Let a = 4i - j + k, b = 11i - j + k and c be a vector such that (a + b) × c = c × (-2a + 3b). If (2a

Q: Wait, why do we reject the negative root or just assume $\lambda$ is a single value?

Great question! We aren't rejecting anything. The equation $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670$ is a linear equation in terms of $\lambda$. When you substitute $\vec{c} = \lambda(4\vec{b} - \vec{a})$, you get a simple scalar equation of the form $K\lambda = 1670$. Since $K$ is a non-zero constant, there is only one unique solution for $\lambda$.

Q: Why can't we just use components to solve for $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ directly?

You absolutely could, but it would be a nightmare! You would end up with a system of three linear equations with three variables ($x, y, z$) and the scalar $\lambda$. By using vector properties like anti-commutativity and parallelism, we bypass the tedious algebra and solve it in a much more elegant, conceptual way.

Q: How do I know for sure that $\vec{u} \times \vec{v} = 0$ means they are parallel?

The magnitude of the cross product is $|\vec{u}| |\vec{v}| \sin(\theta)$, where $\theta$ is the angle between the vectors. If the cross product is zero, then $\sin(\theta)$ must be zero (assuming the vectors are non-zero). This only happens when $\theta = 0^\circ$ or $180^\circ$, which is the definition of parallel or anti-parallel vectors.

Question

Let $\vec{a} = 4\hat{i} - \hat{j} + \hat{k}, \vec{b} = 11\hat{i} - \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} + \vec{b}) 	imes \vec{c} = \vec{c} 	imes (-2\vec{a} + 3\vec{b})$. If $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670$, then $|\vec{c}|^2$ is equal to :

Options:

(A) 1609

(B) 1618

(C) 1600

(D) 1627

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system. You have two fixed vectors, $\vec{a} = 4\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 11\hat{i} - \hat{j} + \hat{k}$, acting as your anchors.

There is a mysterious vector $\vec{c}$ defined by the relationship:
$$(\vec{a} + \vec{b}) 	imes \vec{c} = \vec{c} 	imes (-2\vec{a} + 3\vec{b})$$

At first glance, this looks like a daunting algebraic mess. However, in the world of JEE Advanced mathematics, complexity is often just a mask for an underlying simplicity.

### The Cross Product Dance

The first hurdle is the cross product. We have $\vec{c}$ on the right side of the cross product in the first term, but on the left side in the second.

To solve this, we invoke the **anti-commutative property**: $\vec{u} 	imes \vec{v} = -(\vec{v} 	imes \vec{u})$. By applying this to the right-hand side, we transform the equation:
$$(\vec{a} + \vec{b}) 	imes \vec{c} = -(-2\vec{a} + 3\vec{b}) 	imes \vec{c}$$

Now, both terms have $\vec{c}$ on the right. We move everything to the left-hand side:
$$(\vec{a} + \vec{b}) 	imes \vec{c} + (-2\vec{a} + 3\vec{b}) 	imes \vec{c} = \vec{0}$$

Factoring out the cross product with $\vec{c}$, we obtain:
$$((\vec{a} + \vec{b}) + (-2\vec{a} + 3\vec{b})) 	imes \vec{c} = \vec{0}$$

Simplifying the vector inside the parenthesis, we arrive at the breakthrough:
$$(4\vec{b} - \vec{a}) 	imes \vec{c} = \vec{0}$$

### The Parallelism Revelation

When the cross product of two vectors is the zero vector, it implies they are **collinear**. This means $\vec{c}$ must be a scalar multiple of the vector $(4\vec{b} - \vec{a})$.

We define this as $\vec{c} = \lambda(4\vec{b} - \vec{a})$. Now, let us calculate this vector:
$$4\vec{b} - \vec{a} = 4(11\hat{i} - \hat{j} + \hat{k}) - (4\hat{i} - \hat{j} + \hat{k})$$
$$4\vec{b} - \vec{a} = (44\hat{i} - 4\hat{j} + 4\hat{k}) - (4\hat{i} - \hat{j} + \hat{k}) = 40\hat{i} - 3\hat{j} + 3\hat{k}$$

Thus, our mysterious vector is $\vec{c} = \lambda(40\hat{i} - 3\hat{j} + 3\hat{k})$.

### The Dot Product Climax

We are given one final clue: $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670$. First, we find the vector $(2\vec{a} + 3\vec{b})$:
$$2(4\hat{i} - \hat{j} + \hat{k}) + 3(11\hat{i} - \hat{j} + \hat{k}) = (8\hat{i} - 2\hat{j} + 2\hat{k}) + (33\hat{i} - 3\hat{j} + 3\hat{k}) = 41\hat{i} - 5\hat{j} + 5\hat{k}$$

Now, we take the dot product with $\vec{c}$:
$$(41\hat{i} - 5\hat{j} + 5\hat{k}) \cdot \lambda(40\hat{i} - 3\hat{j} + 3\hat{k}) = 1670$$

Expanding this, we get:
$$\lambda[(41 	imes 40) + (-5 	imes -3) + (5 	imes 3)] = 1670$$
$$\lambda[1640 + 15 + 15] = 1670 \Rightarrow 1670\lambda = 1670$$

This yields $\lambda = 1$. Therefore, our vector is $\vec{c} = 40\hat{i} - 3\hat{j} + 3\hat{k}$.

### Final Calculation

The final step is to find $|\vec{c}|^2$:
$$|\vec{c}|^2 = 40^2 + (-3)^2 + 3^2$$
$$|\vec{c}|^2 = 1600 + 9 + 9 = 1618$$

The final answer is **1618**.

Animated Solution for Mathematics - Vector Algebra: Let $a = 4 \hat{i} - \hat{j} + \hat{k}, b = 11 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + b) \times c = c \times (- 2 a + 3 b)$ . If $(2 a + 3 b) \cdot c = 1670$ , then $∣ c ∣^{2}$ is equal to :

Select Answer:

Visualized Solution

Analyze the Given Vectors

Apply Anti-commutative Property

Rearrange the Equation

Simplify the Resultant Vector

Establish Parallelism

Calculate $4 b - a$

Define Vector $c$

Calculate $2 a + 3 b$

Set up the Dot Product Equation

Solve for $λ$

Final Vector $c$

Calculate $∣ c ∣^{2}$

Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Dance

The Parallelism Revelation

The Dot Product Climax

Final Calculation

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Select Answer:

Visualized Solution

Analyze the Given Vectors

Apply Anti-commutative Property

Rearrange the Equation

Simplify the Resultant Vector

Establish Parallelism

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4b−a

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

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