Let a = i + j + 2k, b = 2i - 3j + k and c = i - j + k be three given vectors. Let v be a vector in t

Q: Wait, why can we just write $\vec{v} = \lambda \vec{a} + \mu \vec{b}$? Is that always true?

That is a fantastic question! Yes, it is always true for any vector $\vec{v}$ that lies in the plane spanned by two non-collinear vectors $\vec{a}$ and $\vec{b}$. By definition, any vector in that plane can be reached by scaling $\vec{a}$ and $\vec{b}$ and adding them together. Think of $\lambda$ and $\mu$ as the 'coordinates' of $\vec{v}$ within that specific plane.

Q: Does the order of vectors in the projection formula matter? Can I write $\frac{\vec{c} \cdot \vec{v}}{|\vec{c}|}$ instead?

Absolutely! The dot product is commutative, meaning $\vec{v} \cdot \vec{c} = \vec{c} \cdot \vec{v}$. So, the order in the numerator does not change the result. However, always remember that the denominator must be the magnitude of the vector you are projecting *onto*—in this case, $\vec{c}$.

Q: Why did we get a system of linear equations? Is there a faster way?

In vector geometry, we often translate geometric constraints (like coplanarity and projection) into algebraic ones. While there are sometimes geometric shortcuts, setting up a system of equations is the most robust and reliable method for JEE Advanced. It minimizes the chance of missing a subtle detail and ensures you arrive at the correct values for $\lambda$ and $\mu$ systematically.

Question

Let $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}, \vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ be three given vectors. Let $\vec{v}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\vec{c}$ is $\frac{2}{\sqrt{3}}$. If $\vec{v} \cdot \hat{j} = 7$, then $\vec{v} \cdot (\hat{i} + \hat{k})$ is equal to :

Options:

(A) 6

(B) 7

(C) 8

(D) 9

Sigma Percentile · Accepted Answer

### The Geometry of the Plane

Imagine you are standing in a vast, empty three-dimensional space. You have two vectors, $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$, anchored at the origin. These two vectors define a flat, infinite sheet—a plane—stretching out in every direction.

We are looking for a mysterious vector $\vec{v}$ that is trapped on this very sheet. Because $\vec{v}$ is confined to this plane, it is not free to roam the entire 3D space; it is bound by the geometry of $\vec{a}$ and $\vec{b}$.

Mathematically, this means $\vec{v}$ must be a linear combination of these two vectors. We express this as $\vec{v} = \lambda \vec{a} + \mu \vec{b}$, where $\lambda$ and $\mu$ are the scalar keys that unlock the exact position of $\vec{v}$.

### Unlocking the Components

To find $\vec{v}$, we first need to see what it looks like in terms of its components. By substituting the known values of $\vec{a}$ and $\vec{b}$ into our linear combination, we get:

$$\vec{v} = \lambda(\hat{i} + \hat{j} + 2\hat{k}) + \mu(2\hat{i} - 3\hat{j} + \hat{k})$$

Now, we group the terms by their unit vectors:

$$\vec{v} = (\lambda + 2\mu)\hat{i} + (\lambda - 3\mu)\hat{j} + (2\lambda + \mu)\hat{k}$$

This is our master equation. It tells us exactly how the components of $\vec{v}$ depend on our unknown scalars $\lambda$ and $\mu$.

### The First Constraint: The Dot Product

The problem gives us a gift: $\vec{v} \cdot \hat{j} = 7$. When you take the dot product of any vector with $\hat{j}$, you are essentially filtering out everything except the $y$-component.

Looking at our master equation, the coefficient of $\hat{j}$ is $(\lambda - 3\mu)$. Therefore, we immediately have our first equation:

$$\lambda - 3\mu = 7$$

### The Shadow of the Vector

Next, we consider the vector $\vec{c} = \hat{i} - \hat{j} + \hat{k}$. We are told that the projection of $\vec{v}$ onto $\vec{c}$ is $\frac{2}{\sqrt{3}}$. The formula for this projection is $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}$.

First, we calculate the magnitude of $\vec{c}$:

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

Next, we compute the dot product $\vec{v} \cdot \vec{c}$. By multiplying the corresponding components, we get:

$$\vec{v} \cdot \vec{c} = (\lambda + 2\mu)(1) + (\lambda - 3\mu)(-1) + (2\lambda + \mu)(1) = 2\lambda + 6\mu$$

Setting this into our projection formula, we have:

$$\frac{2\lambda + 6\mu}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$

The $\sqrt{3}$ cancels out, and dividing by $2$ gives us our second equation:

$$\lambda + 3\mu = 1$$

### The Final Resolution

We now have a simple system: $\lambda - 3\mu = 7$ and $\lambda + 3\mu = 1$. Adding these equations is a moment of pure mathematical elegance—the $\mu$ terms vanish, leaving $2\lambda = 8$, so $\lambda = 4$.

Substituting this back, we find $4 + 3\mu = 1$, which means $3\mu = -3$, so $\mu = -1$. With $\lambda$ and $\mu$ in hand, we find the vector:

$$\vec{v} = 4(\hat{i} + \hat{j} + 2\hat{k}) - 1(2\hat{i} - 3\hat{j} + \hat{k}) = 2\hat{i} + 7\hat{j} + 7\hat{k}$$

Finally, the dot product $\vec{v} \cdot (\hat{i} + \hat{k})$ is:

$$(2)(1) + (7)(0) + (7)(1) = 2 + 7 = 9$$

We have navigated the plane, resolved the constraints, and arrived at the truth. The final answer is **9**.

Select Answer:

Visualized Solution

Visualizing the Plane of $a$ and $b$

Expressing $v$ in Component Form

Applying the First Constraint: $v \cdot \hat{j} = 7$

Understanding Projection on $c$

Calculating Magnitude of $c$

Calculating the Dot Product $v \cdot c$

Forming the Second Equation

Solving for $λ$ and $μ$

Finding the Exact Vector $v$

Final Calculation: $v \cdot (\hat{i} + \hat{k})$

The Sigma Insight: Scalar (Dot) Product

The Geometry of the Plane

Unlocking the Components

The First Constraint

The Shadow of the Vector

The Final Resolution

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Visualized Solution

Visualizing the Plane of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v in Component Form

Applying the First Constraint: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v⋅j^​=7

Understanding Projection on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

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

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