Let a = 6i + j - k and b = i + j. If c is a vector such that |c| ≥ 6, a · c = 6|c|, |c - a| = 2√2 an

Q: Wait, why do we square the magnitude $|\vec{c} - \vec{a}|$? Can't we just solve it directly?

Great question! In vector algebra, magnitudes are defined by square roots, which are notoriously difficult to manipulate directly. Squaring both sides is the standard 'key' to unlock the dot product hidden inside. It turns a geometric constraint into a manageable algebraic equation.

Q: Why do we reject one of the roots for $|\vec{c}|$? Isn't math supposed to be absolute?

Math is absolute, but context is king! The problem explicitly states $|\vec{c}| \ge 6$. When you solve the quadratic, you get two mathematical possibilities, but only one fits the physical reality defined by the problem. Always check your constraints—they are the 'guardrails' of the problem.

Q: Why use $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$ instead of calculating the cross product components?

You *could* calculate the components, but that would be a massive amount of work! Since we already know the magnitudes and the angle, the geometric definition is the elegant, 'JEE-smart' way to solve it. Always look for the shortcut that saves time and reduces calculation errors.

Question

Let $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}| \ge 6, \vec{a} \cdot \vec{c} = 6|\vec{c}|, |\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} 	imes \vec{b}$ and $\vec{c}$ is $60^\circ$, then $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}|$ is equal to:

Options:

(A) $\frac{9}{2}(6 - \sqrt{6})$

(B) $\frac{3}{2}\sqrt{6}$

(C) $\frac{9}{2}(6 + \sqrt{6})$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a vector problem that might look intimidating at first glance, but is actually a masterclass in elegance.

Imagine you are standing in a three-dimensional coordinate system. You have vector $\vec{a}$ and vector $\vec{b}$ fixed in space, and a mysterious vector $\vec{c}$ constrained by rules that pull it in different directions. Our goal is to find the magnitude of the cross product of $(\vec{a} 	imes \vec{b})$ and $\vec{c}$.

### The Mystery of Vector $\vec{c}$

First, we tackle the mystery of vector $\vec{c}$. We are given $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Whenever you see a magnitude of a difference, think of it as a distance between two points.

To extract the information hidden inside, we square both sides:
$$|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$$

Expanding this, we get:
$$|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c} = 8$$

We know $|\vec{a}|^2 = 6^2 + 1^2 + (-1)^2 = 38$. Substituting this and the given dot product $\vec{a} \cdot \vec{c} = 6|\vec{c}|$ into the equation, we arrive at a quadratic equation:
$$|\vec{c}|^2 - 12|\vec{c}| + 30 = 0$$

### The Quadratic Trap

Solving this quadratic equation gives us two roots: $|\vec{c}| = 6 \pm \sqrt{6}$.

However, the problem provides a crucial constraint: $|\vec{c}| \ge 6$. Since $\sqrt{6} \approx 2.45$, the root $6 - \sqrt{6}$ is less than 6. We must discard it.

Therefore, the magnitude is fixed at:
$$|\vec{c}| = 6 + \sqrt{6}$$
This is a classic JEE trap—always check your constraints!

### The Final Act

Now, for the final act, we need to calculate $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}|$. Instead of brute-forcing the cross product, we use the geometric definition: $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}|\sin	heta$.

First, we calculate the magnitude of the cross product of $\vec{a}$ and $\vec{b}$:
$$|\vec{a} 	imes \vec{b}| = |(6\hat{i} + \hat{j} - \hat{k}) 	imes (\hat{i} + \hat{j})| = |\hat{i} - \hat{j} + 5\hat{k}|$$
$$|\vec{a} 	imes \vec{b}| = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{27} = 3\sqrt{3}$$

With the angle between the vectors given as $	heta = 60^\circ$, the final calculation is:
$$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = 3\sqrt{3} 	imes (6 + \sqrt{6}) 	imes \sin(60^\circ)$$
$$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = 3\sqrt{3} 	imes (6 + \sqrt{6}) 	imes \frac{\sqrt{3}}{2}$$

Simplifying the expression, we arrive at the final result:
$$\mathbf{\frac{9}{2}(6 + \sqrt{6})}$$

It is beautiful, isn't it? The way the algebra simplifies and the way the constraints guide us is the heart of JEE mathematics. Keep this logic in your toolkit, and you will conquer any problem.

Select Answer:

Visualized Solution

Analyze Given Vectors $a$ and $b$

Square the Difference Magnitude

Expand Using Vector Properties

Calculate $∣ a ∣^{2}$

Substitute Known Values

Form the Quadratic Equation

Solve for $∣ c ∣$

Apply Constraint $∣ c ∣ \geq 6$

Calculate $a \times b$

Magnitude of $a \times b$

Magnitude of Double Cross Product

Substitute and Calculate

Final Conclusion

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Mystery of Vector $c$

The Quadratic Trap

The Final Act

Similar Questions

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Let $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{j} - \hat{k}$ . If $c$ is a vector such that $a \times c = b$ and $a \cdot c = 3$ , then $a \cdot (b \times c)$ is equal to :

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Let $x, y$ and $z$ be three vectors each of magnitude $2$ and the angle between each pair of them is $\frac{π}{3}$ . If $a$ is a non-zero vector perpendicular to $x$ and $y \times z$ and $b$ is a non-zero vector perpendicular to $y$ and $z \times x$ , then

Select Answer:

Visualized Solution

Analyze Given 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

Square the Difference Magnitude

Expand Using Vector Properties

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