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.

Concept: Mutual Perpendicularity — three lines are mutually perpendicular if the dot product of the direction cosines of any two distinct lines is zero.

Step 1: Let the three direction vectors be

$\vec{a} = \left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right)$,

$\vec{b} = \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$,

$\vec{c} = \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$.

Step 2: Compute $\vec{a} \cdot \vec{b}$:

$\frac{12}{13} \cdot \frac{4}{13} + \frac{-3}{13} \cdot \frac{12}{13} + \frac{-4}{13} \cdot \frac{3}{13} = \frac{48 - 36 - 12}{169} = 0$.

Step 3: Compute $\vec{b} \cdot \vec{c}$:

$\frac{4}{13} \cdot \frac{3}{13} + \frac{12}{13} \cdot \frac{-4}{13} + \frac{3}{13} \cdot \frac{12}{13} = \frac{12 - 48 + 36}{169} = 0$.

Step 4: Compute $\vec{c} \cdot \vec{a}$:

$\frac{3}{13} \cdot \frac{12}{13} + \frac{-4}{13} \cdot \frac{-3}{13} + \frac{12}{13} \cdot \frac{-4}{13} = \frac{36 + 12 - 48}{169} = 0$.

Since each dot product is zero, the three lines are mutually perpendicular.

We need to show that each pair of these lines is perpendicular. For lines given by direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, the condition for perpendicularity is:

Let’s label the three lines:

• Line A: $\left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right)$
• Line B: $\left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$
• Line C: $\left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$

We check all three pairs.

1. Check A and B Compute the dot product:

So A ⟂ B.

2. Check B and C Compute:

So B ⟂ C.

3. Check C and A Compute:

So C ⟂ A.

Since every pair gives a dot product of zero, the three lines are mutually perpendicular.

Method: Prove mutual perpendicularity by all pairwise dot products

"Mutually perpendicular" means every pair among the lines meets at a right angle. For three lines that is three separate conditions — checking one or two pairs is not enough.

Steps

Step 1: Get a direction vector for each line. If direction cosines $(l,m,n)$ are given they are already unit direction vectors; otherwise use the direction ratios.

Step 2: Form every distinct pair. With three lines $A,B,C$ the pairs are $A\text{–}B$, $B\text{–}C$, $C\text{–}A$ — three in all.

Step 3: Test each pair with the dot product. Two directions are perpendicular exactly when

Compute this for all three pairs.

Step 4: Conclude only if all three dot products vanish. A single non-zero result means the set is not mutually perpendicular.

For $n$ lines this generalises to all $\binom{n}{2}$ pairs; the per-pair test is always the same dot-product-equals-zero condition.

Common Mistakes

Mistake 1: Checking only one or two pairs of lines.

Why it's wrong: "mutually perpendicular" requires every pair to be perpendicular — for three lines that is three dot products ($A$–$B$, $B$–$C$, $C$–$A$). Correct approach: verify all three vanish, not just the first.

Mistake 2: Re-normalising the given direction cosines before dotting.

Why it's wrong: direction cosines are already unit vectors ($l^2+m^2+n^2=1$), so dividing again is needless and error-prone. Correct approach: dot them directly; perpendicular means $l_1l_2+m_1m_2+n_1n_2=0$.

Mistake 3: A sign slip inside a dot product (e.g. mishandling $\frac{-3}{13}\cdot\frac{12}{13}$).

Why it's wrong: one wrong sign can hide a true zero. Correct approach: keep the common denominator $169$ and add the numerators carefully: $48-36-12=0$.