Let a = 2i + j - 2k and b = i + j. If c is a vector such that a · c = |c|, |c - a| = 2√2 and the ang

Q: Why can't we just assume $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve for $x, y, z$?

That is a common instinct, but it leads to a dead end! You have three unknowns ($x, y, z$) but only two conditions. This means the system is underdetermined—there are infinitely many vectors $\vec{c}$ that satisfy these conditions. However, the question asks for a specific scalar value, which implies that the magnitude $|\vec{c}|$ and the final cross product are invariant. Always look for properties of magnitudes and dot products before diving into component algebra.

Q: Why does squaring the magnitude equation work so well?

Squaring is the bridge between vector geometry and scalar algebra. When you have $|\vec{c} - \vec{a}|$, you are dealing with a geometric distance. By squaring it, you invoke the identity $|\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})$. This allows you to replace the vector dot product with the scalar magnitude, effectively collapsing the vector complexity into a simple quadratic equation.

Q: Is the angle $\frac{\pi}{6}$ always the angle between the vectors in the cross product formula?

Yes, absolutely! The definition of the magnitude of a cross product is $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$. The problem gives you this angle directly. You don't need to find the components of $\vec{c}$ or the cross product vector itself; you only need their magnitudes and the sine of the angle between them.

Question

Let $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $(\vec{a} 	imes \vec{b})$ and $\vec{c}$ is $\frac{\pi}{6}$, then the value of $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}|$ is :

Options:

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

(B) 4

(C) 3

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a problem; we are choreographing a dance between vectors.

We have two known actors on our stage: $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. There is a mysterious third actor, $\vec{c}$, whose properties we must uncover to solve the problem.

### Phase 1: The Mystery of the Magnitude

Before we interact with $\vec{c}$, let us understand our anchor, $\vec{a}$. We calculate its magnitude, $|\vec{a}|$, using the standard distance formula:

$$|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

Now, consider the conditions given for $\vec{c}$: $\vec{a} \cdot \vec{c} = |\vec{c}|$ and $|\vec{c} - \vec{a}| = 2\sqrt{2}$. To solve this, we use the 'JEE Toolkit'—the power of squaring.

Expanding $|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2$, we obtain:

$$|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c}) = 8$$

Substituting $|\vec{a}|^2 = 9$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$ into the equation, we get a quadratic in terms of $|\vec{c}|$:

$$|\vec{c}|^2 + 9 - 2|\vec{c}| = 8$$

Rearranging this, we find $|\vec{c}|^2 - 2|\vec{c}| + 1 = 0$, which simplifies to $(|\vec{c}| - 1)^2 = 0$. Thus, **$|\vec{c}| = 1$**, confirming that $\vec{c}$ is a unit vector.

### Phase 2: The Cross Product Geometry

Next, we calculate the cross product $\vec{a} 	imes \vec{b}$, which represents the normal to the plane containing $\vec{a}$ and $\vec{b}$:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$$

We calculate the magnitude of this resulting vector:

$$|\vec{a} 	imes \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$$

### Phase 3: The Grand Finale

We are asked to find the magnitude of the cross product between $(\vec{a} 	imes \vec{b})$ and $\vec{c}$. Let $\vec{u} = \vec{a} 	imes \vec{b}$ and $\vec{v} = \vec{c}$.

The geometric definition of the cross product magnitude is $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}|\sin	heta$, where $	heta$ is the angle between them. Given $	heta = \frac{\pi}{6}$, we have:

1. $|\vec{u}| = 3$
2. $|\vec{v}| = 1$
3. $\sin\left(\frac{\pi}{6}ight) = \frac{1}{2}$

Putting it all together, the final value is:

$$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = 3 \cdot 1 \cdot \frac{1}{2} = \mathbf{\frac{3}{2}}$$

We did not need to know the exact components of $\vec{c}$. By understanding its relationship to the other vectors, we arrived at the solution through elegant principles.

Select Answer:

Visualized Solution

Visualize the Vectors $a$ and $b$

Calculate Magnitude of $a$

Analyze the Condition for $c$

Expand the Magnitude Square

Substitute Known Values

Solve for $∣ c ∣$

Compute Cross Product $a \times b$

Magnitude of $a \times b$

The Final Cross Product Formula

Substitute and Calculate

Final Answer

The Sigma Insight: Vector Triple Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualize the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

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

Analyze the Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c