Let a = 2i + 5j - k, b = 2i - 2j + 2k and c be three vectors such that (c + i) × (a + b + i) = a × (

Q: Wait, why can't we just expand the cross product directly?

You absolutely could, but it would be a nightmare of terms! In JEE Advanced, the goal is often to find the 'hidden structure.' By substituting $\vec{p} = \vec{c} + \hat{i}$, we collapse a complex equation into a simple, elegant form. Always look for repeating patterns before diving into brute-force expansion.

Q: How do we know $\vec{p}$ is parallel to the resultant vector?

That is the beauty of the cross product! The property $\vec{A} \times \vec{B} = \vec{0}$ is the fundamental definition of collinearity for non-zero vectors. Since $\vec{p} \times (2\vec{a} + \vec{b} + \hat{i}) = \vec{0}$, it forces $\vec{p}$ to be a scalar multiple of that resultant vector.

Q: What if $\lambda$ was zero?

If $\lambda$ were zero, $\vec{p}$ would be the zero vector, meaning $\vec{c} = -\hat{i}$. You can always check if this satisfies the original equation. In this problem, the dot product condition $\vec{a} \cdot \vec{c} = -29$ prevents $\lambda$ from being zero, confirming our non-trivial solution.

Question

Let $\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$, $\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c} + \hat{i}) 	imes (\vec{a} + \vec{b} + \hat{i}) = \vec{a} 	imes (\vec{c} + \hat{i})$. If $\vec{a} \cdot \vec{c} = -29$, then $\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k})$ is equal to:

Options:

(A) 15

(B) 12

(C) 10

(D) 5

Sigma Percentile · Accepted Answer

### The Vector Landscape: A Journey of Simplification

Welcome, fellow explorer of the mathematical universe. Today, we are going to dismantle a vector problem that, at first glance, looks like a tangled mess of cross products and unknown variables.

But as we peel back the layers, you will see that it is actually a beautifully orchestrated dance of geometry and algebra. Let us begin by looking at the equation:

$$(\vec{c} + \hat{i}) 	imes (\vec{a} + \vec{b} + \hat{i}) = \vec{a} 	imes (\vec{c} + \hat{i})$$

### Phase 1: The Power of Substitution

Whenever you see a repeating term in a complex equation, your intuition should scream 'substitution!' Notice how $(\vec{c} + \hat{i})$ appears on both sides?

Let us define a new vector $\vec{p} = \vec{c} + \hat{i}$. Suddenly, the equation transforms into:

$$\vec{p} 	imes (\vec{a} + \vec{b} + \hat{i}) = \vec{a} 	imes \vec{p}$$

This is much cleaner, isn't it? We have effectively reduced the noise, allowing us to focus on the core relationship between these vectors.

### Phase 2: The Cross Product Dance

Now, we need to bring all our terms to one side to set the equation to zero. Remember the anti-commutative property of the cross product: $\vec{a} 	imes \vec{p} = -(\vec{p} 	imes \vec{a})$.

By applying this, our equation becomes:

$$\vec{p} 	imes (\vec{a} + \vec{b} + \hat{i}) + (\vec{p} 	imes \vec{a}) = \vec{0}$$

Because the cross product is distributive, we can combine these terms:

$$\vec{p} 	imes (\vec{a} + \vec{b} + \hat{i} + \vec{a}) = \vec{0}$$

This simplifies to:

$$\vec{p} 	imes (2\vec{a} + \vec{b} + \hat{i}) = \vec{0}$$

### Phase 3: The Geometric Insight

Here is the moment of truth. If the cross product of two vectors is the zero vector, they must be parallel. This means $\vec{p}$ is a scalar multiple of the vector $(2\vec{a} + \vec{b} + \hat{i})$.

Let us call this scalar $\lambda$. So, $\vec{p} = \lambda(2\vec{a} + \vec{b} + \hat{i})$.

Now, let us calculate that resultant vector. Given $\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$, we compute:

$$2\vec{a} + \vec{b} + \hat{i} = 2(2\hat{i} + 5\hat{j} - \hat{k}) + (2\hat{i} - 2\hat{j} + 2\hat{k}) + \hat{i}$$

Summing the components, we get:

$$(4+2+1)\hat{i} + (10-2)\hat{j} + (-2+2)\hat{k} = 7\hat{i} + 8\hat{j}$$

### Phase 4: Solving for the Unknown

We now have $\vec{c} + \hat{i} = \lambda(7\hat{i} + 8\hat{j})$, which means $\vec{c} = (7\lambda - 1)\hat{i} + 8\lambda\hat{j}$. We are given $\vec{a} \cdot \vec{c} = -29$.

Substituting our expressions, we get:

$$(2\hat{i} + 5\hat{j} - \hat{k}) \cdot ((7\lambda - 1)\hat{i} + 8\lambda\hat{j}) = -29$$

This simplifies to $2(7\lambda - 1) + 5(8\lambda) = -29$. Solving this linear equation:

$$14\lambda - 2 + 40\lambda = -29 \Rightarrow 54\lambda = -27 \Rightarrow \lambda = -\frac{1}{2}$$

### The Final Victory

With $\lambda = -\frac{1}{2}$, we find:

$$\vec{c} = (7(-\frac{1}{2}) - 1)\hat{i} + 8(-\frac{1}{2})\hat{j} = -\frac{9}{2}\hat{i} - 4\hat{j}$$

Finally, we calculate the dot product $\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k})$:

$$(-\frac{9}{2})(-2) + (-4)(1) + (0)(1) = 9 - 4 = 5$$

We have arrived at our answer: **5**. You have successfully navigated the vector landscape!

Select Answer:

Visualized Solution

Define Substitution $p$

Simplify Cross Product

The Parallel Condition

Calculate Sum Vector

Express $c$ in terms of $λ$

Apply Dot Product Condition

Solve for $λ$

Determine Vector $c$

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Vector Landscape

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

Similar Questions

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Select Answer:

Visualized Solution

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

Simplify Cross Product

The Parallel Condition

Calculate Sum Vector

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Apply Dot Product Condition

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

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