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

Q: Wait, why do we square $|\vec{c} - \vec{a}| = 3$ instead of trying to find the components of $\vec{c}$?

Great question! If you try to find the components of $\vec{c}$ (say, $c_x, c_y, c_z$), you end up with three unknowns and only two conditions, which is mathematically insufficient. Squaring the magnitude is a powerful vector identity that directly introduces the dot product $\vec{a} \cdot \vec{c}$, which is exactly what the question asks for. It is a shortcut that bypasses the need to know the individual components of $\vec{c}$.

Q: Is the cross product of $\vec{a}$ and $\vec{b}$ really necessary for the final answer?

Absolutely! The condition $|(\vec{a} \times \vec{b}) \times \vec{c}| = 3$ treats the vector $(\vec{a} \times \vec{b})$ as a single entity. To use the magnitude formula $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$, you need the magnitude of the vector $(\vec{a} \times \vec{b})$. Without calculating this cross product, you would be stuck with an unknown magnitude in your equation.

Q: Can we assume $\vec{c}$ is in a specific direction relative to $\vec{a}$?

No, you cannot. The problem does not define the orientation of $\vec{c}$ relative to $\vec{a}$ or $\vec{b}$ beyond the angle with $\vec{a} \times \vec{b}$. By using scalar magnitudes and dot product properties, we avoid making any assumptions about the direction of $\vec{c}$, which is the most robust way to solve this.

Question

Let $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. Let $\vec{c}$ be a vector such that $|\vec{c} - \vec{a}| = 3$, $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = 3$ and the angle between $\vec{c}$ and $\vec{a} 	imes \vec{b}$ be $30^\circ$. Then $\vec{a} \cdot \vec{c}$ is equal to:

Options:

(A) $\frac{25}{8}$

(B) $2$

(C) $5$

(D) $\frac{1}{8}$

Sigma Percentile · Accepted Answer

### The Vector Odyssey: Unlocking the Mystery of $\vec{c}$

Welcome, future engineers! Today, we are going to embark on a journey through the elegant landscape of 3D vector algebra.

Imagine you are standing in a coordinate system, looking at two vectors, $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. These aren't just arrows on a page; they are the fundamental building blocks of the space around you.

Our mission is to find the dot product $\vec{a} \cdot \vec{c}$, where $\vec{c}$ is a mysterious vector defined by two specific conditions. This problem is a masterclass in using vector identities to simplify complex geometric constraints.

### Phase 1: The First Clue and the Squaring Trick

The first condition given is $|\vec{c} - \vec{a}| = 3$. When you see a magnitude of a difference, do not panic! Your first instinct should be to square both sides.

Squaring transforms the geometric magnitude into the algebraic dot product. By expanding $|\vec{c} - \vec{a}|^2 = 3^2$, we use the identity $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$.

This gives us:
$$|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c} = 9$$

Now, let's calculate the magnitude of $\vec{a}$. We have $|\vec{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$.

Substituting this back into our equation, we get $|\vec{c}|^2 + 9 - 2\vec{a} \cdot \vec{c} = 9$. The $9$s cancel out perfectly, leaving us with the beautiful, simplified relationship:
$$|\vec{c}|^2 = 2\vec{a} \cdot \vec{c}$$

This is our golden key. Keep it safe; we will need it at the very end.

### Phase 2: The Cross Product and the Determinant

Now, let's look at the second condition: $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = 3$. To use this, we first need to find the vector $\vec{a} 	imes \vec{b}$.

We set up the determinant:
$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix}$$

Expanding this, we get $\hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1)$, which simplifies to $2\hat{i} - 2\hat{j} + \hat{k}$. This vector is perpendicular to both $\vec{a}$ and $\vec{b}$.

Now, we find its magnitude:
$$|\vec{a} 	imes \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

### Phase 3: The Synthesis

We are now ready to use the second condition. The magnitude of a cross product is given by $|\vec{u} 	imes \vec{v}| = |\vec{u}| |\vec{v}| \sin 	heta$.

Here, let $\vec{u} = \vec{a} 	imes \vec{b}$ and $\vec{v} = \vec{c}$. We are given that the angle $	heta$ between them is $30^\circ$.

So, we have:
$$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = |\vec{a} 	imes \vec{b}| |\vec{c}| \sin 30^\circ = 3$$

Substituting our known values, we have:
$$3 \cdot |\vec{c}| \cdot \frac{1}{2} = 3$$

Dividing both sides by $3$ and multiplying by $2$, we find $|\vec{c}| = 2$.

### The Grand Finale

Finally, we return to our golden key from Phase 1: $|\vec{c}|^2 = 2\vec{a} \cdot \vec{c}$. Since $|\vec{c}| = 2$, then $|\vec{c}|^2 = 4$.

Substituting this in, we get $4 = 2\vec{a} \cdot \vec{c}$, which leads us directly to:
$$\vec{a} \cdot \vec{c} = 2$$

You have successfully navigated the vector space! By breaking the problem into manageable pieces and using the right identities, you have conquered the challenge. Keep this logical flow in your toolkit, and you will be ready for any JEE Advanced problem that comes your way.

Select Answer:

Visualized Solution

Given Vectors $a$ and $b$

The First Condition

Squaring the Equation

Simplifying the Relation

The Cross Product Vector

Evaluating $a \times b$

Magnitude of $a \times b$

The Second Condition

Applying the Formula

Solving for $∣ c ∣$

Final Calculation

The Sigma Insight: Vector Triple Product

The Vector Odyssey

Phase 1

Phase 2

Phase 3

The Grand Finale

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