If $\vec{x}$ and $\vec{y}$ be two non-zero vectors such that $|\vec{x} + \vec{y}| = |\vec{x}|$ and $2\vec{x} + \lambda\vec{y}$ is perpendicular to $\vec{y}$, then the value of $\lambda$ is

### Analyzing the Setup We are given two non-zero vectors, $\vec{x}$ and $\vec{y}$. We must translate the provided geometric constraints into algebraic expressions to solve for the scalar $\lambda$. ### Phase 1: Breaking the Modulus The first condition is $|\vec{x} + \vec{y}| = |\vec{x}|$. To eliminate the modulus, we square both sides: $$|\vec{x} + \vec{y}|^2 = |\vec{x}|^2$$ Using the identity $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$, we expand the left side: $$|\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y}) = |\vec{x}|^2$$ Subtracting $|\vec{x}|^2$ from both sides yields our first vital equation: $$|\vec{y}|^2 + 2(\vec{x} \cdot \vec{y}) = 0 \quad \text{--- (Equation 1)}$$ ### Phase 2: The Power of Perpendicularity The second condition states that the vector $(2\vec{x} + \lambda\vec{y})$ is perpendicular to $\vec{y}$. In vector algebra, this implies their dot product must be zero: $$(2\vec{x} + \lambda\vec{y}) \cdot \vec{y} = 0$$ Distributing the dot product across the sum, we obtain: $$2(\vec{x} \cdot \vec{y}) + \lambda(\vec{y} \cdot \vec{y}) = 0$$ Since $\vec{y} \cdot \vec{y} = |\vec{y}|^2$, this simplifies to: $$2(\vec{x} \cdot \vec{y}) + \lambda|\vec{y}|^2 = 0 \quad \text{--- (Equation 2)}$$ ### Phase 3: The Synthesis We now compare Equation (1) and Equation (2). From Equation (1), we have: $$2(\vec{x} \cdot \vec{y}) = -|\vec{y}|^2$$ Substituting this into Equation (2), we get: $$-|\vec{y}|^2 + \lambda|\vec{y}|^2 = 0$$ Factoring out $|\vec{y}|^2$, we have: $$(\lambda - 1)|\vec{y}|^2 = 0$$ Since $\vec{y}$ is a non-zero vector, $|\vec{y}|^2 \neq 0$. Therefore, we can safely divide by $|\vec{y}|^2$ to arrive at the final result: $$\mathbf{\lambda = 1}$$

JEE Main 2020 - 6 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: If $x$ and $y$ be two non-zero vectors such that $∣ x + y ∣ = ∣ x ∣$ and $2 x + λ y$ is perpendicular to $y$ , then the value of $λ$ is

Q: Is it possible that $\vec{y} = 0$? The problem says they are non-zero vectors, but does that matter?

It matters immensely! If $\vec{y}$ were the zero vector, the second condition $(2\vec{x} + \lambda\vec{y}) \perp \vec{y}$ would be undefined or trivial. Because the problem explicitly states they are non-zero, we are mathematically justified in dividing by $|\vec{y}|^2$ or comparing coefficients, knowing that $|\vec{y}|^2 \neq 0$.

Enter Numerical Value:

Visualized Solution

$Visualizing x and y$

Let $x$ and $y$ be two non-zero vectors.
We represent them originating from the same point.

$The First Condition$

Given: $∣ x + y ∣ = ∣ x ∣$
The magnitude of the sum vector equals the magnitude of $x$ .

$Squaring the Magnitudes$

To eliminate the modulus, we square both sides:
$∣ x + y ∣^{2} = ∣ x ∣^{2}$

$Expanding the Dot Product$

Using the property $∣ a ∣^{2} = a \cdot a$ :
$∣ x ∣^{2} + ∣ y ∣^{2} + 2 x \cdot y = ∣ x ∣^{2}$

$Simplifying to Equation 1$

Subtracting $∣ x ∣^{2}$ from both sides:
$∣ y ∣^{2} + 2 x \cdot y = 0$
Let's call this Equation (1).

$The Second Condition$

Given: $(2 x + λ y) ⊥ y$
A new vector is formed which is perpendicular to $y$ .

$Applying Perpendicularity$

If two vectors are perpendicular, their dot product is zero.
$(2 x + λ y) \cdot y = 0$

$Expanding the Second Dot Product$

Distributing the dot product:
$2 x \cdot y + λ (y \cdot y) = 0$
$2 x \cdot y + λ ∣ y ∣^{2} = 0$
Let's call this Equation (2).

$Comparing Equations$

Eq (1): $2 x \cdot y + 1 \cdot ∣ y ∣^{2} = 0$
Eq (2): $2 x \cdot y + λ \cdot ∣ y ∣^{2} = 0$
Comparing the coefficients of $∣ y ∣^{2}$ :
$λ = 1$

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The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

We are given two non-zero vectors,

x

and

y

. We must translate the provided geometric constraints into algebraic expressions to solve for the scalar

λ

Phase 1

Breaking the Modulus

The first condition is

∣ x + y ∣ = ∣ x ∣

. To eliminate the modulus, we square both sides:

∣ x + y ∣^{2} = ∣ x ∣^{2}

Using the identity

∣ a ∣^{2} = a \cdot a

, we expand the left side:

∣ x ∣^{2} + ∣ y ∣^{2} + 2 (x \cdot y) = ∣ x ∣^{2}

Subtracting

∣ x ∣^{2}

from both sides yields our first vital equation:

∣ y ∣^{2} + 2 (x \cdot y) = 0 — (Equation 1)

Phase 2

The Power of Perpendicularity

The second condition states that the vector

(2 x + λ y)

is perpendicular to

y

. In vector algebra, this implies their dot product must be zero:

(2 x + λ y) \cdot y = 0

Distributing the dot product across the sum, we obtain:

2 (x \cdot y) + λ (y \cdot y) = 0

Since

y \cdot y = ∣ y ∣^{2}

, this simplifies to:

2 (x \cdot y) + λ ∣ y ∣^{2} = 0 — (Equation 2)

Phase 3

The Synthesis

We now compare Equation (1) and Equation (2). From Equation (1), we have:

2 (x \cdot y) = - ∣ y ∣^{2}

Substituting this into Equation (2), we get:

- ∣ y ∣^{2} + λ ∣ y ∣^{2} = 0

Factoring out

∣ y ∣^{2}

, we have:

(λ - 1) ∣ y ∣^{2} = 0

Since

y

is a non-zero vector, $|\vec{y}|^2 eq 0$. Therefore, we can safely divide by

∣ y ∣^{2}

to arrive at the final result:

λ = 1

Common Traps & FAQs

Wait, why do we square the magnitude equation? Can't we just take the square root of both sides?

Great question! Taking the square root of a magnitude equation like

∣ x + y ∣ = ∣ x ∣

doesn't actually help us simplify the vector components. Squaring is the standard 'bridge' in vector algebra because it transforms the magnitude into a dot product (

∣ a ∣^{2} = a \cdot a

), which allows us to use the distributive property and solve for unknown coefficients.

Is it possible that

y = 0

? The problem says they are non-zero vectors, but does that matter?

It matters immensely! If

y

were the zero vector, the second condition

(2 x + λ y) ⊥ y

would be undefined or trivial. Because the problem explicitly states they are non-zero, we are mathematically justified in dividing by

∣ y ∣^{2}

or comparing coefficients, knowing that

∣ y ∣^{2} \neq = 0

Why can't we just use the Law of Cosines here?

You absolutely could! The Law of Cosines is essentially the geometric version of the dot product expansion. However, in JEE Advanced, using the dot product is often faster and less prone to sign errors than managing trigonometric functions. Stick to the dot product for a more robust algebraic approach.

Animated Solution for Mathematics - Vector Algebra: If $x$ and $y$ be two non-zero vectors such that $∣ x + y ∣ = ∣ x ∣$ and $2 x + λ y$ is perpendicular to $y$ , then the value of $λ$ is

Enter Numerical Value:

Visualized Solution

$Visualizing x and y$

$The First Condition$

$Squaring the Magnitudes$

$Expanding the Dot Product$

$Simplifying to Equation 1$

$The Second Condition$

$Applying Perpendicularity$

$Expanding the Second Dot Product$

$Comparing Equations$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

If the projection of the vector $\hat{i} + 2 \hat{j} + \hat{k}$ on the sum of the two vectors $2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $- λ \hat{i} + 2 \hat{j} + 3 \hat{k}$ is 1, then $λ$ is equal to .

Let $a = 2 \hat{i} + λ_{1} \hat{j} + 3 \hat{k}$ , $b = 4 \hat{i} + (3 - λ_{2}) \hat{j} + 6 \hat{k}$ and $c = 3 \hat{i} + 6 \hat{j} + (λ_{3} - 1) \hat{k}$ be three vectors such that $b = 2 a$ and $a$ is perpendicular to $c$ . Then a possible value of $(λ_{1}, λ_{2}, λ_{3})$ is :-

Let $a$ and $b$ be two unit vectors such that the angle between them is $\frac{π}{3}$ If $λ a + 2 b$ and $3 a - λ b$ are perpendicular to each other, then the number of values of $λ$ in $[- 1, 3]$ is:

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals

Let $a$ and $b$ be the vectors of the same magnitude such that $\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1$ . Then $\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}}$ is :

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $c = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $r$ be a unit vector along $b + c$ . If $r \cdot a = 3$ , then $3 λ$ is equal to:

If $a$ and $b$ are two unit vectors such that $a + 2 b$ and $5 a - 4 b$ are perpendicular to each other then the angle between $a$ and $b$ is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing x and y​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The First Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Squaring the Magnitudes

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Dot Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying to Equation 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Second Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Perpendicularity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Second Dot Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Equations

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the projection of the vector i^+2j^​+k^ on the sum of the two vectors 2i^+4j^​−5k^ and −λi^+2j^​+3k^ is 1, then λ is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a and b be the vectors of the same magnitude such that ∣a+b∣−∣a−b∣∣a+b∣+∣a−b∣​=2​+1. Then ∣a∣2∣a+b∣2​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​+3k^,b=2i^+3j^​−5k^ and c=3i^−j^​+λk^ be three vectors. Let r be a unit vector along b+c. If r⋅a=3, then 3λ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a and b are two unit vectors such that a+2b and 5a−4b are perpendicular to each other then the angle between a and b is

$Visualizing x and y$

$The First Condition$

$Squaring the Magnitudes$

$Expanding the Dot Product$

$Simplifying to Equation 1$

$The Second Condition$

$Applying Perpendicularity$

$Expanding the Second Dot Product$

$Comparing Equations$

If the projection of the vector $\hat{i} + 2 \hat{j} + \hat{k}$ on the sum of the two vectors $2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $- λ \hat{i} + 2 \hat{j} + 3 \hat{k}$ is 1, then $λ$ is equal to .

Let $a$ and $b$ be the vectors of the same magnitude such that $\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1$ . Then $\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}}$ is :

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $c = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $r$ be a unit vector along $b + c$ . If $r \cdot a = 3$ , then $3 λ$ is equal to:

If $a$ and $b$ are two unit vectors such that $a + 2 b$ and $5 a - 4 b$ are perpendicular to each other then the angle between $a$ and $b$ is