Q.Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular. OR Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ and is in the direction $\hat{i} + 2\hat{j} - \hat{k}$.

Perpendicular Vectors: The Dot-Product Test

Two vectors are perpendicular (orthogonal) when they meet at a right angle — like the east and north directions. But you cannot reach for a protractor in 3D, so you need an algebraic test.

The key idea: when two vectors are perpendicular, neither has any "shadow" along the other. Walk along one and you make zero progress in the direction of the other. The dot product measures exactly this overlap, so perpendicularity means the dot product vanishes.

Why? The dot product has two equal forms:

When $\theta = 90^\circ$, $\cos 90^\circ = 0$, so the product is zero regardless of the vectors' lengths.

Example. For $\vec{a}=(1,2,3)$ and $\vec{b}=(2,-1,0)$:

so they are perpendicular. By contrast $(2,1)\cdot(1,3) = 2+3 = 5 \neq 0$, so those two are not.