Q.Show that the straight lines whose direction cosines are given by $2l + 2m - n = 0$ and $mn + nl + lm = 0$ are at right angles.

Concept: Angle Between Lines — two lines are perpendicular if the sum of the products of their corresponding direction cosines is zero.

We are given:

Step 1: From the first equation, $n = 2l + 2m$. Substitute into the second:

Step 2: Simplify:

Step 3: Divide by $m^2$ (assuming $m \neq 0$; the case $m=0$ can be checked separately and yields the same conclusion):

This quadratic in $l/m$ gives two ratios, say $l_1/m_1$ and $l_2/m_2$, whose product is $1$ (constant term/coefficient of $t^2$). Hence:

$\frac{l_1}{m_1} \cdot \frac{l_2}{m_2} = 1 \quad\Rightarrow\quad l_1 l_2 = m_1 m_2.$ …

Why This Approach Works

When two lines are perpendicular in space, the dot product of their direction vectors is zero. For direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, this means:

The problem gives us two equations that both pairs of direction cosines satisfy. Instead of solving for individual values (which would be messy), we can work with the ratios of the direction cosines. The key insight: if we treat $l/m$ or $m/n$ as unknowns, the given equations become quadratic in these ratios. The two roots of that quadratic correspond to the two lines, and we can use sum and product of roots to directly check the perpendicular condition.

Step-by-Step Solution

1. Set up the equations

We have two lines whose direction cosines satisfy:

Let $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ be the direction cosines of the two lines.

2. Eliminate $n$ using equation (1)

From (1): $n = 2l + 2m$

Substitute into (2):

Expand:

Combine like terms:

3. Form a quadratic in the ratio $l/m$

Divide through by $m^2$ (assuming $m \neq 0$; we'll check the edge case later):

Let $t = l/m$. Then:

The two roots $t_1$ and $t_2$ correspond to $l_1/m_1$ and $l_2/m_2$ respectively.

Here, $a=2$, $b=5$, $c=2$, so:

4. Express the perpendicular condition in terms of ratios

We need to check: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$

From $n = 2l + 2m$, we have:

Substitute into the dot product:

Expand the last term:

Combine:

5. Factor using the ratios

Divide the entire expression by $m_1 m_2$ (again assuming $m_1, m_2 \neq 0$):

Let $t_1 = l_1/m_1$ and $t_2 = l_2/m_2$. Then the expression inside brackets becomes:

6. Plug in the sum and product

We have $t_1 t_2 = 1$ and $t_1 + t_2 = -\frac{5}{2}$.

Substitute:

Therefore $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$, proving the lines are perpendicular. …

Method: Testing perpendicularity of two lines defined by direction-cosine equations

Use this when two lines are described only by two equations their direction cosines satisfy (one linear, one quadratic) and you must show they are perpendicular — without finding the lines separately.

Steps

Step 1: Reduce to a quadratic in a ratio.

Eliminate one letter using the linear equation, substitute into the quadratic, and divide through by $m^2$ (or $l^2$) to get a quadratic in a single ratio $t=l/m$. Its two roots $t_1,t_2$ correspond to the two lines.

Step 2: Read off sum and product of roots.

For $at^2+bt+c=0$, use $t_1+t_2=-\tfrac{b}{a}$ and $t_1t_2=\tfrac{c}{a}$ — you never need the roots individually. …

Common Mistakes

Mistake 1: Solving for the individual direction cosines.

Why it's wrong: it is long and unnecessary — the test $l_1l_2+m_1m_2+n_1n_2=0$ can be checked straight from the sum and product of the ratio-roots. Correct approach: reduce to $2t^2+5t+2=0$ and use $t_1t_2$ and $t_1+t_2$.

Mistake 2: Ignoring the edge case $m=0$ before dividing by $m^2$.

Why it's wrong: dividing by $m^2$ silently assumes $m\neq0$. Correct approach: check $m=0$ separately (it forces all cosines to zero, which is impossible), so the division is safe. …