Three Dimensional Geometry

Every point you have ever plotted lived on a flat page — two coordinates and you were done. This chapter breaks out of the page. A point in space needs three coordinates, a direction needs three numbers too (not two angles but three linked cosines), and the two basic objects of solid geometry — the straight line and the flat plane — each acquire a compact vector recipe and an equally compact Cartesian one. The whole chapter is a translation exercise: everything you already know about lines and points in the plane gets re-expressed with one extra coordinate, using the vector algebra of the previous chapter as the working language. Three ideas carry the entire chapter. First, direction: a line's direction is captured by its direction cosines $l, m, n$ (with $l^2+m^2+n^2=1$ always) or, more conveniently, by any direction ratios proportional to them. Second, the line and the plane each have a vector equation — a point plus a direction, or a point plus a normal — that collapses into one clean Cartesian equation. Third, once you have two lines or two planes, the same four questions repeat throughout: are they parallel, are they perpendicular, what angle do they make, and how far apart are they? Skew lines — lines that are neither parallel nor intersecting, a genuinely three-dimensional situation with no analogue in plane geometry — force you to ask what "distance between two lines" even means, and the answer (project the join of any two points on them onto their common perpendicular direction) is one of the most elegant derivations in the whole CBSE syllabus. This is a formula-dense chapter, and every formula in it descends from just two vector operations — the dot product, which answers every question about angles and projections, and the cross product, which builds perpendicular directions and measures shortest distances. Learn the central derivations properly and the rest of the formula list becomes recoverable rather than memorised — precisely the standard CBSE boards, JEE and NEET expect.

Key takeaways

Direction cosines $l, m, n$ of a line satisfy $l^2+m^2+n^2=1$ ; any numbers $a, b, c$ proportional to them are its direction ratios, and they are not unique.
Direction ratios of the line joining $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ are $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ , with $PQ = \sqrt{\sum(x_2-x_1)^2}$ .
A line through a point with position vector $\vec a$ , parallel to $\vec b$ , has vector equation $\vec r=\vec a+\lambda\vec b$ and Cartesian form $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$ .
The angle between two lines uses $\cos\theta=\left|\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{\sum a_1^2}\sqrt{\sum a_2^2}}\right|$ ; the lines are perpendicular when the numerator is zero, parallel when the ratios match.
Skew lines (neither parallel nor intersecting) have a unique shortest distance along their common perpendicular, $d=\left|\dfrac{(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)}{|\vec b_1\times\vec b_2|}\right|$ ; parallel lines use a simpler cross-product formula instead.
A plane is fixed by a point and a normal vector, $(\vec r-\vec a)\cdot\vec N=0$ , or by three non-collinear points, or in intercept form $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ .
The distance of a point $(x_1,y_1,z_1)$ from the plane $Ax+By+Cz=D$ is $\dfrac{|Ax_1+By_1+Cz_1-D|}{\sqrt{A^2+B^2+C^2}}$ ; the angle between a line and a plane uses sine, not cosine.
Two lines are coplanar exactly when a $3\times3$ determinant built from their direction ratios and the join of their base points vanishes — the same determinant whose value, when nonzero, feeds the shortest-distance formula.
Key idea: every line is a point plus a direction and every plane is a point plus a normal, and the dot and cross products of those direction and normal vectors answer every question of angle, parallelism, perpendicularity, coplanarity and distance in the chapter.

The core insight

A line in space is a point plus a direction ( $\vec r=\vec a+\lambda\vec b$ , Cartesian $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$ ) and a plane is a point plus a normal ( $(\vec r-\vec a)\cdot\vec N=0$ ), so that every angle, parallelism, perpendicularity, coplanarity and distance question in this chapter reduces to a dot product or a cross product of these direction and normal vectors.

What this chapter covers

The chapter builds outward in a fixed order, and it helps to hold that order in mind while studying it. It opens with direction cosines and direction ratios, the numeric fingerprint of a direction in space. It then writes the equation of a line, first through a point and parallel to a given vector, then through two given points. Next come the angle between two lines and the special cases of parallel and perpendicular lines, followed by the shortest distance between two lines — first the general skew-line case, then the simpler parallel-line case. The second half of the chapter turns to the plane: its equation in normal form, in point-plus-normal form, through three non-collinear points, in intercept form, and as a family of planes through the intersection of two given planes. It closes with the angle between two planes, the angle between a line and a plane, the distance of a point from a plane, and finally the coplanarity of two lines, which ties the line half of the chapter to the plane half. Every idea here is available in both a vector form (compact, coordinate-free) and a Cartesian form (the one you actually compute with) — learning to translate fluently between the two is itself a chapter skill, and both forms are given for every result below.

The core ideas

Direction cosines and direction ratios of a line

If a directed line $L$ makes angles $\alpha, \beta, \gamma$ with the positive $x, y, z$ axes, these are its direction angles, and their cosines are its direction cosines, always written $l, m, n$ :

l=\cos\alpha,\quad m=\cos\beta,\quad n=\cos\gamma. \qquad (1)

where: $\alpha, \beta, \gamma$ are measured from the positive direction of each axis to the line. Reversing the line's direction replaces each angle by its supplement, flipping the sign of all three direction cosines — so a genuinely unique triple $(l, m, n)$ exists only once the line has been given a direction, that is, turned into a directed line. Because $l, m, n$ are the projections of a unit vector along $L$ onto the three axes,

l^2+m^2+n^2=1. \qquad (2)

This single identity is used constantly to check or to fix the sign of a missing direction cosine, and it is the chapter's most-quoted formula. Any three numbers $a, b, c$ proportional to $l, m, n$ — that is, $a=\lambda l$ , $b=\lambda m$ , $c=\lambda n$ for some nonzero $\lambda$ — are called direction ratios of the line. Squaring and adding, $\lambda^2(a^2+b^2+c^2)=1$ , so

l=\pm\dfrac{a}{\sqrt{a^2+b^2+c^2}},\quad m=\pm\dfrac{b}{\sqrt{a^2+b^2+c^2}},\quad n=\pm\dfrac{c}{\sqrt{a^2+b^2+c^2}}. \qquad (3)

where: the sign choice, all plus or all minus, fixes which of the line's two directions is meant. Unlike direction cosines, direction ratios are not unique even up to sign — $ka, kb, kc$ for any nonzero $k$ is an equally valid set — which is exactly why they are easier to work with in practice: normalising is postponed to the last step, if it is needed at all. For two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ , the direction ratios of line $PQ$ are simply $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ (or the negatives, for the opposite direction), and dividing each by

PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

gives the direction cosines of the directed line from $P$ to $Q$ . This is the workhorse formula for turning any two given points into a usable direction.

Figure

A three dimensional coordinate frame with x y and z axes meeting at the origin and a directed line rising from the origin making angles alpha beta and gamma with the three axes, labelled with the direction cosines l equals cos alpha, m equals cos beta, n equals cos gamma

Equation of a line in space

A line is fixed uniquely either by a point and a direction, or by two points. Let $A$ have position vector $\vec a$ and let the line pass through $A$ parallel to a vector $\vec b$ . For any point $P$ on the line with position vector $\vec r$ , the vector $\vec{AP}=\vec r-\vec a$ is parallel to $\vec b$ , so $\vec r-\vec a=\lambda\vec b$ for some scalar $\lambda$ , giving the vector equation

\vec r=\vec a+\lambda\vec b. \qquad (4)

where: $\lambda$ is a free real parameter sweeping out every point of the line as it varies, and $\vec b=a\hat i+b\hat j+c\hat k$ carries the direction ratios $a, b, c$ of the line directly as its components. Writing $\vec r=x\hat i+y\hat j+z\hat k$ and $\vec a=x_1\hat i+y_1\hat j+z_1\hat k$ and equating components gives the parametric equations $x=x_1+\lambda a,\ y=y_1+\lambda b,\ z=z_1+\lambda c$ ; eliminating $\lambda$ produces the Cartesian equation

\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}. \qquad (5)

If the direction cosines $l, m, n$ are used in place of $a, b, c$ , the same equation reads $\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}$ , with the advantage that any point on the line at signed distance $r$ from $A$ is exactly $(x_1+lr,\ y_1+mr,\ z_1+nr)$ . When the line is fixed by two points $A(\vec a)$ and $B(\vec b)$ instead, the direction vector is simply $\vec b-\vec a$ , giving

\vec r=\vec a+\lambda(\vec b-\vec a),\qquad \dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}. \qquad (6)

where: $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are the two given points; setting $\lambda=0$ recovers $A$ and $\lambda=1$ recovers $B$ .

Angle and shortest distance between two lines

Two lines with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ meet — or would meet, if translated to a common point — at an angle $\theta$ given by

\cos\theta=\left|\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\ \sqrt{a_2^2+b_2^2+c_2^2}}\right|,\qquad \sin\theta=\dfrac{\sqrt{(a_1b_2-a_2b_1)^2+(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2}}{\sqrt{a_1^2+b_1^2+c_1^2}\ \sqrt{a_2^2+b_2^2+c_2^2}}. \qquad (7)

where: the modulus keeps $\theta$ the acute angle between the lines, since a line has no preferred direction; in vector form the same statement is $\cos\theta=\left|\dfrac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}\right|$ . The lines are perpendicular exactly when $a_1a_2+b_1b_2+c_1c_2=0$ , and parallel exactly when $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ . Two lines in space either intersect, are parallel, or are skew — neither parallel nor intersecting, lying in no common plane, a case with no analogue in two-dimensional geometry (in a plane, two distinct non-parallel lines always meet). The classic picture is a rectangular room: a diagonal drawn across the ceiling and a diagonal drawn down an adjacent wall never meet and are never parallel — they are skew. Skew lines still have a well-defined, unique shortest distance between them, achieved along the one line segment perpendicular to both simultaneously; this is derived in full below. When the two lines happen to be parallel instead, the shortest distance reduces to the ordinary perpendicular distance from any point of one line to the other:

d=\left|\dfrac{\vec b\times(\vec a_2-\vec a_1)}{|\vec b|}\right|. \qquad (8)

Figure

A rectangular room drawn as a box in three dimensions with two diagonal line segments, one crossing the ceiling and one running from a top back corner down to a front floor corner, that pass near each other without touching, with a short dashed segment marked d joining their closest points to show the shortest distance between skew lines

The plane — standard forms of its equation

A plane, like a line, needs a point and one more piece of geometric data — but for a plane that extra data is a normal vector, a vector perpendicular to every line lying in the plane. If $\vec N$ is a normal and $A$ (position vector $\vec a$ ) is a known point of the plane, then $P(\vec r)$ lies in the plane exactly when $\vec{AP}=\vec r-\vec a$ is perpendicular to $\vec N$ :

(\vec r-\vec a)\cdot\vec N=0,\qquad\text{i.e.}\qquad A(x-x_1)+B(y-y_1)+C(z-z_1)=0, \qquad (9)

where: $\vec N=A\hat i+B\hat j+C\hat k$ , so $A, B, C$ are the direction ratios of the normal — and, on expanding, they are exactly the coefficients of $x, y, z$ in the plane's Cartesian equation $Ax+By+Cz=D$ . A special, very useful case is the normal form: if $\hat n$ is the unit normal and $d\ge0$ is the perpendicular distance of the plane from the origin, the plane is $\vec r\cdot\hat n=d$ , or in Cartesian form $lx+my+nz=d$ with $l, m, n$ the direction cosines of the normal. A plane through three non-collinear points is fixed without ever naming a normal explicitly — the normal is manufactured from the points themselves by a cross product, and the whole construction collapses into a single determinant, derived in full below:

\begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}=0. \qquad (10)

If a plane cuts off intercepts $a, b, c$ on the $x, y, z$ axes respectively, its equation takes the tidy intercept form

\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1. \qquad (11)

Finally, every plane through the line of intersection of two given planes $P_1: A_1x+B_1y+C_1z-D_1=0$ and $P_2: A_2x+B_2y+C_2z-D_2=0$ , except $P_2$ itself, is captured by the one-parameter family of planes

P_1+\lambda P_2=0,\qquad\text{i.e.}\qquad (A_1x+B_1y+C_1z-D_1)+\lambda(A_2x+B_2y+C_2z-D_2)=0. \qquad (12)

where: $\lambda$ is chosen afterwards to satisfy one further condition, most often that the plane passes through a named third point — this is the standard shortcut for "find the plane through this line of intersection and this extra point," and it avoids ever finding the line of intersection explicitly.

Angle between planes, a line and a plane, and the distance from a point

Two planes with normals $\vec N_1=(A_1,B_1,C_1)$ and $\vec N_2=(A_2,B_2,C_2)$ meet at a dihedral angle $\theta$ equal to the angle between their normals:

\cos\theta=\left|\dfrac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\ \sqrt{A_2^2+B_2^2+C_2^2}}\right|. \qquad (13)

The planes are perpendicular when $A_1A_2+B_1B_2+C_1C_2=0$ and parallel when $\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}$ . A line with direction ratios $a, b, c$ meets a plane with normal direction ratios $A, B, C$ at an angle $\varphi$ measured from the plane itself, not from the normal, so $\varphi$ is complementary to the angle between the two direction vectors:

\sin\varphi=\left|\dfrac{aA+bB+cC}{\sqrt{a^2+b^2+c^2}\ \sqrt{A^2+B^2+C^2}}\right|. \qquad (14)

where: the line is parallel to the plane precisely when $aA+bB+cC=0$ , its direction vector perpendicular to the normal, and perpendicular to the plane precisely when $\dfrac{a}{A}=\dfrac{b}{B}=\dfrac{c}{C}$ , its direction vector parallel to the normal — the reverse pairing from the plane-plane case, and a frequent source of confusion. The perpendicular distance of a point $Q(x_1,y_1,z_1)$ from the plane $Ax+By+Cz=D$ is obtained by projecting the join of $Q$ and any point of the plane onto the unit normal direction, derived in full below:

d=\dfrac{|Ax_1+By_1+Cz_1-D|}{\sqrt{A^2+B^2+C^2}}. \qquad (15)

Coplanarity of two lines

Two lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ lie in a common plane exactly when the vector from a point of the first to a point of the second, $\vec a_2-\vec a_1$ , is coplanar with $\vec b_1$ and $\vec b_2$ — equivalently, when the scalar triple product vanishes:

(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=0. \qquad (16)

In Cartesian form, with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ and points $(x_1,y_1,z_1)$ , $(x_2,y_2,z_2)$ , the condition is the vanishing of exactly the same numerator determinant that appears in the shortest-distance formula:

\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{vmatrix}=0. \qquad (17)

where: this is exactly the statement that the shortest distance between the two lines is zero — coplanar lines are precisely the lines for which the general skew-line formula collapses, either because they intersect ( $\theta\ne0$ ) or because they are parallel. Checking coplanarity is therefore the natural first step of any shortest-distance calculation, not a separate topic.

Derivation AShortest distance between two skew lines

The goal is to find the length of the one line segment that meets both of two skew lines at right angles — the shortest possible join between them — starting only from a point and a direction vector on each line, using nothing beyond the dot and cross product built in the vector algebra chapter. Skew lines never meet and are never parallel, so no elementary distance formula applies directly; instead the direction of the common perpendicular is manufactured first, and only then is the actual separation of the two lines measured along it.

Step 1. Let the two skew lines be $l_1:\vec r=\vec a_1+\lambda\vec b_1$ and $l_2:\vec r=\vec a_2+\mu\vec b_2$ , with $S$ a chosen point on $l_1$ (position vector $\vec a_1$ ) and $T$ a chosen point on $l_2$ (position vector $\vec a_2$ ). The vector $\vec{ST}=\vec a_2-\vec a_1$ joins the two lines, but it is not, in general, the shortest such join — it depends entirely on which points $S$ and $T$ happened to be picked.

Step 2. Construct the one direction that does not depend on the choice of $S$ and $T$ : the line of shortest distance $PQ$ must be perpendicular to both $\vec b_1$ and $\vec b_2$ simultaneously, since it is the unique segment realising the minimum separation of the two lines. The only vector, up to scale, perpendicular to both $\vec b_1$ and $\vec b_2$ is their cross product, so the unit vector along $PQ$ is $\hat n=\dfrac{\vec b_1\times\vec b_2}{|\vec b_1\times\vec b_2|}$ .

Step 3. Relate the fixed vector $\vec{ST}$ to the unknown shortest distance. Whatever the true feet $P$ (on $l_1$ ) and $Q$ (on $l_2$ ) of the common perpendicular turn out to be, the length $PQ$ is exactly the projection of $ST$ onto the fixed direction $\hat n$ , because moving from $S$ to $P$ or from $T$ to $Q$ along the lines only ever adds components parallel to $\vec b_1$ or $\vec b_2$ , both of which vanish upon dotting with $\hat n$ , since $\hat n$ is perpendicular to both. So, if $d$ is the unsigned shortest distance, $d=PQ=|\vec{ST}\cdot\hat n|$ .

Step 4. Substitute Steps 1 and 2 into Step 3 to reach the working formula. Since $\vec{ST}=\vec a_2-\vec a_1$ and $\hat n=\dfrac{\vec b_1\times\vec b_2}{|\vec b_1\times\vec b_2|}$ ,

d=\left|\dfrac{(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)}{|\vec b_1\times\vec b_2|}\right|.

Step 5. Confirm the formula degrades sensibly at its edges. If the lines actually intersect, $\vec a_2-\vec a_1$ lies in the plane spanned by $\vec b_1,\vec b_2$ , the numerator vanishes, and $d=0$ — exactly the coplanarity condition of equation (16). If the lines are parallel instead, $\vec b_1\times\vec b_2=\vec 0$ and the formula breaks down as a division by zero, which is why the parallel case needs its own formula, equation (8), obtained by the same projection argument but onto a normal to the single shared direction $\vec b$ .

Go deeper · Derivation B · 2 minEquation of a plane through three non-collinear points

Derivation BEquation of a plane through three non-collinear points

The goal is to write down the equation of the unique plane through three given non-collinear points $A(x_1,y_1,z_1)$ , $B(x_2,y_2,z_2)$ , $C(x_3,y_3,z_3)$ without ever being handed a normal vector directly — the three points must manufacture their own normal, and the final answer should reduce, in a single stroke, to a determinant equation that can be written straight from the coordinates.

Step 1. Since $A, B, C$ are non-collinear, the two vectors $\vec{AB}=(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$ and $\vec{AC}=(x_3-x_1,\,y_3-y_1,\,z_3-z_1)$ are not parallel, so they span a genuine plane, and their cross product $\vec N=\vec{AB}\times\vec{AC}$ is a nonzero vector perpendicular to both — hence perpendicular to the whole plane through $A, B, C$ . This $\vec N$ is exactly the normal vector the point-plus-normal formula, equation (9), needs.

Step 2. Let $P$ , with position vector $\vec r=(x,y,z)$ , be any point of the plane. Because $\vec N\perp\vec{AB}$ and $\vec N\perp\vec{AC}$ , and $P$ lies in the plane spanned by $A, B, C$ , the vector $\vec{AP}=\vec r-\vec a$ is a linear combination of $\vec{AB}$ and $\vec{AC}$ , so it is automatically perpendicular to $\vec N$ as well: $\vec{AP}\cdot\vec N=0$ . This is exactly equation (9) with $\vec N=\vec{AB}\times\vec{AC}$ substituted in.

Step 3. Expand the condition using the scalar triple product identity: a vector $\vec u$ is perpendicular to $\vec v\times\vec w$ exactly when the scalar triple product $\vec u\cdot(\vec v\times\vec w)$ vanishes, and this triple product is precisely the determinant with $\vec u,\vec v,\vec w$ as its three rows. Taking $\vec u=\vec{AP}=(x-x_1,y-y_1,z-z_1)$ , $\vec v=\vec{AB}$ , $\vec w=\vec{AC}$ gives directly

\begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}=0,

which is the required equation, and it is manifestly satisfied at $(x,y,z)=A,B,C$ in turn — each substitution collapses the determinant to zero, either by repeating a row or by making the first row a combination of the other two.

Step 4. Read off the coplanarity link. The determinant in Step 3 is structurally the same object as the numerator of the coplanarity test for two lines, equation (17): both express "the vector from a base point to a target point lies in the plane spanned by two given directions" as the vanishing of a scalar triple product. Once expanded along the first row, the determinant is linear in $x, y, z$ , so it always reduces to the Cartesian form $Ax+By+Cz=D$ with $A, B, C$ exactly the components of $\vec N=\vec{AB}\times\vec{AC}$ .

Go deeper · Derivation C · 1 minDistance of a point from a plane

Derivation CDistance of a point from a plane

The goal is to derive the compact Cartesian formula $d=\dfrac{|Ax_1+By_1+Cz_1-D|}{\sqrt{A^2+B^2+C^2}}$ for the perpendicular distance of a point $Q(x_1,y_1,z_1)$ from the plane $Ax+By+Cz=D$ , starting from nothing more than the definition of "distance from a plane" as the length of the perpendicular dropped onto it, together with the point-plus-normal equation of a plane already established.

Step 1. Let $\vec N=(A,B,C)$ be a normal to the plane and $\hat N=\vec N/|\vec N|$ the corresponding unit normal. Pick any point $R$ of the plane with position vector $\vec p=(p,q,s)$ , so $Ap+Bq+Cs=D$ simply because $R$ lies on the plane. Let $Q$ have position vector $\vec q_0=(x_1,y_1,z_1)$ .

Step 2. The perpendicular distance from $Q$ to the plane is, by definition, the length of the projection of $\vec{RQ}=\vec q_0-\vec p$ onto the normal direction $\hat N$ — every component of $\vec{RQ}$ that runs along the plane contributes nothing to how far $Q$ sits off the plane, only the component along $\hat N$ does. So $d=|\vec{RQ}\cdot\hat N|=\left|\dfrac{(\vec q_0-\vec p)\cdot\vec N}{|\vec N|}\right|$ .

Step 3. Expand the dot product in coordinates: $(\vec q_0-\vec p)\cdot\vec N=A(x_1-p)+B(y_1-q)+C(z_1-s)=(Ax_1+By_1+Cz_1)-(Ap+Bq+Cs)$ . The bracketed second term is exactly $D$ , by Step 1, so the unknown reference point $R$ drops out entirely: $(\vec q_0-\vec p)\cdot\vec N=Ax_1+By_1+Cz_1-D$ .

Step 4. Substitute back into Step 2, with $|\vec N|=\sqrt{A^2+B^2+C^2}$ , to get the final formula, valid for any choice of the reference point $R$ , which is exactly why $R$ never needs to appear in the answer:

d=\dfrac{|Ax_1+By_1+Cz_1-D|}{\sqrt{A^2+B^2+C^2}}.

Step 5. Specialise to the origin, $(x_1,y_1,z_1)=(0,0,0)$ : $d=\dfrac{|-D|}{\sqrt{A^2+B^2+C^2}}=\dfrac{|D|}{\sqrt{A^2+B^2+C^2}}$ , which matches the normal-form distance $d$ in $\vec r\cdot\hat n=d$ once the plane's equation is divided through by $|\vec N|$ to make the normal a unit vector — confirming that the normal form of equation (9) is simply this general formula applied at the origin.

Worked examples

Board level - direction cosines and the equation of a line through two given points

Find the direction cosines of the line joining $P(-2,4,-5)$ and $Q(1,2,3)$ , and hence write the vector and Cartesian equations of the line $PQ$ .

Step 1. Direction ratios of $PQ$ are $x_2-x_1,\,y_2-y_1,\,z_2-z_1=1-(-2),\,2-4,\,3-(-5)=3,\,-2,\,8$ .

Step 2. Length $PQ=\sqrt{3^2+(-2)^2+8^2}=\sqrt{9+4+64}=\sqrt{77}$ , so the direction cosines are $\dfrac{3}{\sqrt{77}},\ \dfrac{-2}{\sqrt{77}},\ \dfrac{8}{\sqrt{77}}$ .

Step 3. Taking $A=P$ as the base point with position vector $\vec a=-2\hat i+4\hat j-5\hat k$ and direction vector $\vec b=3\hat i-2\hat j+8\hat k$ from Step 1, the vector equation of line $PQ$ is $\vec r=(-2\hat i+4\hat j-5\hat k)+\lambda(3\hat i-2\hat j+8\hat k)$ .

Step 4. Eliminating $\lambda$ gives the Cartesian form directly from equation (5): $\dfrac{x+2}{3}=\dfrac{y-4}{-2}=\dfrac{z+5}{8}$ . Answer: direction cosines $\left(\dfrac{3}{\sqrt{77}},\dfrac{-2}{\sqrt{77}},\dfrac{8}{\sqrt{77}}\right)$ ; line $\vec r=(-2\hat i+4\hat j-5\hat k)+\lambda(3\hat i-2\hat j+8\hat k)$ , that is $\dfrac{x+2}{3}=\dfrac{y-4}{-2}=\dfrac{z+5}{8}$ .

Board level - equation of a plane through three points and its distance from the origin

Find the equation of the plane through the points $A(1,0,0)$ , $B(2,1,0)$ and $C(1,1,1)$ , and find its distance from the origin.

Step 1. Form two vectors lying in the plane: $\vec{AB}=(2-1,\,1-0,\,0-0)=(1,1,0)$ and $\vec{AC}=(1-1,\,1-0,\,1-0)=(0,1,1)$ .

Step 2. Compute the normal, $\vec N=\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&0\\0&1&1\end{vmatrix}=\hat i(1\cdot1-0\cdot1)-\hat j(1\cdot1-0\cdot0)+\hat k(1\cdot1-1\cdot0)=\hat i-\hat j+\hat k$ , so $A=1,\ B=-1,\ C=1$ .

Step 3. Using the point $A(1,0,0)$ in equation (9): $1(x-1)-1(y-0)+1(z-0)=0$ , which simplifies to $x-y+z=1$ .

Step 4. Distance from the origin, using equation (15) with $(x_1,y_1,z_1)=(0,0,0)$ and $D=1$ : $d=\dfrac{|1(0)-1(0)+1(0)-1|}{\sqrt{1^2+(-1)^2+1^2}}=\dfrac{1}{\sqrt3}$ . Answer: the plane is $x-y+z=1$ , at perpendicular distance $\dfrac{1}{\sqrt3}$ from the origin.

Competitive level (JEE / NEET) - shortest distance between skew lines and the angle a line makes with a plane

Two lines are given by $l_1:\vec r=(\hat i+\hat j)+\lambda(2\hat i-\hat j+\hat k)$ and $l_2:\vec r=(2\hat i+\hat j-\hat k)+\mu(3\hat i-5\hat j+2\hat k)$ . Find the shortest distance between $l_1$ and $l_2$ , and then find the angle that $l_1$ makes with the plane $x-y+z=1$ found in the previous example.

Step 1. Read off $\vec a_1=\hat i+\hat j$ , $\vec b_1=2\hat i-\hat j+\hat k$ , $\vec a_2=2\hat i+\hat j-\hat k$ , $\vec b_2=3\hat i-5\hat j+2\hat k$ , so $\vec a_2-\vec a_1=\hat i-\hat k$ .

Step 2. Compute $\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\3&-5&2\end{vmatrix}=\hat i(-2+5)-\hat j(4-3)+\hat k(-10+3)=3\hat i-\hat j-7\hat k$ , with magnitude $|\vec b_1\times\vec b_2|=\sqrt{9+1+49}=\sqrt{59}$ .

Step 3. Shortest distance, using the derivation above: $d=\left|\dfrac{(3\hat i-\hat j-7\hat k)\cdot(\hat i-\hat k)}{\sqrt{59}}\right|=\left|\dfrac{3-0+7}{\sqrt{59}}\right|=\dfrac{10}{\sqrt{59}}$ .

Step 4. For the line-plane angle, take $\vec b_1=(2,-1,1)$ as the line direction and $\vec N=(1,-1,1)$ as the plane's normal from the previous example. By equation (14), $\sin\varphi=\left|\dfrac{(2)(1)+(-1)(-1)+(1)(1)}{\sqrt{4+1+1}\sqrt{1+1+1}}\right|=\left|\dfrac{2+1+1}{\sqrt6\sqrt3}\right|=\dfrac{4}{\sqrt{18}}=\dfrac{4}{3\sqrt2}=\dfrac{2\sqrt2}{3}$ .

Step 5. Since $\dfrac{2\sqrt2}{3}\approx0.943$ , $\varphi=\sin^{-1}\!\left(\dfrac{2\sqrt2}{3}\right)\approx70.5^\circ$ . Answer: shortest distance $=\dfrac{10}{\sqrt{59}}$ ; angle between $l_1$ and the plane is $\sin^{-1}\dfrac{2\sqrt2}{3}\approx70.5^\circ$ . Timing note: this two-part JEE-style question should take under four minutes once the cross-product mechanics are automatic. Compute $\vec b_1\times\vec b_2$ once and reuse it for the shortest-distance dot product — do not recompute it for the second part. The commonest slip under time pressure is plugging the line's direction ratios into the plane-plane angle formula (cosine, no reference to sine) instead of equation (14); say out loud which formula applies — line versus plane means sine, plane versus plane or line versus line means cosine — before writing a single number.

Common traps

Forgetting the modulus in the angle-between-lines formula. Raw direction ratios have no fixed sign, so $\cos\theta$ from the plain dot-product formula can come out negative; always take the absolute value so $\theta$ stays the acute angle between the lines, as in equation (7).
Confusing direction ratios with direction cosines. Direction ratios $a, b, c$ are not unique and do not satisfy $a^2+b^2+c^2=1$ in general — only the normalised $l, m, n$ do. Never plug raw direction ratios into a formula that expects direction cosines without dividing by their magnitude first.
Swapping the plane-plane and line-plane angle formulas. The angle between two planes uses cosine on the two normals, equation (13); the angle between a line and a plane uses sine, comparing the line's direction to the plane's normal, equation (14) — because the angle worth reporting for a line meeting a plane is measured from the plane, not from the normal.
Using the skew-line shortest-distance formula on parallel lines. If $\vec b_1\times\vec b_2=\vec 0$ , the general formula divides by zero; switch to the dedicated parallel-line formula, equation (8), instead.
Sign inconsistency in the plane's constant. Some sources write the plane as $Ax+By+Cz+D=0$ , others as $Ax+By+Cz=D$ ; the distance formula's sign convention changes accordingly. Fix one convention — this article uses $Ax+By+Cz=D$ — and keep it for an entire problem.
Treating coplanar as automatically parallel. Coplanar lines can equally well intersect; coplanarity only rules out the skew case. Always check the determinant condition, equation (17), rather than assuming parallelism.

Common trapSkew lines are not simply non-parallel lines - they share no common plane at all, and that is the idea students get wrong most often

The single most common conceptual slip in this chapter is picturing skew lines as just "two lines that happen not to be parallel," and then expecting them to intersect somewhere if extended far enough, the way two non-parallel lines in a plane always do. In three dimensions that guarantee disappears completely: two lines can fail to be parallel and still never meet, because they live in different planes with no shared plane containing both of them. Every angle formula between two lines, equation (7), is silently computed as if the lines had been translated to intersect at a common point — it measures the angle between their directions, not any real point of intersection — which is exactly why skew lines still have a well-defined angle between them even though they never actually cross. Keep the two questions separate in your head: "what angle do these directions make" is always answerable, while "do these lines actually meet" is answerable only through the coplanarity determinant, equation (17), never by inspecting the angle alone.

You should be able to…

Find the direction cosines and direction ratios of a line from two given points, and verify the identity $l^2+m^2+n^2=1$ .
Write the vector and Cartesian equations of a line through a point and a direction, or through two points, and translate fluently between the two forms.
Find the angle between two lines from their direction ratios or direction cosines, and decide when they are parallel or perpendicular.
Distinguish skew lines from intersecting and parallel lines, and compute the shortest distance between two skew lines using the cross-product formula.
Compute the distance between two parallel lines.
Write the equation of a plane in normal form, in point-plus-normal form, through three non-collinear points, and in intercept form.
Find the angle between two planes and the angle between a line and a plane, without confusing which formula needs sine and which needs cosine.
Find the perpendicular distance of a given point from a plane.
Test whether two lines are coplanar using the determinant condition, and connect a zero result to the shortest-distance formula.
Build the family of planes through the intersection of two given planes and pin down the parameter using an extra condition.

Ask the tutor: if you are stuck on whether two lines in a problem are skew, parallel or intersecting, ask the tutor to walk through the coplanarity determinant with your exact numbers.

What is the actual difference between direction cosines and direction ratios?

Direction cosines $l, m, n$ are the cosines of the angles a line makes with the $x$ , $y$ , $z$ axes, and they always satisfy $l^2+m^2+n^2=1$ — one such triple per direction. Direction ratios $a, b, c$ are any numbers proportional to $l, m, n$ , so they need not square-sum to $1$ and are never unique: $(3,-2,8)$ and $(6,-4,16)$ describe the same direction. In practice, compute direction ratios first (subtraction of coordinates gives them for free), and normalise into direction cosines only when a formula specifically needs them.

How do I tell that two lines are skew without trying to visualise them in 3D?

Do not rely on a mental picture — use the coplanarity determinant, equation (17). Take the two lines' direction ratios as the second and third rows and the vector joining a point on each line as the first row; if that determinant is zero, the lines are coplanar (either intersecting or parallel), and if it is nonzero, they are skew. As a quick pre-check, see whether the direction ratios are proportional first — if they are, the lines are parallel, not skew, and the determinant is not even needed.

Why does the shortest-distance formula use a cross product instead of just subtracting the two given points?

Because the vector joining any arbitrarily chosen point on each line is not, by itself, the shortest possible join — it depends on which two points you picked, and only one particular segment, perpendicular to both lines at once, is actually the minimum. The cross product $\vec b_1\times\vec b_2$ is the unique direction perpendicular to both lines' directions simultaneously, so projecting the join of any two chosen points onto that fixed direction always gives the same answer regardless of which points were chosen — exactly what the derivation above proves.

Is the plane equation Ax + By + Cz = D the same thing as the normal form?

Yes, up to one rescaling: the normal form $lx+my+nz=d$ is the special case where the coefficients have been divided by $\sqrt{A^2+B^2+C^2}$ so that $l, m, n$ become the direction cosines of the normal (satisfying $l^2+m^2+n^2=1$ ) rather than arbitrary direction ratios, and $d\ge0$ is then automatically the plane's perpendicular distance from the origin. Dividing any plane $Ax+By+Cz=D$ by $\sqrt{A^2+B^2+C^2}$ , with the sign chosen so the right side is non-negative, converts it to normal form directly.

How much of this chapter actually shows up in JEE Main and Advanced compared to boards?

Boards test the line formulas heavily — direction cosines, the equations of a line, the angle between two lines, and the shortest distance between skew or parallel lines. JEE and NEET go further into the plane half: equations of a plane through three points or via the family-of-planes trick, the angle between a line and a plane, distance of a point from a plane, and especially coplanarity of two lines combined with finding an unknown constant, a favourite multi-step JEE pattern. Both levels expect fluent, hesitation-free translation between vector and Cartesian forms.