Let a = i - j + 2k and b be a vector such that a × b = 2i - k and a · b = 3. Then the projection of

Q: Wait, why don't we just solve for the components of $\vec{b}$? Can't we assume $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve the system of equations?

That is a very natural instinct! However, in JEE Advanced, we must be strategic. If you try to solve for the components of $\vec{b}$, you will find that the system is underdetermined. The cross product $\vec{a} \times \vec{b}$ only provides two independent constraints, not three. This means there is an infinite family of vectors $\vec{b}$ that satisfy the given conditions. The beauty of this problem is that the projection is invariant—it doesn't depend on the specific $\vec{b}$, only on the scalar quantities provided. Always look for the scalar path first; it is the path of least resistance and highest elegance.

Q: What is Lagrange's Identity and why is it useful here?

Lagrange's Identity, $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$, is the bridge between the 'perpendicular' world (cross product) and the 'parallel' world (dot product). In this problem, we were given the 'perpendicular' info and the 'parallel' info, but we were missing the 'size' (magnitude) of $\vec{b}$. This identity is the only tool that connects these three distinct geometric properties. Whenever you see both a cross product and a dot product in the same problem, let Lagrange's Identity be the first thing that comes to your mind.

Q: Is the projection always positive? What if the dot product is negative?

Excellent question! The projection formula $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$ is a scalar, not a magnitude. If the angle between $\vec{u}$ and $\vec{v}$ is obtuse (greater than $90^\circ$), the dot product $\vec{u} \cdot \vec{v}$ will be negative, and thus the projection will be negative. This simply means the shadow falls in the opposite direction of the vector $\vec{v}$. In this specific problem, our result is positive, but always keep the sign in mind—it carries physical meaning!

Question

Let $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} 	imes \vec{b} = 2\hat{i} - \hat{k}$ and $\vec{a} \cdot \vec{b} = 3$. Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b}$ is :-

Options:

(A) $\frac{2}{\sqrt{21}}$

(B) $2\sqrt{\frac{3}{7}}$

(C) $\frac{2}{3}\sqrt{\frac{7}{3}}$

(D) $\frac{2}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Shadows: A Journey Through Vector Space

Welcome, fellow traveler in the realm of physics and mathematics. Today, we are not just solving a problem; we are embarking on a journey to understand the hidden relationships between vectors.

When you look at a problem involving $\vec{a}$ and $\vec{b}$, it is easy to get bogged down in the components—the $i, j, k$ coordinates. But I want you to pause. Before you start writing down equations, visualize the scene.

We have two vectors, $\vec{a}$ and $\vec{b}$. We are asked to find the projection of $\vec{b}$ onto the vector $(\vec{a} - \vec{b})$. The projection is simply the shadow that $\vec{b}$ casts onto $(\vec{a} - \vec{b})$. It is a measure of how much of $\vec{b}$ 'aligns' with that difference vector.

### Phase 1: The Algebraic Blueprint

Let us start with the definition of projection. The projection of a vector $\vec{u}$ onto a vector $\vec{v}$ is defined as $P = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$. In our case, $\vec{u} = \vec{b}$ and $\vec{v} = \vec{a} - \vec{b}$.

So, our goal is to calculate:

$$P = \frac{\vec{b} \cdot (\vec{a} - \vec{b})}{|\vec{a} - \vec{b}|}$$

Now, let us expand the numerator. Using the distributive property of the dot product, we get:

$$\vec{b} \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{b} - |\vec{b}|^2$$

We already know $\vec{a} \cdot \vec{b} = 3$. But what about $|\vec{b}|^2$? This is our first missing puzzle piece. We cannot solve the problem without it. This is where the detective work begins.

### Phase 2: The Detective Work with Lagrange's Identity

We are given $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{a} 	imes \vec{b} = 2\hat{i} - \hat{k}$. We have the dot product and the cross product. We need the magnitude of $\vec{b}$.

This is the moment to invoke the powerful **Lagrange's Identity**:

$$|\vec{a} 	imes \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$

This identity is the bridge between the perpendicular and parallel components of the vectors. Let us calculate the pieces we need. First, the magnitude of $\vec{a}$ squared:

$$|\vec{a}|^2 = (1)^2 + (-1)^2 + (2)^2 = 1 + 1 + 4 = 6$$

Next, the magnitude of the cross product squared:

$$|\vec{a} 	imes \vec{b}|^2 = (2)^2 + (0)^2 + (-1)^2 = 4 + 0 + 1 = 5$$

Now, substitute these into Lagrange's Identity:

$$5 + (3)^2 = 6 \cdot |\vec{b}|^2$$
$$5 + 9 = 6 |\vec{b}|^2$$
$$14 = 6 |\vec{b}|^2 \implies |\vec{b}|^2 = \frac{14}{6} = \frac{7}{3}$$

We have found it! The magnitude of $\vec{b}$ squared is $\frac{7}{3}$.

### Phase 3: The Final Assembly

Now that we have $|\vec{b}|^2$, we can find the denominator of our projection formula, which is $|\vec{a} - \vec{b}|$. We use the expansion of the magnitude of a difference:

$$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$$

Substituting our known values:

$$|\vec{a} - \vec{b}|^2 = 6 + \frac{7}{3} - 2(3)$$
$$|\vec{a} - \vec{b}|^2 = 6 + \frac{7}{3} - 6 = \frac{7}{3}$$

So, $|\vec{a} - \vec{b}| = \sqrt{\frac{7}{3}}$. Finally, let us return to our numerator: $\vec{a} \cdot \vec{b} - |\vec{b}|^2$.

Substituting the values:

$$	ext{Numerator} = 3 - \frac{7}{3} = \frac{9 - 7}{3} = \frac{2}{3}$$

Putting it all together into our projection formula:

$$P = \frac{\frac{2}{3}}{\sqrt{\frac{7}{3}}} = \frac{2}{3} \cdot \frac{\sqrt{3}}{\sqrt{7}}$$

We can simplify $\frac{\sqrt{3}}{3}$ as $\frac{1}{\sqrt{3}}$. Thus:

$$P = \frac{2}{\sqrt{3} \cdot \sqrt{7}} = \frac{2}{\sqrt{21}}$$

And there we have it. The shadow cast by $\vec{b}$ onto the difference vector is exactly **$\frac{2}{\sqrt{21}}$**. Notice how we never had to find the individual components of $\vec{b}$? We treated the vectors as geometric objects, using their properties rather than their coordinates.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} - \hat{j} + 2 \hat{k}$ and $b$ be a vector such that $a \times b = 2 \hat{i} - \hat{k}$ and $a \cdot b = 3$ . Then the projection of $b$ on the vector $a - b$ is :-

Select Answer:

Visualized Solution

Visualizing the Goal

Expanding the Projection Formula

Calculating $∣ a ∣^{2}$

Calculating $∣ a \times b ∣^{2}$

Lagrange's Identity

Solving for $∣ b ∣^{2}$

Finding $∣ a - b ∣^{2}$

Calculating $∣ a - b ∣^{2}$

Calculating the Numerator

Final Substitution

The Sigma Insight: Scalar (Dot) Product

The Geometry of Shadows

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualizing the Goal

Expanding the Projection Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a∣2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a×b∣2