Let ABC be a triangle such that BC = a, CA = b, AB = c, |a| = 6√2, |b| = 2√3 and b · c = 12. Conside

Q: Wait, why is $\vec{a} + \vec{b} + \vec{c} = \vec{0}$? Shouldn't it be $\vec{a} + \vec{b} = \vec{c}$?

That is a classic point of confusion! The equation $\vec{a} + \vec{b} = \vec{c}$ only holds if the vectors are arranged head-to-tail in a specific way. In a closed triangle, if you follow the perimeter, you return to your starting point, which means the net displacement is zero. Thus, $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ is the correct representation of a closed loop.

Q: Why did we get a negative value for (S1)? Is the magnitude of a vector ever negative?

Great catch! Remember that the expression in (S1) is $|(\vec{a} \times \vec{b}) + (\vec{c} \times \vec{b})| - |\vec{c}|$. The first part, the magnitude of the cross product, is always non-negative. However, the entire expression evaluates to $0 - |\vec{c}|$, which is $-6$. We are not saying the magnitude of a vector is negative; we are saying the *result of the entire expression* is negative.

Q: Can I use the Sine Rule for (S2)?

You absolutely could, but it would be much more tedious! The Sine Rule involves $\sin C$, and you would need to calculate the area of the triangle first using the cross product. The Cosine Rule is far more direct here because you already have all the side lengths and the dot product information.

Question

Let ABC be a triangle such that $\vec{BC} = \vec{a}$, $\vec{CA} = \vec{b}$, $\vec{AB} = \vec{c}$, $|\vec{a}| = 6\sqrt{2}$, $|\vec{b}| = 2\sqrt{3}$ and $\vec{b} \cdot \vec{c} = 12$. Consider the statements :
(S1) : $|(\vec{a} 	imes \vec{b}) + (\vec{c} 	imes \vec{b})| - |\vec{c}| = 6(2\sqrt{2} - 1)$
(S2) : $\angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}ight)$. Then

Options:

(A) both (S1) and (S2) are true

(B) only (S1) is true

(C) only (S2) is true

(D) both (S1) and (S2) are false

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into Triangle ABC

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden architecture of a triangle.

When we look at vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ representing the sides of a triangle, we are looking at the fundamental building blocks of spatial geometry. Let us embark on this journey together.

### Phase 1: The Closed Loop

Imagine you are standing at vertex $A$ of triangle $ABC$. You walk along $\vec{AB}$, then $\vec{BC}$, and finally $\vec{CA}$. You have returned to where you started.

In the language of vectors, this is the Triangle Law of Addition. Because you have returned to your origin, the sum of these displacements must be zero:

$$\vec{a} + \vec{b} + \vec{c} = \vec{0}$$

This simple, elegant truth is the key that unlocks the entire problem. By rearranging this to $\vec{b} + \vec{c} = -\vec{a}$, we have created a bridge between the vectors we know and the one we need to find.

### Phase 2: The Magnitude Hunt

We are given the magnitudes $|\vec{a}| = 6\sqrt{2}$ and $|\vec{b}| = 2\sqrt{3}$, and the dot product $\vec{b} \cdot \vec{c} = 12$. We need to find $|\vec{c}|$.

How do we extract the magnitude of a vector from a sum? We square it! By taking the dot product of the equation $\vec{b} + \vec{c} = -\vec{a}$ with itself, we get:

$$|\vec{b} + \vec{c}|^2 = |-\vec{a}|^2$$

Expanding this using the distributive property of the dot product, we arrive at:

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

Substituting our known values, we find that $12 + |\vec{c}|^2 + 24 = 72$. With a quick algebraic breath, we see that $|\vec{c}|^2 = 36$, which means $|\vec{c}| = 6$. We have successfully conquered the first obstacle.

### Phase 3: The Cross Product Trap (S1)

Now, let us look at statement (S1): $|(\vec{a} 	imes \vec{b}) + (\vec{c} 	imes \vec{b})| - |\vec{c}|$. At first glance, this looks like a nightmare of cross products.

But wait! Look at the structure. We can factor out the $	imes \vec{b}$ term: $|(\vec{a} + \vec{c}) 	imes \vec{b}| - |\vec{c}|$.

Recall our closed loop: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, which implies $\vec{a} + \vec{c} = -\vec{b}$. Substituting this, the expression becomes:

$$|(-\vec{b}) 	imes \vec{b}| - |\vec{c}|$$

Since the cross product of any vector with itself is the zero vector, this simplifies beautifully to $0 - |\vec{c}|$, which is $-6$. Statement (S1) claims the value is $6(2\sqrt{2} - 1)$, which is clearly positive. Thus, **(S1) is false**.

### Phase 4: The Cosine Rule (S2)

Finally, we turn to (S2), which asks about $\angle ABC$. We have all the side lengths: $|\vec{a}| = 6\sqrt{2}$, $|\vec{b}| = 2\sqrt{3}$, and $|\vec{c}| = 6$.

The Cosine Rule is our most reliable guide here:

$$\cos C = \frac{|\vec{a}|^2 + |\vec{b}|^2 - |\vec{c}|^2}{2|\vec{a}||\vec{b}|}$$

Substituting our values, we get:

$$\cos C = \frac{(6\sqrt{2})^2 + (2\sqrt{3})^2 - (6)^2}{2(6\sqrt{2})(2\sqrt{3})} = \frac{72 + 12 - 36}{24\sqrt{6}} = \frac{48}{24\sqrt{6}} = \frac{2}{\sqrt{6}}$$

Simplifying further, $\cos C = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}$. This matches statement (S2) perfectly!

### Conclusion

We have navigated the vector space, simplified the cross products, and applied the laws of trigonometry. We found that (S1) is false and (S2) is true.

The correct option is that **only (S2) is true**. Remember, in JEE Advanced, it is not just about the calculation; it is about seeing the symmetry and the relationships between the vectors. You have done well today.

Select Answer:

Visualized Solution

Visualizing the Triangle Vectors

The Triangle Law of Addition

Finding the Magnitude of $c$

Substituting Known Values

Solving for $∣ c ∣$

Evaluating Statement (S1)

Simplifying the Cross Product

Conclusion for (S1)

Evaluating Statement (S2)

Applying the Cosine Rule

Calculating $cos C$

Final Verdict

The Sigma Insight: Scalar (Dot) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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