Let a = i + 2j + k, b = i - j + k and c = i + j - k. A vector in the plane of a and b whose projecti

Q: Wait, why do we use $\vec{u} = \vec{a} + \lambda \vec{b}$? Can't we use $\vec{u} = x\vec{a} + y\vec{b}$?

That is a brilliant question! You absolutely can use $\vec{u} = x\vec{a} + y\vec{b}$. In fact, that is the most general form for any vector in the plane. However, in many JEE problems, we can simplify the algebra by assuming one coefficient is 1 (unless the vector is parallel to $\vec{b}$), or by using a single parameter $\lambda$ to represent the ratio of the two scalars. It reduces the number of variables and makes the calculation much cleaner without losing any generality.

Q: Why did we use the absolute value $|2 - \lambda| = 1$ instead of just $2 - \lambda = 1$?

Great catch! The scalar projection $\frac{\vec{u} \cdot \vec{c}}{|\vec{c}|}$ is a signed value. If the problem asks for a projection of a specific value, it usually implies the signed value. However, if the projection is given as a magnitude or if the context allows for both directions, we must check both cases. In this specific problem, checking both cases ensures we find the vector that actually appears in the options, as the projection could be in the direction of $\vec{c}$ or opposite to it.

Q: Is there a geometric way to solve this without all this algebra?

Geometry is the soul of physics, but algebra is the language of precision. You could visualize the intersection of the plane and the locus of points having a constant projection on $\vec{c}$ (which is a plane perpendicular to $\vec{c}$), but solving the system of equations is the most robust and error-proof method for JEE. It minimizes the risk of missing a solution or misinterpreting a geometric orientation.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$. A vector in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$, is

Options:

(A) $4\hat{i} - \hat{j} + 4\hat{k}$

(B) $3\hat{i} + \hat{j} - 3\hat{k}$

(C) $2\hat{i} + \hat{j} - 2\hat{k}$

(D) $4\hat{i} + \hat{j} - 4\hat{k}$

Sigma Percentile · Accepted Answer

### The Geometry of Planes and Projections: A Journey into Vector Space

Welcome, fellow traveler of the JEE Advanced landscape. Today, we are not just solving a problem; we are exploring the architecture of 3D space.

Imagine you are standing in a vast, empty room. You have three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. These aren't just arrows; they are the building blocks of a coordinate system.

We are tasked with finding a mystery vector $\vec{u}$ that lives on the plane defined by $\vec{a}$ and $\vec{b}$, such that its projection onto $\vec{c}$ is exactly $\frac{1}{\sqrt{3}}$. Let's break this down.

### Phase 1: The Coplanar Canvas

When we say $\vec{u}$ is in the plane of $\vec{a}$ and $\vec{b}$, we are saying that $\vec{u}$ is a linear combination of these two vectors. Think of $\vec{a}$ and $\vec{b}$ as the basis vectors for a 2D sheet floating in our 3D room.

Any point on that sheet can be reached by walking along $\vec{a}$ and then shifting by some multiple of $\vec{b}$. Mathematically, we express this as $\vec{u} = \vec{a} + \lambda \vec{b}$.

Here, $\lambda$ is our degree of freedom—our scalar parameter. By varying $\lambda$, we can sweep across every single point on that plane. It is a powerful, elegant way to capture an infinite set of vectors with a single variable.

### Phase 2: Constructing the Mystery Vector

Now, let's get our hands dirty with the components. We are given $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Substituting these into our equation $\vec{u} = \vec{a} + \lambda \vec{b}$, we get:

$$\vec{u} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$$

Grouping the components is where the clarity emerges. We collect the $\hat{i}$, $\hat{j}$, and $\hat{k}$ terms to see the structure of $\vec{u}$ clearly:

$$\vec{u} = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + \lambda)\hat{k}$$

This is our general vector $\vec{u}$. It is a function of $\lambda$. Every value of $\lambda$ gives us a different vector on that plane.

### Phase 3: The Shadow of the Vector

What is a projection? Geometrically, it is the shadow cast by $\vec{u}$ onto the line of $\vec{c}$. If you shine a light perpendicular to $\vec{c}$, the length of the shadow is the scalar projection.

The formula is given by:

$$	ext{Projection} = \frac{\vec{u} \cdot \vec{c}}{|\vec{c}|}$$

Let's calculate the dot product $\vec{u} \cdot \vec{c}$ first. With $\vec{c} = \hat{i} + \hat{j} - \hat{k}$, we have:

$$\vec{u} \cdot \vec{c} = (1 + \lambda)(1) + (2 - \lambda)(1) + (1 + \lambda)(-1)$$

Watch the signs carefully here—this is where many students stumble. Expanding this, we get $1 + \lambda + 2 - \lambda - 1 - \lambda$.

The $\lambda$ terms and the constants interact beautifully. The $1$ and $-1$ cancel, and the $\lambda$ and $-\lambda$ cancel, leaving us with $2 - \lambda$.

Now, the magnitude of $\vec{c}$ is:

$$|\vec{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$$

### Phase 4: The Final Resolution

We are told the projection is $\frac{1}{\sqrt{3}}$. So, we set up our equation:

$$\frac{|2 - \lambda|}{\sqrt{3}} = \frac{1}{\sqrt{3}}$$

The $\sqrt{3}$ terms cancel out, leaving us with $|2 - \lambda| = 1$. This absolute value tells us there are two possible scenarios: $2 - \lambda = 1$ or $2 - \lambda = -1$.

Solving these gives $\lambda = 1$ or $\lambda = 3$. Substituting $\lambda = 3$ back into our expression for $\vec{u}$, we get:

$$\vec{u} = 4\hat{i} - \hat{j} + 4\hat{k}$$

This matches our options perfectly! You have successfully navigated the vector space, constrained the mystery vector, and solved for the unknown.

This is the essence of JEE Advanced mathematics—taking a complex, abstract problem and systematically reducing it to a clear, logical conclusion. Keep this confidence with you; you are capable of mastering any problem that comes your way.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

Select Answer:

Visualized Solution

Visualizing the Vector Space

Defining Coplanar Vector $u$

Substituting Vector Values

Grouping Components

The Scalar Projection Formula

Calculating the Dot Product

Simplifying the Dot Product

Magnitude of Vector $c$

Setting up the Projection Equation

Solving for $λ$

Finding the Final Vector

The Sigma Insight: Scalar (Dot) Product

The Geometry of Planes and Projections

Phase 1

Phase 2

Phase 3

Phase 4

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