expansion

Using the identity

∣ u + v ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2} + 2 (u \cdot v)

, we expand both sides of the equation:

4∣ a ∣^{2} + 9∣ b ∣^{2} + 12 (a \cdot b) = 9∣ a ∣^{2} + ∣ b ∣^{2} + 6 (a \cdot b)

By rearranging the terms to one side, we simplify the expression:

8∣ b ∣^{2} + 6 (a \cdot b) - 5∣ a ∣^{2} = 0

The Final Bridge

The Dot Product

We resolve the dot product using the definition

a \cdot b = ∣ a ∣∣ b ∣ cos θ

. Given

∣ a ∣ = 8

and

θ = 6 0^{\circ}

a \cdot b = 8 \cdot ∣ b ∣ \cdot cos (6 0^{\circ}) = 8 \cdot ∣ b ∣ \cdot \frac{1}{2} = 4∣ b ∣

Substituting this value back into our simplified equation:

8∣ b ∣^{2} + 6 (4∣ b ∣) - 5 (8^{2}) = 0

8∣ b ∣^{2} + 24∣ b ∣ - 320 = 0

Dividing the entire equation by

8

, we obtain the quadratic equation:

∣ b ∣^{2} + 3∣ b ∣ - 40 = 0

Factoring the quadratic yields:

(∣ b ∣ + 8) (∣ b ∣ - 5) = 0

Since the magnitude of a vector must be non-negative, we discard the negative root. Thus, the final result is:

$∣ b ∣ = 5$

Common Traps & FAQs

Wait, why do we square both sides? Can't we just solve it using the triangle law of vector addition?

That is a fantastic question! While the triangle law is geometrically beautiful, it becomes incredibly tedious when you have coefficients like 2, 3, and 1 attached to the vectors. Squaring both sides is the 'JEE-standard' algebraic shortcut. It allows us to convert the modulus (which is hard to manipulate) into dot products (which are easy to manipulate). It is the most efficient path to the solution.

Why do we reject the negative root when solving the quadratic equation for

∣ b ∣

Remember,

∣ b ∣

represents the magnitude of a vector. By definition, magnitude is the length of a vector, and length is a scalar quantity that measures distance. Distance, in our physical world, cannot be negative. Even if the algebra gives you a negative result, it is an 'extraneous solution' that doesn't fit the physical reality of the problem.

I got confused with the cross terms. Why is it

2 (u \cdot v)

and not just

2 u v

Ah, the classic trap! Remember that vectors do not multiply like scalar numbers. When you expand

(2 a + 3 b) \cdot (2 a + 3 b)

, you are performing a dot product. The distributive property gives you

4 (a \cdot a) + 6 (a \cdot b) + 6 (b \cdot a) + 9 (b \cdot b)

. Since the dot product is commutative,

6 (a \cdot b) + 6 (b \cdot a)

becomes

12 (a \cdot b)

. Always treat the dot product as the 'multiplication' operator for vectors.

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :

Select Answer:

Visualized Solution

Geometry of the Vectors

Magnitude of Vector $a$

The Magnitude Equation

Expanding the Left Hand Side

Expanding the Right Hand Side

Equating and Rearranging

Evaluating the Dot Product

Forming the Quadratic Equation

Simplifying the Quadratic

Final Solution for $∣ b ∣$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Squaring Strategy

The Algebraic Expansion

The Final Bridge

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Visualized Solution

Geometry of the Vectors

Magnitude of Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Magnitude Equation

Expanding the Left Hand Side

Expanding the Right Hand Side

Equating and Rearranging

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Forming the Quadratic Equation

Simplifying the Quadratic

Final Solution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b∣

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Squaring Strategy

The Algebraic Expansion

The Final Bridge

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