Line — Equation, Direction Cosines, and Vector Form

For a line with direction ratios $(a, b, c)$:

• The direction cosines are $l = \dfrac{a}{\sqrt{a^2+b^2+c^2}}$, $m = \dfrac{b}{\sqrt{a^2+b^2+c^2}}$, $n = \dfrac{c}{\sqrt{a^2+b^2+c^2}}$, where $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$ and $\alpha, \beta, \gamma$ are the angles with the X, Y, Z axes.
• They always satisfy the identity $l^2 + m^2 + n^2 = 1$, i.e. $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

So from any two of the three axis-angles you can recover the third, and a sign choice ($\pm$) decides which of the two supplementary angles the line makes.

A line through point $A(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$:

• Vector form: $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ is the position vector of $A$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is the direction.
• Symmetric (Cartesian) form: $\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$.

To convert: the constants in the numerators give the point, the denominators give the direction. The two forms describe exactly the same line.

The line through points $A(\vec{a})$ and $B(\vec{b})$ has:

• Direction $\vec{b} - \vec{a}$ (head minus tail).
• Vector form $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.

A line parallel to $AB$ but passing through a different point $P(\vec{p})$ keeps the same direction: $\vec{r} = \vec{p} + \lambda(\vec{b} - \vec{a})$.

Given $px - p_0 = qy - q_0 = rz - r_0$ (the $x, y, z$ coefficients $p, q, r$ are not 1):

• Factor each piece: $p\left(x - \tfrac{p_0}{p}\right) = q\left(y - \tfrac{q_0}{q}\right) = r\left(z - \tfrac{r_0}{r}\right)$.
• Divide through to symmetric form: $\dfrac{x - p_0/p}{1/p} = \dfrac{y - q_0/q}{1/q} = \dfrac{z - r_0/r}{1/r}$.

So the point is $\left(\tfrac{p_0}{p}, \tfrac{q_0}{q}, \tfrac{r_0}{r}\right)$ and the direction ratios are $\left(\tfrac{1}{p}, \tfrac{1}{q}, \tfrac{1}{r}\right)$ — multiply by the LCM to make them integers.

Compute the direction via the $3 \times 3$ determinant:

• the two lines' direction vectors (for a line $\perp$ both lines, or perpendicular to two given vectors), OR
• the two planes' normal vectors $\vec{n}_1, \vec{n}_2$ (for a line parallel to both planes, or the intersection of the two planes).

For a plane $ax + by + cz = d$, the normal is $(a, b, c)$. The resulting $\vec{d}$ gives the direction ratios directly; divide by $|\vec{d}|$ for direction cosines or a unit vector.

The unit vector perpendicular to two lines with directions $\vec{d}_1, \vec{d}_2$ is

For a line with direction $\vec{d} = (a, b, c)$:

• Angle $\alpha$ with the X-axis: $\cos\alpha = \dfrac{a}{\sqrt{a^2 + b^2 + c^2}}$. Similarly $\cos\beta = \dfrac{b}{|\vec{d}|}$, $\cos\gamma = \dfrac{c}{|\vec{d}|}$.
• Equally inclined to all axes: $|a| = |b| = |c|$; each $\cos = \dfrac{1}{\sqrt{3}}$, so the angle is $\cos^{-1}\tfrac{1}{\sqrt 3}$.

When a line is the intersection of two planes, first get its direction $\vec{n}_1 \times \vec{n}_2$, then take $\cos\alpha = a/|\vec{d}|$.

Given a linear relation (e.g. $l - 5m + 3n = 0$) and a quadratic relation (e.g. $7l^2 + 5m^2 - 3n^2 = 0$):

• Solve the linear relation for one variable, say $l = 5m - 3n$.
• Substitute into the quadratic → a homogeneous quadratic in $m, n$ → factor for the ratio $m : n$ (two cases).
• Back-substitute to get the full ratio $l : m : n$ for each case, then normalise (divide by $\sqrt{l^2 + m^2 + n^2}$) to get genuine direction cosines.

Two ratios usually emerge, so there are two valid lines — the answer often lists both.