Key Concepts and Formulas

• Magnitude of a Vector: For a vector $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$, its magnitude is given by $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
• Properties of Squares: For any real number $x$, $x^2 \geq 0$.
• Definition of Maximum: For a set of numbers, the maximum is the largest value in the set. For $M = \max\{|a_1|, |a_2|, |a_3|\}$, it implies $|a_i| \leq M$ for $i=1,2,3$.

Step-by-Step Solution

We are given a vector $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$ and two statements to evaluate. The condition $10\left|a_{i}\right|<1$ for $i=1,2,3$ implies $|a_i| < 0.1$, which means the components are small. However, the truthfulness of the statements will depend on the general properties of vectors and their magnitudes, not this specific condition.

Analysis of Statement (A): $\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\} \leq|\vec{a}|$

Step 1: Define the maximum absolute component. Let $M = \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$. By definition of the maximum, this means $M \geq |a_1|$, $M \geq |a_2|$, and $M \geq |a_3|$.

Step 2: Relate the magnitude of the vector to its components. The magnitude of the vector is $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.

Step 3: Establish an inequality using the maximum component. Since $M \geq |a_i|$, it follows that $M^2 \geq a_i^2$ for each $i=1, 2, 3$. Consider the square of the magnitude: $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$ Since $M^2 \geq a_1^2$, $M^2 \geq a_2^2$, and $M^2 \geq a_3^2$, we can write: $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \leq M^2 + M^2 + M^2 = 3M^2$ This inequality, however, is for statement (B). Let's re-evaluate statement (A) directly.

Step 3 (Revised): Express the magnitude squared in terms of the maximum component. Let's assume, without loss of generality, that $M = |a_1|$. This means $|a_1|$ is the largest among $|a_1|, |a_2|, |a_3|$. We need to check if $M \leq |\vec{a}|$, which is equivalent to checking if $M^2 \leq |\vec{a}|^2$ since both are non-negative. Substituting $M = |a_1|$, we need to check: $|a_1|^2 \leq a_1^2 + a_2^2 + a_3^2$ $a_1^2 \leq a_1^2 + a_2^2 + a_3^2$

Step 4: Simplify the inequality. Subtracting $a_1^2$ from both sides, we get: $0 \leq a_2^2 + a_3^2$

Step 5: Conclude the truthfulness of Statement (A). Since $a_2^2 \geq 0$ and $a_3^2 \geq 0$ for any real numbers $a_2$ and $a_3$, their sum $a_2^2 + a_3^2$ is always non-negative. Thus, $0 \leq a_2^2 + a_3^2$ is always true. This implies that the original inequality $M \leq |\vec{a}|$ is always true.

Conclusion for (A): Statement (A) is TRUE.

Analysis of Statement (B): $|\vec{a}| \leq 3 \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$

Step 1: Define the maximum absolute component. Let $M = \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$. This implies that $|a_1| \leq M$, $|a_2| \leq M$, and $|a_3| \leq M$.

Step 2: Square the inequalities for each component. Since absolute values are non-negative, squaring preserves the inequalities: $a_1^2 \leq M^2$ $a_2^2 \leq M^2$ $a_3^2 \leq M^2$

Step 3: Express the magnitude squared in terms of the maximum component. The square of the magnitude of the vector is $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$. Using the inequalities from Step 2, we can bound the magnitude squared: $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \leq M^2 + M^2 + M^2$ $|\vec{a}|^2 \leq 3M^2$

Step 4: Take the square root of both sides. Since $|\vec{a}|^2$ and $3M^2$ are non-negative, we can take the square root of both sides: $\sqrt{|\vec{a}|^2} \leq \sqrt{3M^2}$ $|\vec{a}| \leq \sqrt{3}M$

Step 5: Compare the derived inequality with Statement (B). We have derived that $|\vec{a}| \leq \sqrt{3}M$. Statement (B) is $|\vec{a}| \leq 3M$. We know that $\sqrt{3} \approx 1.732$. Since $\sqrt{3} < 3$, it follows that $\sqrt{3}M \leq 3M$ for any non-negative $M$. Therefore, if $|\vec{a}| \leq \sqrt{3}M$, then it must also be true that $|\vec{a}| \leq 3M$.

Conclusion for (B): Statement (B) is TRUE.

Common Mistakes & Tips

• Over-reliance on specific conditions: The condition $10|a_i|<1$ is a distraction. The inequalities hold true for all real components $a_i$. Focus on proving the general case.
• Incorrectly squaring inequalities: Always ensure that both sides of an inequality are non-negative before squaring. Magnitudes and squares of real numbers are always non-negative, so this is safe here.
• Confusing upper and lower bounds: Understand that $\max\{|a_i|\}$ is an upper bound for each $|a_i|$, but $|a_i|$ is not necessarily an upper bound for $\max\{|a_i|\}$.

Summary

Statement (A) is proven by showing that the square of the maximum absolute component is less than or equal to the sum of squares of all components. Statement (B) is proven by bounding each component's square by the square of the maximum absolute component and then summing these bounds. Both statements are fundamental inequalities relating a vector's magnitude to its components and hold true for any vector. The specific condition $10|a_i|<1$ is not essential for the validity of these statements.

The final answer is $\boxed{A}$.