Q.Let $\vec{a} = 3\hat{i} + 2\hat{j} \text{ and } \vec{b} = 2\hat{i} + 3\hat{j}$ Is $|\vec{a}| = |\vec{b}|$? Is $\vec{a} = \vec{b}$?

When you're given two vectors and asked about their relationship, you need to check three fundamental properties: parallelism (same or opposite direction), perpendicularity (right angles), and magnitude (length). Each has a clear algebraic test.

Two vectors are parallel if one is a scalar multiple of the other: $\vec{a} = \lambda \vec{b}$ for some real $\lambda$. Two vectors are perpendicular (orthogonal) if their dot product vanishes: $\vec{a} \cdot \vec{b} = 0$. The magnitude of a vector $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ is $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$.

Let's systematically check each possibility.

Checking the options

1. Are the vectors parallel?

For $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ to be parallel, we'd need

The first ratio gives $3$, the second gives $-2$, and the third gives $-1$. These are all different, so the vectors are not parallel. Option (A) is incorrect.

2. Are the vectors perpendicular?

Compute the dot product:

The dot product is zero, which means the vectors are perpendicular. Option (B) looks promising.

3. Compare the magnitudes …

When two matrices are equal, every entry in the same position must match exactly. This is the fundamental idea of vector equality extended to matrices — think of it as comparing two tables of numbers cell by cell. No shortcuts, no tricks: if the top-left corner of the first matrix is $x+y$, and the top-left corner of the second matrix is $6$, then $x+y$ must equal $6$. The same logic applies to every other position.

Here, both matrices are $2\times 2$, so we get four equations. But notice that two of the entries are already identical ($2$ and $5$), so they give no new information. The real constraints come from the other two positions.

1. Equate the (1,1) entries — the top-left corner:

   $x + y = 6$

2. Equate the (2,2) entries — the bottom-right corner:

   $xy = 8$

We now have two equations in two unknowns. You could solve for $x$ and $y$ individually (they are the roots of $t^2 - 6t + 8 = 0$, so $t = 2$ or $4$), but the question doesn't ask for $x$ and $y$ separately — it asks for $\frac{24}{x} + \frac{24}{y}$.