Three Dimensional Geometry

Two lines. Neither is parallel to the other. Find where they meet.

$L_1:\frac{x}{1}=\frac{y}{1}=\frac{z}{1},L_2:\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{3}$

Their direction ratios are $(1,1,1)$ and $(1,2,3)$, which are not proportional, so they are certainly not parallel. In a plane that would settle it: non-parallel lines meet, always, exactly once.

Write a general point on each and set them equal.

$t=1+s,t=1+2s,t=3s$

The first two give $s=0$ and $t=1$ immediately. Substitute into the third.

$1=0$

The lines do not meet. They are skew: not parallel, and yet with no point in common.

That leftover equation is the whole chapter. Two unknowns can satisfy two of the three equations always; whether they also satisfy the third is a condition, not a certainty. When it holds, the lines are coplanar and meet. When it fails, they miss, and the amount by which the equation fails measures how far apart they pass.

$\text{shortest distance}=\frac{|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|}{|{\vec{b}}_1\times {\vec{b}}_2|}=\frac{|-1|}{\sqrt{6}}=\frac{1}{\sqrt{6}}$

Everything below is the machinery for writing lines and planes in space, and every condition in it — perpendicularity, coplanarity, a line lying in a plane — is one equation being satisfied rather than assumed.

1. Coordinates in Space

A point needs three coordinates. Distance extends the plane formula by one term, and the section formula extends componentwise.

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

$\text{dividing }P,Q\text{ in }m:n\Rightarrow \left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)$

For external division, replace $n$ by $-n$. The midpoint is the case $m=n$.

Illustration 1

Find the point dividing $A(1,-2,3)$ and $B(3,4,-5)$ internally in the ratio $2:3$.

Apply the formula to each coordinate separately; there is nothing three-dimensional about the work.

$x=\frac{2(3)+3(1)}{5}=\frac{9}{5},y=\frac{2(4)+3(-2)}{5}=\frac{2}{5},z=\frac{2(-5)+3(3)}{5}=-\frac{1}{5}$

$\left(\frac{9}{5},\frac{2}{5},-\frac{1}{5}\right)$

Check the ratio holds by comparing one coordinate's two gaps: from $1$ to $9/5$ is $4/5$, and from $9/5$ to $3$ is $6/5$, in the ratio $2:3$ as required.

2. Direction Cosines and Direction Ratios

Direction cosines $(l,m,n)$ are the cosines of the angles a line makes with the positive $x$, $y$ and $z$ axes.

$l^2+m^2+n^2=1$

Direction ratios $(a,b,c)$ are any numbers proportional to them. A line has infinitely many sets of direction ratios but only two sets of direction cosines, opposite in sign, one for each direction along it.

$l=\frac{a}{\sqrt{a^2+b^2+c^2}},\text{and similarly for }m,n$

Illustration 2

A line makes $45^{\circ }$ with the $x$-axis and $60^{\circ }$ with the $y$-axis. What angle does it make with the $z$-axis?

The three angles are not independent, so the third is forced by the identity.

${\cos }^{2}45^{\circ }+{\cos }^{2}60^{\circ }+{\cos }^2\gamma =1$

$\frac{1}{2}+\frac{1}{4}+{\cos }^2\gamma =1\Rightarrow {\cos }^2\gamma =\frac{1}{4}\Rightarrow \cos \gamma =\pm \frac{1}{2}$

$\gamma =60^{\circ }\text{ or }120^{\circ }$

Two answers, and both are genuine. The identity fixes ${\cos }^2\gamma$, not $\cos \gamma$, and the two signs correspond to the two directions along the same line. Quoting only $60^{\circ }$ is the standard omission here.

Trap. Never write $\cos \alpha +\cos \beta +\cos \gamma =1$. It is the squares that sum to $1$, and the un-squared version is false for almost every line.

3. Angle Between Two Lines

With direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$:

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

The modulus in the numerator forces the acute angle, which is what "the angle between two lines" always means. Two skew lines still have an angle between them: translate one until they meet and measure there.

Illustration 3

Find the angle between the lines with direction ratios $(1,2,2)$ and $(2,-2,1)$.

Compute the dot product of the ratios first, because a zero there ends the question.

$1(2)+2(-2)+2(1)=2-4+2=0$

$\theta =90^{\circ }$

No denominators were needed. Both vectors happen to have length $3$, so the formula would have given $0/9$, but the numerator alone settles perpendicularity and it is always worth computing first.

4. Equation of a Line

A line is fixed by one point on it and one direction.

$\text{vector form: }\vec{r}=\vec{a}+\lambda \vec{b}$

$\text{Cartesian: }\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

The denominators are direction ratios and the numerators locate the fixed point. Through two points, the direction ratios are the coordinate differences.

Trap. A zero denominator does not mean division by zero; it means that coordinate is constant. Writing $\frac{x-1}{0}=\frac{y-2}{3}=\frac{z}{4}$ is standard shorthand for the line $x=1$ with the other two coordinates varying.

5. Skew Lines and the Shortest Distance

Two lines in space are parallel, intersecting, or skew. The third case has no analogue in the plane, and it is the default: two lines picked at random in space are almost certainly skew.

$\text{coplanar}⟺({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)=0$

That expression is the scalar triple product of the joining vector and the two directions. It is the volume of the box they span, and a flat box means everything lies in one plane.

$d=\frac{|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|}{|{\vec{b}}_1\times {\vec{b}}_2|}$

For parallel lines the cross product vanishes and this formula is useless. Use the other one.

$d_{\text{parallel}}=\frac{|({\vec{a}}_2-{\vec{a}}_1)\times \vec{b}|}{|\vec{b}|}$

Illustration 4

Compute the shortest distance between the two lines from the opening.

${\vec{a}}_1=\vec{0},{\vec{b}}_1=(1,1,1);{\vec{a}}_2=(1,1,0),{\vec{b}}_2=(1,2,3)$

Take the cross product of the directions first; it is needed in both numerator and denominator.

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}i& j& k\\ 1& 1& 1\\ 1& 2& 3\end{array}\right|=(3-2)i-(3-1)j+(2-1)k=(1,-2,1)$

$({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)=(1)(1)+(1)(-2)+(0)(1)=-1$

Non-zero, confirming the lines are skew, which is what the failed intersection already told us.

$d=\frac{|-1|}{\sqrt{1+4+1}}=\frac{1}{\sqrt{6}}\approx 0.408$

Notice that the numerator is exactly the inconsistency in the third equation. The algebra that refused to solve and the geometry that refuses to meet are the same fact.

Illustration 5

Find $k$ so that the lines below are coplanar.

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4},\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$

Coplanarity is the triple product vanishing, so set it up and solve for $k$.

${\vec{a}}_2-{\vec{a}}_1=(2,k+1,-1),{\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}i& j& k\\ 2& 3& 4\\ 1& 2& 1\end{array}\right|=(-5,2,1)$

$2(-5)+(k+1)(2)+(-1)(1)=-10+2k+2-1=2k-9$

$2k-9=0\Rightarrow k=\frac{9}{2}$

For every other value of $k$ the lines are skew, and the shortest distance is $|2k-9|/\sqrt{30}$. One parameter, one condition, one answer: that is the shape of most coplanarity questions.

6. Equation of a Plane

A plane is fixed by one point and one normal direction. Everything else is a rearrangement of that.

The coefficients are the normal. That single reading answers most plane questions without any further formula: parallel planes share $(a,b,c)$, perpendicular planes have normals with zero dot product, and a line lies parallel to a plane when its direction is perpendicular to the normal.

Illustration 6

Find the plane through $(2,1,0)$, $(3,-2,-2)$ and $(3,1,7)$.

Build two vectors in the plane, then cross them to get the normal.

${\vec{v}}_1=(1,-3,-2),{\vec{v}}_2=(1,0,7)$

${\vec{v}}_1\times {\vec{v}}_2=\left(-21,-9,3\right)=-3(7,3,-1)$

Scale away the common factor, since only the direction of the normal matters.

$7(x-2)+3(y-1)-1(z-0)=0\Rightarrow 7x+3y-z=17$

Check the two points not used to anchor it: $21-6+2=17$ and $21+3-7=17$. Both lie on it, so all three do, and the plane is right.

The family of planes through a line

Two intersecting planes meet in a line, and every plane containing that line is a combination of the two.

$P_1+\lambda P_2=0$

This is the exact analogue of the family of lines through a point of intersection, and it earns its place for the same reason: it produces the answer without ever finding the line of intersection.

Illustration 7

Find the plane through the line of intersection of $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to $x-y+z=0$.

Write the family, so the line of intersection never has to be found.

$(x+y+z-1)+\lambda (2x+3y+4z-5)=0$

Collect the coefficients, because they are the normal to whichever member you end up choosing.

$(1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z-(1+5\lambda )=0$

Perpendicular planes have normals with zero dot product, and the given plane's normal is $(1,-1,1)$.

$(1+2\lambda )-(1+3\lambda )+(1+4\lambda )=1+3\lambda =0\Rightarrow \lambda =-\frac{1}{3}$

Substituting gives coefficients $\frac{1}{3},0,-\frac{1}{3}$ and constant $\frac{2}{3}$. Multiplying through by $3$ clears the fractions.

$x-z+2=0$

Verify the condition rather than trusting the arithmetic: the normal $(1,0,-1)$ dotted with $(1,-1,1)$ gives $1-0-1=0$. The $y$ coefficient vanishing is the visible sign that $\lambda$ was chosen correctly.

7. Angles Between Planes, and Between a Line and a Plane

Because the coefficients are the normal, the angle between two planes is the angle between their normals.

$\cos \theta =\frac{|{\vec{n}}_1\cdot {\vec{n}}_2|}{|{\vec{n}}_1||{\vec{n}}_2|}$

For a line and a plane the formula uses sine, not cosine, and the reason is worth holding on to: the normal sticks out of the plane, so the angle to the normal is the complement of the angle to the plane.

$\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$

Trap. Using cosine here gives the angle to the normal, which is $90^{\circ }$ minus the answer. It is the most frequently set trap in the unit, because the formula looks identical to the two-line one.

Check it against the extremes. A line lying in the plane is perpendicular to the normal, so the dot product is zero and the sine formula correctly returns $0^{\circ }$. A line along the normal gives sine equal to $1$, and $90^{\circ }$.

Illustration 8

Find the angle between the line with direction ratios $(2,3,6)$ and the plane $10x+2y-11z=3$.

The normal is read straight off the coefficients: $\vec{n}=(10,2,-11)$.

$\vec{b}\cdot \vec{n}=20+6-66=-40,|\vec{b}|=7,|\vec{n}|=\sqrt{100+4+121}=15$

$\sin \theta =\frac{40}{105}=\frac{8}{21}\Rightarrow \theta ={\sin }^{-1}\frac{8}{21}\approx 22.4^{\circ }$

Had cosine been used, the answer would have come out near $67.6^{\circ }$, the angle to the normal. The two are complementary, and both appear among the options in exam questions of this type.

8. Line and Plane: The Three Cases

Substitute the line's parametric point into the plane and watch what happens to the parameter.

The coefficient of $\lambda$ in that substitution is exactly $\vec{b}\cdot \vec{n}$. Non-zero gives the first row; zero sends you to the other two, which one point of the line then separates.

Illustration 9

Where does $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ meet the plane $x-y+z=5$?

Parametrise the line, then substitute into the plane. That is the whole method.

$x=2+3t,y=-1+4t,z=2+2t$

$(2+3t)-(-1+4t)+(2+2t)=5\Rightarrow 5+t=5\Rightarrow t=0$

$\text{point}=(2,-1,2)$

The coefficient of $t$ came out as $1$, which is $\vec{b}\cdot \vec{n}=3-4+2$. Non-zero, so a unique meeting point was guaranteed before the constant was even collected.

Illustration 10

Show that the line through $(1,1,1)$ with direction $(2,3,4)$ is parallel to the plane $3x+2y-3z=5$, and find how far apart they are.

Test the direction against the normal first, because that decides which of the three rows applies.

$\vec{b}\cdot \vec{n}=2(3)+3(2)+4(-3)=6+6-12=0$

Parallel, so the only question left is whether the line lies inside the plane.

$3(1)+2(1)-3(1)=2\ne 5$

It does not, so the line runs parallel and outside. Every point on it is then the same distance from the plane, and one point is enough to measure it.

$d=\frac{|3+2-3-5|}{\sqrt{9+4+9}}=\frac{3}{\sqrt{22}}$

9. Distance from a Point to a Plane

$d=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$

It is the two-dimensional line formula with one more term, for the same reason: divide by the length of the normal.

Before the modulus, the sign tells you which side. Two points give the same sign when they lie on the same side of the plane, and opposite signs when the plane separates them.

For two parallel planes, make the coefficients match and then take the difference of the constants over the normal's length.

Illustration 11

Find the distance from $(1,2,3)$ to $2x-y+2z=5$, and the distance between $2x-2y+z=3$ and $2x-2y+z=9$.

$d=\frac{|2(1)-2+2(3)-5|}{\sqrt{4+1+4}}=\frac{1}{3}$

For the parallel planes the coefficients already match, so only the constants differ.

$d=\frac{|9-3|}{\sqrt{4+4+1}}=\frac{6}{3}=2$

Trap. Match the coefficients before subtracting. The planes $2x-2y+z=3$ and $4x-4y+2z=9$ are parallel, but the gap is $|4.5-3|/3=0.5$, not $6/3$. Halving the second equation first is the safe habit.

Foot of the perpendicular, and the image of a point

Both come from the same idea: walk from the point along the normal until you hit the plane, then keep going the same distance again.

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}=k$

The common value $k$ lands you at the foot; doubling it lands you at the image.

Illustration 12

Find the foot of the perpendicular from $P(1,2,3)$ to the plane $2x-y+2z=5$, and the image of $P$ in it.

Write the plane as $2x-y+2z-5=0$ so the constant has the right sign, then compute the common value.

$k=-\frac{2(1)-2+2(3)-5}{4+1+4}=-\frac{1}{9}$

The foot is reached by stepping $k$ times the normal from $P$.

$F=\left(1+2k,2-k,3+2k\right)=\left(1-\frac{2}{9},2+\frac{1}{9},3-\frac{2}{9}\right)=\left(\frac{7}{9},\frac{19}{9},\frac{25}{9}\right)$

Check that $F$ lies on the plane: $\frac{14}{9}-\frac{19}{9}+\frac{50}{9}=\frac{45}{9}=5$.

The image needs the same step taken twice.

$P^{\prime }=(1+4k,2-2k,3+4k)=\left(\frac{5}{9},\frac{20}{9},\frac{23}{9}\right)$

The distance $PF$ is $|k|\sqrt{a^2+b^2+c^2}=\frac{1}{9}\times 3=\frac{1}{3}$, agreeing with the distance formula computed earlier for this same point and plane.

Summary

In a plane, non-parallel lines always meet. In space they usually do not, because equating two lines gives three equations for two unknowns and the leftover one is a condition rather than a certainty.

That leftover condition is coplanarity, $({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)=0$, and the amount by which it fails is the shortest distance.

Distance and the section formula extend from the plane by adding one term and one coordinate.

Direction cosines satisfy $l^2+m^2+n^2=1$; direction ratios are any proportional triple. Two angles to the axes fix the third only up to sign, so such questions have two answers.

The angle between lines uses the modulus of the dot product, so it is always the acute one, and a zero numerator settles perpendicularity without any denominator.

A line is a point plus a direction, and a zero denominator in the symmetric form means that coordinate is constant, not that anything is divided by zero.

Parallel lines need the other distance formula, since the cross product of their directions vanishes.

For a plane, the coefficients are the normal, and that single reading answers parallelism, perpendicularity and most of the rest.

The angle between a line and a plane uses sine, because the normal is perpendicular to the plane; using cosine gives the complement.

Substituting a line into a plane decides all three cases from the parameter: a unique value, no solution, or every value.

Every plane through the line where two planes meet is $P_1+\lambda P_2=0$, which answers the question without ever finding that line.

The foot of a perpendicular and the image of a point are one walk along the normal and the same walk twice.

The distance from a point to a plane divides by the length of the normal, and its sign before the modulus tells you which side. For parallel planes, match the coefficients before differencing the constants.