Angles — Line, Plane, and Direction Conditions

The four building blocks every later formula uses:

• Direction ratios (d.r.s): the components $(a, b, c)$ of any vector $\vec{d}$ along the line; for $\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$ they are the denominators.
• Direction cosines (d.c.s): $l = \dfrac{a}{|\vec{d}|}$, $m = \dfrac{b}{|\vec{d}|}$, $n = \dfrac{c}{|\vec{d}|}$, with $|\vec{d}| = \sqrt{a^2 + b^2 + c^2}$. They obey $l^2 + m^2 + n^2 = 1$.
• Plane normal: for $Ax + By + Cz = d$, the normal vector is $\vec{n} = (A, B, C)$.
• Dot and cross: $\vec{u}\cdot\vec{v} = u_1v_1 + u_2v_2 + u_3v_3$ measures alignment (zero ⟹ perpendicular); $\vec{n_1}\times\vec{n_2}$ gives a vector perpendicular to both — the direction of the line where two planes meet.

For lines with direction vectors $\vec{d_1} = (a_1, b_1, c_1)$ and $\vec{d_2} = (a_2, b_2, c_2)$, the acute angle $\theta$ between them satisfies:

For planes $A_1x + B_1y + C_1z = d_1$ and $A_2x + B_2y + C_2z = d_2$ with normals $\vec{n_1}, \vec{n_2}$:

• Solve-for-the-unknown: if one normal has an unknown (say $\alpha$) and $\theta$ is given, equate and square; you usually get a **quadratic in $\alpha$** with two roots.
• Difference of the values: for a quadratic $a\alpha^2 + b\alpha + c = 0$, the gap between the roots is $\dfrac{\sqrt{b^2 - 4ac}}{|a|}$ — no need to find each root.

For a line with direction $\vec{d}$ and a plane with normal $\vec{n}$, the line–plane angle $\theta$ is:

• Why sine: the line makes angle $(90^\circ - \theta)$ with the normal, and $\cos(90^\circ - \theta) = \sin\theta$.
• **Convert a given $\cos^{-1}$:** if $\theta = \cos^{-1}\!\left(\tfrac{2\sqrt2}{3}\right)$, then $\sin\theta = \sqrt{1 - \tfrac{8}{9}} = \tfrac{1}{3}$ — plug that into the formula.
• Solve: set the formula equal to $\sin\theta$, square, and solve the resulting equation for $\lambda$.

Three conditions, all from one idea (perpendicular ⟺ dot product zero):

• Two lines perpendicular: $\vec{d_1}\cdot\vec{d_2} = a_1a_2 + b_1b_2 + c_1c_2 = 0$.
• Two lines parallel: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ (proportional d.r.s).
• Line parallel to a plane: the line's direction lies IN the plane, so it is perpendicular to the normal: $\vec{d}\cdot\vec{n} = 0$. (For a point $P$ off the line and $Q$ on it, $\vec{PQ}$ parallel to the plane likewise needs $\vec{PQ}\cdot\vec{n} = 0$.)

Solve-for-the-unknown: when a coefficient $p$ or parameter $\mu$ appears, the single equation $\vec{d_1}\cdot\vec{d_2} = 0$ (or $\vec{PQ}\cdot\vec{n} = 0$) is linear — solve directly.

A line $\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$ lies in the plane $Ax + By + Cz = d$ iff:

• Point on plane: $(x_1, y_1, z_1)$ satisfies $Ax_1 + By_1 + Cz_1 = d$.
• Direction perpendicular to normal: $\vec{d}\cdot\vec{n} = aA + bB + cC = 0$.

Single-unknown version: if only the line's point carries an unknown $m$ (and the direction already satisfies condition 2), the point-on-plane equation alone gives $m$. Two-unknown version: the two conditions form a linear system in (say) $l, m$; solve for both, then read off whatever the question asks (e.g. $l^2 + m^2$).

• Equal-angle line: if a line makes $45^\circ$ with the $x$-axis and equal angles $\beta$ with $y$- and $z$-axes, then $\cos^2 45^\circ + 2\cos^2\beta = 1 \Rightarrow \cos^2\beta = \tfrac{1}{4} \Rightarrow \beta = 60^\circ$. The sum of the three angles is then $45^\circ + 60^\circ + 60^\circ = 165^\circ$.
• Two-constraint system: given $l + m + n = 0$ (linear) and a quadratic like $2l^2 + m^2 - n^2 = 0$: substitute $n = -(l+m)$ into the quadratic, factor to get two direction sets, then apply the line–line angle formula $\cos\theta = \dfrac{|l_1l_2 + m_1m_2 + n_1n_2|}{1\cdot 1}$ (d.c.s are already unit).