Let a = i + j + k, b = i - j + k and c = i - j - k be three vectors. A vector v in the plane of a an

Q: Wait, why do we have two unknowns ($\lambda$ and $\mu$) but only one equation? Does this mean the problem is unsolvable?

That is a brilliant observation! You are correct that one equation with two variables usually implies an infinite number of solutions. However, in the context of a multiple-choice question, we are looking for a specific vector that satisfies the condition. The algebra reveals that the $\hat{j}$ component of our vector $\vec{v}$ is fixed, which allows us to identify the correct option even without solving for $\lambda$ and $\mu$ individually.

Q: Why is the projection formula $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}$? I always get confused between the scalar projection and the vector projection.

It is a common point of confusion! Remember, the 'scalar projection' is just a length—a number. To get a length, we take the dot product (which gives a scalar) and divide by the magnitude of the reference vector to 'normalize' it. If you were asked for the 'vector projection,' you would multiply that scalar result by the unit vector $\hat{c}$.

Q: How did we know to look at the $\hat{j}$ component to eliminate options? Is that a standard trick?

It is not just a trick; it is mathematical efficiency! When you simplify the expression for $\vec{v}$, you notice that the $\hat{j}$ component simplifies to $\lambda - \mu$. Since our projection equation gave us $\mu - \lambda = 1$, we know $\lambda - \mu = -1$. This means the $\hat{j}$ component is *always* $-1$ for any valid vector $\vec{v}$ in this plane. Checking the options for this constant value is the fastest way to the finish line.

Question

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

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### The Geometry of Planes and Projections

Welcome, fellow traveler in the world of JEE mathematics. Today, we are not just solving a vector problem; we are exploring the architecture of three-dimensional space.

We are given three vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + \hat{k}$, and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$. Our mission is to find a vector $\vec{v}$ that lies in the plane defined by $\vec{a}$ and $\vec{b}$, while satisfying a specific projection condition onto $\vec{c}$.

### Phase 1: The Coplanarity Constraint

Imagine you are standing on a flat surface—a plane—defined by two vectors, $\vec{a}$ and $\vec{b}$. Any vector $\vec{v}$ that lies on this surface can be reached by walking some distance along $\vec{a}$ and some distance along $\vec{b}$.

Mathematically, this is the definition of a linear combination:
$$\vec{v} = \lambda \vec{a} + \mu \vec{b}$$

Let us substitute the components of $\vec{a}$ and $\vec{b}$ into this definition. We have $\vec{v} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$.

By grouping the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$, we obtain:
$$\vec{v} = (\lambda + \mu)\hat{i} + (\lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$$

### Phase 2: The Projection Bridge

The problem gives us a constraint: the projection of $\vec{v}$ onto $\vec{c}$ is $\frac{1}{\sqrt{3}}$. Recall the definition of scalar projection:
$$	ext{Proj}_{\vec{c}} \vec{v} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}$$

First, let us calculate the magnitude of $\vec{c}$. Since $\vec{c} = \hat{i} - \hat{j} - \hat{k}$, its magnitude is:
$$|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$$

Next, we calculate the dot product $\vec{v} \cdot \vec{c}$. Using our grouped components for $\vec{v}$ and the components of $\vec{c}$, we calculate:
$$\vec{v} \cdot \vec{c} = (\lambda + \mu)(1) + (\lambda - \mu)(-1) + (\lambda + \mu)(-1)$$

Expanding this, we get $\lambda + \mu - \lambda + \mu - \lambda - \mu$. The terms simplify elegantly to:
$$\vec{v} \cdot \vec{c} = \mu - \lambda$$

### Phase 3: The Elegant Collapse

Now, we equate our results to the given projection value:
$$\frac{\mu - \lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}}$$

This simplifies beautifully to the constraint:
$$\mu - \lambda = 1$$

Here is where the JEE master's intuition kicks in. We do not need to solve for $\lambda$ and $\mu$ individually. Look back at our expression for $\vec{v}$: the $\hat{j}$ component is $\lambda - \mu$.

Since we know $\mu - \lambda = 1$, it follows that:
$$\lambda - \mu = -1$$

This means that for **any** vector $\vec{v}$ satisfying the condition, the $\hat{j}$ component must be $-1$. We have navigated the algebra, respected the geometry, and identified the invariant property of the vector.

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

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Visualized Solution

Visualizing the Vectors

Vector $v$ in the Plane of $a$ and $b$

Expressing $v$ in Component Form

Grouping the Components

The Projection Formula

Setting up the Dot Product

Evaluating $v \cdot c$

Calculating the Magnitude $∣ c ∣$

Solving for $λ$ and $μ$

The Final Vector $v$

The Sigma Insight: Scalar (Dot) Product

The Geometry of Planes and Projections

Phase 1

Phase 2

Phase 3

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