Animated Solution for Mathematics - Vector Algebra: For any vector $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, with $10|a_i|<1,i=1,2,3$, consider the following statements: (A) $\max \{|a_1|,|a_2|,|a_3|\}\le |\vec{a}|$, (B) $|\vec{a}|\le 3\max \{|a_1|,|a_2|,|a_3|\}$

Imagine you are standing in a vast, empty three-dimensional room. You have a vector $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ floating right in front of you. It is not just a line; it is a displacement, a journey from the origin to a point in space.

To understand its nature, we must break it down into its fundamental building blocks: the components $a_1$, $a_2$, and $a_3$. These components are the projections of our vector onto the $x$, $y$, and $z$ axes.

Together, they define a rectangular parallelepiped—a box—where our vector $\vec{a}$ acts as the main diagonal, stretching from one corner to the opposite corner.

Let us define a value $M=\max \{|a_1|,|a_2|,|a_3|\}$. Think of $M$ as the "dominant" dimension of our box. If you were to look at the box from a distance, $M$ is the length of the longest side.

Our goal is to see how the total length of the diagonal, the magnitude $|\vec{a}|$, relates to this dominant side $M$. We know from the Pythagorean theorem in 3D that the magnitude is given by:

To make our calculations easier, let us square both sides:

The Lower Bound

Now, consider the relationship between the sum of squares and the maximum component $M$. Since $M$ is the maximum of the absolute values of the components, $M^2$ must be equal to one of the terms in our sum (say, $a_1^2$).

The other two terms, $a_2^2$ and $a_3^2$, are squares of real numbers, which means they are always greater than or equal to zero. When we add these non-negative values to $M^2$, the total sum can only increase or stay the same:

Substituting our magnitude equation, we get $|\vec{a}{|}^2\ge M^2$. Taking the square root of both sides, we arrive at $|\vec{a}|\ge M$.

This is exactly what Statement (A) claims: the magnitude of the vector is always greater than or equal to its largest component. Geometrically, the diagonal of a box must be longer than any of its individual edges.

The Upper Bound

Now, let us tackle the upper bound. We want to know how large the magnitude can possibly be relative to $M$. We return to our squared magnitude equation:

Since $M$ is the maximum absolute value of any component, we know that $a_1^2\le M^2$, $a_2^2\le M^2$, and $a_3^2\le M^2$. If we replace each component in our sum with the largest possible value, $M^2$, we create an upper bound for the magnitude:

Taking the square root, we find $|\vec{a}|\le \sqrt{3}M$.

Now, look at Statement (B). It claims $|\vec{a}|\le 3M$. Since $\sqrt{3}\approx 1.732$, and $1.732<3$, it is mathematically certain that if the magnitude is less than or equal to $\sqrt{3}M$, it is also less than or equal to $3M$. Thus, Statement (B) is also true.

We have navigated the geometry of the vector, established the bounds, and verified both statements. The constraint $10|a_i|<1$ was merely a distraction, a test of your confidence in the underlying principles.

You have successfully proven that both (A) and (B) are true. Keep this geometric intuition close; it is the key to mastering vector analysis in JEE Advanced.