Q.If $l_1, m_1, n_1$; $l_2, m_2, n_2$; $l_3, m_3, n_3$ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to $l_1 + l_2 + l_3$, $m_1 + m_2 + m_3$, $n_1 + n_2 + n_3$ makes equal angles with them.

Concept: Mutual Perpendicularity — For three mutually perpendicular lines, the direction cosines satisfy orthogonality conditions:

$l_i l_j + m_i m_j + n_i n_j = 0$ for $i \neq j$, and each set squares to 1.

Step 1: Let the direction ratios of the new line be $(l_1+l_2+l_3,\; m_1+m_2+m_3,\; n_1+n_2+n_3)$. The cosine of the angle $\theta_1$ with the first line is

$\cos\theta_1 = \frac{l_1(l_1+l_2+l_3) + m_1(m_1+m_2+m_3) + n_1(n_1+n_2+n_3)}{\sqrt{(l_1+l_2+l_3)^2 + (m_1+m_2+m_3)^2 + (n_1+n_2+n_3)^2}}.$ …

1. Understand the given and what we need to prove. We have three mutually perpendicular lines with direction cosines $(l_1, m_1, n_1)$, $(l_2, m_2, n_2)$, and $(l_3, m_3, n_3)$. “Mutually perpendicular” means each pair is orthogonal:

Also, since these are direction cosines, each set satisfies $l_i^2 + m_i^2 + n_i^2 = 1$ for $i = 1,2,3$.

We consider a new line whose direction ratios are $(l_1 + l_2 + l_3,\; m_1 + m_2 + m_3,\; n_1 + n_2 + n_3)$.

We need to show that this line makes equal angles with each of the three given lines.

2. What does “makes equal angles” mean in terms of direction cosines? If a line has direction cosines $(L, M, N)$, the cosine of the angle $\theta_i$ between it and the $i$-th given line is

So “equal angles” means $\cos\theta_1 = \cos\theta_2 = \cos\theta_3$.

For our new line, we don’t yet have its direction cosines — we have direction ratios. Let’s denote them as

The actual direction cosines of this line are $\left(\frac{a}{\sqrt{a^2+b^2+c^2}},\; \frac{b}{\sqrt{a^2+b^2+c^2}},\; \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$.

But since the denominator is the same for all three dot products, it’s enough to compare the unnormalised dot products $a l_i + b m_i + c n_i$ — they will all be equal if and only if the actual cosines are equal.

3. Compute the dot product with the first line.

The first bracket is $1$ (since it’s a direction cosine). The second bracket is $0$ (orthogonality of line 1 and line 2). The third bracket is $0$ (orthogonality of line 1 and line 3).

So the dot product equals $1$.

4. Now compute the dot product with the second line.

The first bracket is $0$, the second is $1$, the third is $0$. Again we get $1$.

5. Similarly for the third line.

So all three unnormalised dot products equal $1$.

6. Conclude that the angles are equal. …

Method: Angles made by the 'sum' line with an orthonormal set of directions

Use this proof pattern when three mutually perpendicular lines are given and you must analyse the line whose direction ratios are the sums of their corresponding direction cosines.

Steps

Step 1: Write down the two standing relations.

Mutual perpendicularity gives $l_il_j+m_im_j+n_in_j=0$ for $i\neq j$, and each set being direction cosines gives $l_i^2+m_i^2+n_i^2=1$.

Step 2: Dot the sum-vector with each given direction.

For the vector $(a,b,c)=(\sum l_i,\sum m_i,\sum n_i)$, compute $a\,l_k+b\,m_k+c\,n_k$. Every cross term is $0$ and exactly one square-sum is $1$, so each dot product collapses to $1$. …

Common Mistakes

Mistake 1: Treating $(l_1{+}l_2{+}l_3,\ m_1{+}m_2{+}m_3,\ n_1{+}n_2{+}n_3)$ as direction cosines.

Why it's wrong: they are direction ratios; the true cosine with each line is the dot product divided by $\sqrt{a^2+b^2+c^2}$. Correct approach: note the divisor is the same for all three, so equal raw dot products already prove equal angles.

Mistake 2: Not using the orthonormality relations. …