Let î, ĵ and k̂ be the unit vectors along the three positive coordinate axes. Let a⃗ = 3î + ĵ -

Q: Wait, why is that matrix equal to the cross product? I don't see the connection.

That is a classic JEE trap! If you expand the matrix multiplication $\begin{pmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ b_2 \\ b_3 \end{pmatrix}$, you get components that exactly match the expansion of the determinant for $\vec{c} \times \vec{b}$. It is a standard representation of the cross product operator. Recognizing this saves you minutes of tedious calculation!

Q: Why did we take the dot product with $\vec{b}$ and $\vec{c}$? Is there a rule for this?

It is not a rule, but a strategy! When you have an equation involving cross products, dotting with the vectors involved is the most efficient way to 'kill' terms. Since $\vec{c} \times \vec{b}$ is perpendicular to both $\vec{c}$ and $\vec{b}$, dotting with either vector makes that term vanish, leaving you with a much simpler scalar equation. Always look for ways to simplify the vector equation before jumping into components.

Q: Why is $|\vec{b}| > \sqrt{10}$ and not $\ge \sqrt{10}$?

Great catch! The constraint $b_2 b_3 > 0$ tells us that $b_2$ and $b_3$ must have the same sign. Since $b_3 = b_2 + 3$, this forces $b_2$ to be outside the interval $[-3, 0]$. The minimum value of the parabola $2b_2^2 + 6b_2 + 10$ occurs at $b_2 = -1.5$, which is exactly inside that forbidden zone. Since the function is strictly increasing or decreasing as you move away from the vertex, the value at the boundaries ($b_2 = 0$ or $-3$) is $10$, but since the domain excludes these points, the value must be strictly greater than $10$.

Question

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + b_2\hat{j} + b_3\hat{k}$, $b_2, b_3 \in \mathbb{R}$, $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, $c_1, c_2, c_3 \in \mathbb{R}$ be three vectors such that $b_2b_3 > 0$, $\vec{a} \cdot \vec{b} = 0$ and $\begin{pmatrix} 0 & -c_3 & c_2 \ c_3 & 0 & -c_1 \ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix} 1 \ b_2 \ b_3 \end{pmatrix} = \begin{pmatrix} 3 - c_1 \ 1 - c_2 \ -1 - c_3 \end{pmatrix}$. Then, which of the following is/are TRUE?

Options:

(A) $\vec{a} \cdot \vec{c} = 0$

(B) $\vec{b} \cdot \vec{c} = 0$

(C) $|\vec{b}| > \sqrt{10}$

(D) $|\vec{c}| \le \sqrt{11}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of matrices and variables. When you see a $3 	imes 3$ matrix like

$$
\begin{pmatrix} 0 & -c_3 & c_2 \ c_3 & 0 & -c_1 \ -c_2 & c_1 & 0 \end{pmatrix}
$$

acting on a vector, your heart might skip a beat. But take a deep breath. In the world of JEE Advanced, this is not a matrix problem—it is a vector problem in disguise.

This specific matrix is a skew-symmetric matrix, and it is the mathematical equivalent of the cross product operator. If you perform the multiplication, you will find that the left-hand side is exactly $\vec{c} 	imes \vec{b}$.

Suddenly, the entire equation simplifies to a beautiful, elegant vector identity:

$$
\vec{c} 	imes \vec{b} = \vec{a} - \vec{c}
$$

This is the key that unlocks the entire problem. Never let the notation intimidate you; look for the underlying geometry.

### The Power of the Dot Product

Now that we have our core equation, $\vec{c} 	imes \vec{b} = \vec{a} - \vec{c}$, how do we extract the information we need? We use the scalpel of vector algebra: the dot product.

If we take the dot product of both sides with $\vec{b}$, we get:

$$
(\vec{c} 	imes \vec{b}) \cdot \vec{b} = (\vec{a} - \vec{c}) \cdot \vec{b}
$$

Because $\vec{c} 	imes \vec{b}$ is perpendicular to $\vec{b}$, the left side vanishes to zero! We are left with $0 = \vec{a} \cdot \vec{b} - \vec{c} \cdot \vec{b}$.

Since the problem gives us $\vec{a} \cdot \vec{b} = 0$, we immediately conclude that $\vec{c} \cdot \vec{b} = 0$. This proves that $\vec{b}$ and $\vec{c}$ are perpendicular. Just like that, **Option B is confirmed!**

### The Geometric Revelation

Let us take this further. If we rearrange our core equation to $\vec{a} = \vec{c} + (\vec{c} 	imes \vec{b})$, we see something profound. Since $\vec{c} 	imes \vec{b}$ is perpendicular to $\vec{c}$, these two vectors form the legs of a right-angled triangle, with $\vec{a}$ as the hypotenuse.

This is the 'Aha!' moment. By the Pythagorean theorem, we know that:

$$
|\vec{a}|^2 = |\vec{c}|^2 + |\vec{c} 	imes \vec{b}|^2
$$

Because $\vec{b}$ and $\vec{c}$ are perpendicular, the magnitude of their cross product is simply $|\vec{c}| |\vec{b}|$. Thus:

$$
|\vec{a}|^2 = |\vec{c}|^2 + |\vec{c}|^2 |\vec{b}|^2 = |\vec{c}|^2 (1 + |\vec{b}|^2)
$$

Since $|\vec{a}|^2 = 3^2 + 1^2 + (-1)^2 = 11$, we have $|\vec{c}|^2 = \frac{11}{1 + |\vec{b}|^2}$. This immediately tells us that $|\vec{c}|^2 \le 11$, or $|\vec{c}| \le \sqrt{11}$. **Option D is now proven!**

### The Inequality Trap

Finally, we must address the constraints on $\vec{b}$. We know $\vec{a} \cdot \vec{b} = 0$, which leads to $3(1) + 1(b_2) - 1(b_3) = 0$, or $b_3 = b_2 + 3$. The problem adds a crucial constraint: $b_2 b_3 > 0$.

Substituting $b_3$, we get $b_2(b_2 + 3) > 0$. This inequality forces $b_2$ to be in the domain $(-\infty, -3) \cup (0, \infty)$.

When we calculate $|\vec{b}|^2 = 1 + b_2^2 + b_3^2$, we get a quadratic expression: $2b_2^2 + 6b_2 + 10$. This is an upward-opening parabola. Its vertex is at $b_2 = -1.5$, but our domain strictly excludes the region between $-3$ and $0$.

Therefore, the minimum value of $|\vec{b}|^2$ occurs at the boundaries of our allowed domain, which is $10$. Since our domain is strictly outside these boundaries, $|\vec{b}|^2$ must be strictly greater than $10$. Thus, $|\vec{b}| > \sqrt{10}$. **Option C is confirmed!**

This problem was a journey through matrix algebra, vector orthogonality, geometric visualization, and inequality analysis. You didn't just solve a problem; you navigated the landscape of JEE physics and mathematics. Keep this mindset—always look for the geometry behind the algebra.

Select Answer:

Visualized Solution

The Matrix Equation

Decoding to Vector Form

Dot Product with $b$

Proving Option B

Visualizing the Vectors

Dot Product with $c$

Applying Pythagoras

Bounding the Magnitude of $c$

Proving Option D

Constraints on Vector $b$

Finding the Domain of $b_{2}$

Minimizing $∣ b ∣^{2}$

Proving Option C

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of the Dot Product

The Geometric Revelation

The Inequality Trap

Similar Questions

Let $a, b, c$ be unit vectors such that $a + b + c = 0$ . Which one of the following is correct?

Let $u = \hat{i} + \hat{j}, v = \hat{i} - \hat{j}$ and $w = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . If $\overset{n}{^}$ is a unit vector such that $u \cdot \overset{n}{^} = 0$ and $v \cdot \overset{n}{^} = 0$ , then $∣ w \cdot \overset{n}{^} ∣$ is equal to

Let $a = \hat{i} - 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ be two vectors. If $c$ is a vector such that $b \times c = b \times a$ and $c \cdot a = 0$ then $b \cdot c$ is equal to:

Let $a = \hat{i} + 2 \hat{j} - \hat{k}, b = \hat{i} - \hat{j}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $r \cdot a$ is equal to

Let $a = - \hat{i} - \hat{k}$ , $b = - \hat{i} + \hat{j}$ and $c = \hat{i} + 2 \hat{j} + 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times b = c \times b$ and $r \cdot a = 0$ , then the value of $r \cdot b$ is .........

Let $a = \hat{i} + \hat{j} + \hat{k}, c = \hat{j} - \hat{k}$ and a vector $b$ be such that $a \times b = c$ and $a \cdot b = 3$ . Then $∣ b ∣$ equals :

Let three vectors $a = α \hat{i} + 4 \hat{j} + 2 \hat{k}, b = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}, c = x \hat{i} + y \hat{j} + z \hat{k}$ form a triangle such that $c = a - b$ and the area of the triangle is $56$ . If $α$ is a positive real number, then $∣ c ∣^{2}$ is equal to:

Let $a = \hat{j} - \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $b$ satisfying $a \times b + c = 0$ and $a \cdot b = 3$

Select Answer:

Visualized Solution

The Matrix Equation

Decoding to Vector Form

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Proving Option B

Visualizing the Vectors

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

Applying Pythagoras

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

Proving Option D

Constraints on Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Finding the Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2​

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