Between the following two statements: Statement I : Let a = i + 2j - 3k and b = 2i + j - k. Then the

Q: Wait, why can't I just cancel $\vec{a}$ from both sides of $\vec{a} \times \vec{r} = \vec{a} \times \vec{b}$?

Great question! In vector algebra, the cross product is not invertible like multiplication in arithmetic. If $\vec{a} \times \vec{r} = \vec{a} \times \vec{b}$, it only implies that $\vec{a} \times (\vec{r} - \vec{b}) = \vec{0}$, which means $\vec{r} - \vec{b}$ is parallel to $\vec{a}$, not necessarily equal to $\vec{b}$.

Q: How did we get $\vec{r} = \vec{b} + \lambda \vec{a}$? Is that the only solution?

Yes, because $\vec{r} - \vec{b}$ must be parallel to $\vec{a}$, it must be a scalar multiple of $\vec{a}$. We introduce the parameter $\lambda$ to represent all possible vectors that satisfy this condition. The dot product condition $\vec{a} \cdot \vec{r} = 0$ then uniquely determines the specific $\lambda$ for this problem.

Q: Why is the minimum value of $\cos 2A + \cos 2B + \cos 2C$ related to the equilateral triangle?

In a triangle, the angles are constrained by $A+B+C = 180^\circ$. The expression $\cos 2A + \cos 2B + \cos 2C$ can be transformed into $-1 - 4 \cos A \cos B \cos C$. To minimize the whole expression, we need to maximize the product $\cos A \cos B \cos C$. By symmetry (or Jensen's inequality), this product is maximized when $A=B=C=60^\circ$.

Question

Between the following two statements:
Statement I : Let $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} 	imes \vec{r} = \vec{a} 	imes \vec{b}$ and $\vec{a} \cdot \vec{r} = 0$ is of magnitude $\sqrt{10}$.
Statement II : In a triangle $ABC$, $\cos 2A + \cos 2B + \cos 2C \ge -\frac{3}{2}$.

Options:

(A) Statement I is incorrect but Statement II is correct.

(B) Both Statement I and Statement II are correct.

(C) Statement I is correct but Statement II is incorrect.

(D) Both Statement I and Statement II are incorrect.

Sigma Percentile · Accepted Answer

### Analyzing the Vector Relation

In the realm of vectors, the cross product is a jealous guardian of information. Given the relation $\vec{a} 	imes \vec{r} = \vec{a} 	imes \vec{b}$, we cannot simply cancel $\vec{a}$. Instead, we must bring the terms together:

$$\vec{a} 	imes \vec{r} - \vec{a} 	imes \vec{b} = \vec{0}$$

By the distributive property, this simplifies to $\vec{a} 	imes (\vec{r} - \vec{b}) = \vec{0}$. This implies that the vector $(\vec{r} - \vec{b})$ must be parallel to $\vec{a}$.

We can express this relationship using a scalar parameter $\lambda$:

$$\vec{r} - \vec{b} = \lambda \vec{a} \quad \Rightarrow \quad \vec{r} = \vec{b} + \lambda \vec{a}$$

### Solving for the Unknown Vector

We are given the second condition $\vec{a} \cdot \vec{r} = 0$, which indicates that $\vec{r}$ is perpendicular to $\vec{a}$. Substituting our parameterized expression for $\vec{r}$ into this dot product yields:

$$\vec{a} \cdot (\vec{b} + \lambda \vec{a}) = 0$$

Expanding this expression, we obtain:

$$\vec{a} \cdot \vec{b} + \lambda |\vec{a}|^2 = 0$$

Given $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$, we calculate the components:

$$\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (-3)(-1) = 7$$

$$|\vec{a}|^2 = 1^2 + 2^2 + (-3)^2 = 14$$

Substituting these values into our equation, we find $7 + 14\lambda = 0$, which results in $\lambda = -\frac{1}{2}$. Consequently, the vector $\vec{r}$ is:

$$\vec{r} = \left(2 - \frac{1}{2}ight)\hat{i} + \left(1 - 1ight)\hat{j} + \left(-1 + \frac{3}{2}ight)\hat{k} = \frac{3}{2}\hat{i} + 0\hat{j} + \frac{1}{2}\hat{k}$$

The magnitude is $|\vec{r}| = \sqrt{(\frac{3}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}}$. Since this is not $\sqrt{10}$, **Statement I is incorrect**.

### The Trigonometric Identity

We now evaluate the inequality $\cos 2A + \cos 2B + \cos 2C \ge -\frac{3}{2}$ for a triangle. We utilize the standard trigonometric identity:

$$\cos 2A + \cos 2B + \cos 2C = -1 - 4 \cos A \cos B \cos C$$

To minimize this sum, we must maximize the product $\cos A \cos B \cos C$. In any triangle, this product reaches its maximum when the triangle is equilateral, i.e., $A = B = C = 60^\circ$.

The maximum value of the product is:

$$\left(\cos 60^\circight)^3 = \left(\frac{1}{2}ight)^3 = \frac{1}{8}$$

Substituting this back into the identity, the minimum value of the sum is:

$$-1 - 4\left(\frac{1}{8}ight) = -1 - \frac{1}{2} = -\frac{3}{2}$$

Thus, the inequality holds true, and **Statement II is correct**.

Select Answer:

Visualized Solution

Analyzing Statement I: Cross Product

Parallel Vectors Condition

Applying the Dot Product Condition

Calculating Dot Product and Magnitude

Finding Vector $r$

Magnitude of Vector $r$

Analyzing Statement II: Trigonometric Identity

Maximizing the Cosine Product

Evaluating the Minimum Value

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Vector Relation

Solving for the Unknown Vector

The Trigonometric Identity

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