Q.Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4)$, $(-1, 1, 2)$ and $(-5, -5, -2)$.

The direction cosines of a side are the components of its unit vector: form each side vector, then divide by its length.

With $A(3,5,-4)$, $B(-1,1,2)$, $C(-5,-5,-2)$:

$\vec{AB} = (-4,-4,6)$, $\;|\vec{AB}| = \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$

$\vec{BC} = (-4,-6,-4)$, $\;|\vec{BC}| = \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$

$\vec{CA} = (8,10,-2)$, $\;|\vec{CA}| = \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$ …

What direction cosines are

A directed segment in space makes angles $\alpha,\beta,\gamma$ with the $x$-, $y$-, $z$-axes. Its direction cosines $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$ are exactly the components of the corresponding unit vector. So for a side vector $(a,b,c)$ of length $r$,

Side AB

Dividing each component by $2\sqrt{17}$:

Check: $\dfrac{4+4+9}{17} = 1$ ✓

Side BC

Check: $\dfrac{4+9+4}{17} = 1$ ✓

Side CA

$\vec{CA} = A-C = (3-(-5),\;5-(-5),\;-4-(-2)) = (8,10,-2).$ …

Method: Direction cosines of a segment — vector, then normalise

The direction cosines of a side (a directed segment between two vertices) are just the direction cosines of the vector along it. So the task splits into: build the segment vector, then normalise — repeated once per side.

Steps

Step 1: Form each side vector as the difference of its endpoints. For a triangle $A,B,C$:

Never use a vertex's coordinates themselves as direction cosines — positions are not directions.

Step 2: Find each length. $|\vec{AB}|=\sqrt{a^2+b^2+c^2}$, and likewise for the others. …

Common Mistakes

Mistake 1: Using the vertex coordinates as direction cosines.

Why it's wrong: coordinates are positions; a side's direction is the difference of its endpoints. Correct approach: form $\vec{AB}=B-A$ first, then normalise.

Mistake 2: Dividing every side by the same length.

Why it's wrong: each side has its own magnitude — $|\vec{AB}|=|\vec{BC}|=2\sqrt{17}$ here, but $|\vec{CA}|=2\sqrt{42}$. Correct approach: use each side's own length as its divisor. …