Let a = 2i + λ1j + 3k, b = 4i + (3 - λ2)j + 6k and c = 3i + 6j + (λ3 - 1)k be three vectors such tha

Q: Wait, we have three variables ($\lambda_1, \lambda_2, \lambda_3$) but only two equations. How can we possibly find a unique solution?

That is a brilliant observation! You are absolutely right—this is an underdetermined system. In linear algebra, this means there isn't a single unique point solution, but rather a line or a plane of solutions. However, in the context of a JEE multiple-choice question, we are looking for a 'possible' value. By testing the given options, we are essentially checking which of the provided candidates lies on the solution space defined by our two equations.

Q: Why does $\vec{b} = 2\vec{a}$ allow us to equate components directly? Can't we just say the magnitudes are related?

Great question. The vector equation $\vec{b} = 2\vec{a}$ is a vector identity. Because the basis vectors $\hat{i}, \hat{j},$ and $\hat{k}$ are linearly independent, the only way for two vectors to be equal is if their corresponding components are identical. Thus, $b_x = 2a_x$, $b_y = 2a_y$, and $b_z = 2a_z$ must all hold true simultaneously.

Q: Why do we use the dot product for perpendicularity? Is there another way?

You could use the definition $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos(\theta)$, but since $\theta = 90^\circ$ and $\cos(90^\circ) = 0$, the entire right side collapses to zero. The component-wise dot product formula is simply the most efficient algebraic tool to extract the relationship between the variables without needing to calculate magnitudes or angles.

Question

Let $\vec{a} = 2\hat{i} + \lambda_1\hat{j} + 3\hat{k}$, $\vec{b} = 4\hat{i} + (3 - \lambda_2)\hat{j} + 6\hat{k}$ and $\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3 - 1)\hat{k}$ be three vectors such that $\vec{b} = 2\vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$. Then a possible value of $(\lambda_1, \lambda_2, \lambda_3)$ is :-

Options:

(A) (\frac{1}{2}, 4, -2)

(B) (-\frac{1}{2}, 4, 0)

(C) (1, 3, 1)

(D) (1, 5, 1)

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into Alignment and Orthogonality

Welcome, future engineer! Today, we are not just solving a problem; we are decoding the language of space itself.

Vector algebra is the bedrock of physics—from the trajectory of a satellite to the forces acting on a bridge. When you look at vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, do not just see a collection of numbers and symbols. See them as arrows in three-dimensional space, waiting for us to uncover their hidden relationships.

### Phase 1: The Power of Parallelism

We are given the condition $\vec{b} = 2\vec{a}$. In the world of vectors, this is a profound statement. It tells us that $\vec{b}$ is not just any vector; it is a scaled version of $\vec{a}$.

Geometrically, they are collinear—they point in the exact same direction. Algebraically, this is even more powerful. If $\vec{b} = 2\vec{a}$, then every single component of $\vec{b}$ must be exactly twice the corresponding component of $\vec{a}$.

Let us write this out clearly. We have $\vec{a} = 2\hat{i} + \lambda_1\hat{j} + 3\hat{k}$ and $\vec{b} = 4\hat{i} + (3 - \lambda_2)\hat{j} + 6\hat{k}$. When we set $\vec{b} = 2\vec{a}$, we are essentially saying:

$$4\hat{i} + (3 - \lambda_2)\hat{j} + 6\hat{k} = 2(2\hat{i} + \lambda_1\hat{j} + 3\hat{k})$$

Expanding the right side gives us $4\hat{i} + 2\lambda_1\hat{j} + 6\hat{k}$. By comparing the coefficients of $\hat{j}$, we immediately find our first bridge between variables: $3 - \lambda_2 = 2\lambda_1$.

Rearranging this, we get our first master equation:

$$2\lambda_1 + \lambda_2 = 3$$

Keep this safe; it is the key to unlocking the first part of our puzzle.

### Phase 2: The Silent Orthogonality

Now, let us turn to the second condition: $\vec{a} \perp \vec{c}$. The word 'perpendicular' should trigger an immediate reflex in your mind: the dot product must be zero.

Why? Because the dot product is defined as $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos(	heta)$. When the angle $	heta$ is $90^\circ$, the cosine term vanishes, and the entire product becomes zero.

This is the beauty of mathematics—a complex geometric relationship collapses into a simple algebraic sum. We calculate the dot product by multiplying corresponding components:

$$(2)(3) + (\lambda_1)(6) + (3)(\lambda_3 - 1) = 0$$

Let us expand this carefully. We get $6 + 6\lambda_1 + 3\lambda_3 - 3 = 0$. Combining the constants, we are left with $6\lambda_1 + 3\lambda_3 + 3 = 0$.

Notice how the number 3 is a common factor? Let us divide the entire equation by 3 to make it elegant: $2\lambda_1 + \lambda_3 + 1 = 0$. This gives us our second master equation:

$$2\lambda_1 + \lambda_3 = -1$$

### Phase 3: The Art of Verification

We now stand at the threshold of the solution. We have two equations:

1) $2\lambda_1 + \lambda_2 = 3$

2) $2\lambda_1 + \lambda_3 = -1$

We have three variables, which means we have a degree of freedom. In a competitive exam like the JEE, this is where you stop calculating and start strategizing.

We don't need to solve for every variable in terms of a parameter; we simply need to verify which of the given options fits these two constraints. By testing the options, we quickly find that $(-\frac{1}{2}, 4, 0)$ satisfies both equations perfectly.

Remember, physics and math are not about memorizing formulas; they are about understanding the constraints of the universe. You have successfully navigated the collinearity and the orthogonality of these vectors.

Take a moment to appreciate that—you have just mastered the spatial logic of the problem. Keep practicing, stay curious, and keep pushing the boundaries of your understanding!

Select Answer:

Visualized Solution

Visualizing the Vectors

The Collinearity Condition $b = 2 a$

Substituting Components

Relating $λ_{1}$ and $λ_{2}$

The Perpendicularity Condition $a ⊥ c$

Setting up the Dot Product

Simplifying the Dot Product

The Second Relation

Testing the Options

Verifying the Second Equation

The Sigma Insight: Scalar (Dot) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

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