Revision: coordinates in space

Coordinates in space is one of the "easy marks" topics of grade 12. Part I usually has two to four questions that are read straight from an equation: a normal vector, the centre and radius of a sphere, a distance. Part II or Part III usually has one mixed question, often set in a real-life context. Altogether this topic is usually worth about $1$ to $1.5$ marks.

The secret is to know the formulas thoroughly and to watch the signs. This revision lesson covers four areas: vectors, planes, lines, and spheres with real-life models.

The coordinate system

Oxyz

and the point

M(2, 3, 2)

Vectors and coordinates

Theorem Formulas for

\vec a=(a_1, a_2, a_3)

\vec b=(b_1, b_2, b_3)

$\overrightarrow{AB}=(x_B-x_A, y_B-y_A, z_B-z_A)$ and $AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}$ .
The midpoint $I$ of $AB$ : the mean of the coordinates of $A$ and $B$ ; the centroid $G$ of a triangle: the mean of the coordinates of the three vertices.
Scalar product $\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3$ ; $\vec a\perp\vec b\iff\vec a\cdot\vec b=0$ .
$\cos(\vec a, \vec b)=\dfrac{\vec a\cdot\vec b}{|\vec a|\cdot|\vec b|}$ , where $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$ .
The vector product $[\vec a, \vec b]=(a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1)$ is perpendicular to both $\vec a$ and $\vec b$ .

Example Multiple choice: the cosine of the angle between two vectors

In $Oxyz$ , let $\vec a=(1, -2, 2)$ and $\vec b=(2, 1, -2)$ . The value of $\cos(\vec a, \vec b)$ is

A. $\dfrac49$ .

B. $-\dfrac49$ .

C. $-\dfrac43$ .

D. $0$ .

Solution. We compute the scalar product and the two lengths, then use the formula:

$\vec a\cdot\vec b=2-2-4=-4,\qquad |\vec a|=\sqrt{1+4+4}=3,\qquad |\vec b|=\sqrt{4+1+4}=3.$

So $\cos(\vec a, \vec b)=\dfrac{-4}{3\cdot3}=-\dfrac49$ . The answer is B.

Look back. The angle between two vectors can be obtuse, so the cosine may be negative. Option A is the cosine of the angle between two lines (with an absolute value); option D has a sign error in the second term.

Equations of planes

Theorem Planes

The plane through $M(x_0, y_0, z_0)$ with normal vector $\vec n=(A, B, C)$ : $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$ .
The plane $Ax+By+Cz+D=0$ has normal vector $\vec n=(A, B, C)$ .
The plane through three points $A$ , $B$ , $C$ : use $\vec n=[\overrightarrow{AB}, \overrightarrow{AC}]$ . The plane through the intercepts $(a, 0, 0)$ , $(0, b, 0)$ , $(0, 0, c)$ : $\dfrac xa+\dfrac yb+\dfrac zc=1$ .
Distance: $d(M, (P))=\dfrac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ .
$(P)\perp(Q)\iff\vec n_P\cdot\vec n_Q=0$ ; $(P)\parallel(Q)$ when the normal vectors are parallel and the planes do not coincide.

Example True/false: the intercept form of a plane

In $Oxyz$ , consider the points $A(1, 0, 0)$ , $B(0, 2, 0)$ , $C(0, 0, 3)$ .

a) The plane $(ABC)$ has normal vector $\vec n=(6, 3, 2)$ .

b) The origin $O$ lies in the plane $(ABC)$ .

c) The distance from $O$ to the plane $(ABC)$ is $\dfrac67$ .

d) The plane $(ABC)$ is perpendicular to the plane $(Oxy)$ .

Solution. The three points lie on the three axes, so we use the intercept form and multiply through by the common denominator $6$ :

$\frac x1+\frac y2+\frac z3=1\iff 6x+3y+2z-6=0.$

a) True. The coefficients of $x$ , $y$ , $z$ are $6$ , $3$ , $2$ .

b) False. Substituting $O$ : $0+0+0-6=-6\neq0$ .

c) True. $d(O, (ABC))=\dfrac{|-6|}{\sqrt{36+9+4}}=\dfrac67$ .

d) False. The plane $(Oxy)$ has normal vector $\vec k=(0, 0, 1)$ and $\vec n\cdot\vec k=2\neq0$ .

So the answers are a) True, b) False, c) True, d) False.

Equations of lines and angle formulas

Theorem Lines and angles

The line through $M(x_0, y_0, z_0)$ with direction vector $\vec u=(a, b, c)$ : parametric equations $x=x_0+at$ , $y=y_0+bt$ , $z=z_0+ct$ ; if $abc\neq0$ , Cartesian equations $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$ .
The angle between two lines: $\cos\varphi=\dfrac{|\vec u\cdot\vec u'|}{|\vec u|\cdot|\vec u'|}$ .
The angle between a line and a plane: $\sin\varphi=\dfrac{|\vec u\cdot\vec n|}{|\vec u|\cdot|\vec n|}$ .
The angle between two planes: $\cos\varphi=\dfrac{|\vec n\cdot\vec n'|}{|\vec n|\cdot|\vec n'|}$ .

A line perpendicular to a plane takes the plane's normal vector as its direction vector. This idea is behind many multiple-choice questions.

Example Multiple choice: a line perpendicular to a plane

The line through $A(1, 2, -1)$ perpendicular to the plane $(P)\colon 2x-y+3z-5=0$ has parametric equations

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

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

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

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

Solution. The line is perpendicular to $(P)$ , so its direction vector is $\vec u=\vec n_P=(2, -1, 3)$ ; it passes through $A(1, 2, -1)$ :

$x=1+2t,\qquad y=2-t,\qquad z=-1+3t.$

The answer is A. Option B swaps the point and the vector, option C has the wrong sign for the coefficient of $y$ , and option D has the wrong sign for the $z$ -coordinate of $A$ .

Spheres and real-life coordinate models

In a real-life problem we put the origin in a convenient place (a corner of a room, the foot of a tower, a radar station), choose a unit (m or km), and then translate the question into one about points, lines, planes or spheres.

Example Short answer: an aircraft crossing radar coverage

Choose axes $Oxyz$ (unit: km) with a radar station at the origin $O$ . The radar detects every object within $5$ km of it. An aircraft flies in a straight line with position $M(1+2t, 2-t, 2)$ . Find the length of the part of the flight path inside the radar's range (in km).

Solution. Understand. The detection zone is the solid sphere with centre $O$ and radius $5$ ; the required distance is the chord of the flight path inside the sphere.

Plan. Solve the condition $OM\le5$ for $t$ , or use the distance from $O$ to the flight path.

Solve. Squaring the condition $OM\le5$ :

$(1+2t)^2+(2-t)^2+4\le25\iff 5t^2+9\le25\iff t^2\le\frac{16}{5}.$

So $t$ runs from $-\dfrac{4}{\sqrt5}$ to $\dfrac{4}{\sqrt5}$ . The direction vector $\vec u=(2, -1, 0)$ has length $\sqrt5$ , so the distance is

$\sqrt5\cdot\frac{8}{\sqrt5}=8\ \text{(km)}.$

Look back. The point closest to $O$ is at $t=0$ , namely $H(1, 2, 2)$ with $OH=3$ ; by Pythagoras' theorem the chord is $2\sqrt{25-9}=8$ long, which agrees.

Key terms

English	Tiếng Việt
Oxyz coordinate system	hệ trục toạ độ Oxyz
scalar product	tích vô hướng
vector product (cross product)	tích có hướng
normal vector	vectơ pháp tuyến
direction vector	vectơ chỉ phương
parametric equations	phương trình tham số
Cartesian equations (of a line)	phương trình chính tắc
sphere	mặt cầu
tangent	tiếp xúc
distance from a point to a plane	khoảng cách từ điểm đến mặt phẳng

Exercises

The exercises go from easy to hard, and each one states its exam format.

Exercise 1 Multiple choice

The plane through $M(1, -2, 3)$ with normal vector $\vec n=(2, -1, 1)$ has equation

A. $2x-y+z-7=0$ . B. $2x-y+z+7=0$ . C. $x-2y+3z-7=0$ . D. $2x-y+z-3=0$ .

What you need and how to start

What you need: $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$ .

Steps:

Substitute $\vec n$ and the coordinates of $M$ into the formula.
Expand and simplify.

Show solution

$2(x-1)-(y+2)+(z-3)=0\iff 2x-y+z-7=0$ . Option C swaps the point and the normal vector.

Answer. A.

Exercise 2 Multiple choice

The distance from $M(1, 2, 3)$ to the plane $(P)\colon 2x-2y+z+3=0$ is

A. $4$ . B. $\dfrac49$ . C. $\dfrac43$ . D. $\dfrac23$ .

What you need and how to start

What you need: $d(M, (P))=\dfrac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ .

Steps:

Compute the numerator $|2-4+3+3|$ .
Compute the denominator $\sqrt{4+4+1}$ .

Show solution

$d=\dfrac{|2-4+3+3|}{3}=\dfrac43$ . Option B forgets the square root in the denominator.

Answer. C.

Exercise 3 Multiple choice

The sphere $(S)\colon x^2+y^2+z^2-2x+4y-6z-2=0$ has centre $I$ and radius $R$ given by

A. $I(1, -2, 3)$ , $R=4$ . B. $I(-1, 2, -3)$ , $R=4$ . C. $I(1, -2, 3)$ , $R=2\sqrt3$ . D. $I(2, -4, 6)$ , $R=4$ .

What you need and how to start

What you need: $-2a=-2$ , $-2b=4$ , $-2c=-6$ ; $R=\sqrt{a^2+b^2+c^2-d}$ .

Steps:

Find $a$ , $b$ , $c$ .
Compute $R$ with $d=-2$ .

Show solution

$a=1$ , $b=-2$ , $c=3$ and $R=\sqrt{1+4+9+2}=4$ . Option C takes $R^2=14-2$ by mistake; option B has the wrong signs for the centre.

Answer. A.

Exercise 4 True/false

Consider the line $d\colon x=1+t$ , $y=2-t$ , $z=2t$ and the plane $(P)\colon x+y+z-3=0$ .

a) $d$ has direction vector $\vec u=(1, -1, 2)$ .

b) The point $A(2, 1, 2)$ lies on $d$ .

c) $d$ is parallel to $(P)$ .

d) $d$ meets $(P)$ at the point $(1, 2, 0)$ .

What you need and how to start

What you need: if $d\parallel(P)$ then $\vec u\cdot\vec n=0$ ; the intersection is found by substituting $x$ , $y$ , $z$ in terms of $t$ into $(P)$ .

Steps:

Read the direction vector from the coefficients of $t$ .
Compute $\vec u\cdot\vec n$ and solve the equation in $t$ .

Show solution

a) True. The coefficients of $t$ are $1$ , $-1$ , $2$ .

b) True. With $t=1$ we get $(2, 1, 2)$ .

c) False. $\vec u\cdot\vec n=1-1+2=2\neq0$ .

d) True. $(1+t)+(2-t)+2t-3=0\iff 2t=0\iff t=0$ , which gives the point $(1, 2, 0)$ .

Answer. a) True; b) True; c) False; d) True.

Exercise 5 Short answer

Find the angle between the lines $d_1\colon x=t$ , $y=1+t$ , $z=2$ and $d_2\colon x=1$ , $y=2+s$ , $z=-s$ (in degrees).

What you need and how to start

What you need: $\cos\varphi=\dfrac{|\vec u_1\cdot\vec u_2|}{|\vec u_1|\cdot|\vec u_2|}$ .

Steps:

$\vec u_1=(1, 1, 0)$ and $\vec u_2=(0, 1, -1)$ .
Substitute into the formula.

Show solution

$\cos\varphi=\dfrac{|0+1+0|}{\sqrt2\cdot\sqrt2}=\dfrac12$ , so $\varphi=60^\circ$ .

Answer. $60$ .

Exercise 6 Short answer

Let $A(1, 0, 1)$ , $B(2, 1, 0)$ , $C(0, 2, 3)$ . Find the distance from the origin $O$ to the plane $(ABC)$ (to 2 decimal places).

What you need and how to start

What you need: the normal vector $\vec n=[\overrightarrow{AB}, \overrightarrow{AC}]$ .

Steps:

$\overrightarrow{AB}=(1, 1, -1)$ and $\overrightarrow{AC}=(-1, 2, 2)$ ; compute the vector product.
Write the equation of $(ABC)$ , then find the distance.

Show solution

$[\overrightarrow{AB}, \overrightarrow{AC}]=(1\cdot2-(-1)\cdot2,\ (-1)(-1)-1\cdot2,\ 1\cdot2-1\cdot(-1))=(4, -1, 3)$ . The plane $(ABC)$ is $4(x-1)-y+3(z-1)=0\iff 4x-y+3z-7=0$ . Then $d(O, (ABC))=\dfrac{7}{\sqrt{26}}\approx1.37$ .

Answer. $1.37$ .

Exercise 7 Short answer

A room is a cuboid $6$ m long, $4$ m wide and $3$ m high. Choose axes $Oxyz$ with the origin at a corner of the floor and the axes along three edges, in metres. A straight wire is stretched from the point $A(0, 0, 3)$ in a corner of the ceiling to the point $B(6, 4, 0)$ in the opposite corner of the floor. Find the length of the wire (in m, to 2 decimal places).

What you need and how to start

What you need: $AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}$ .

Steps:

Find $\overrightarrow{AB}$ .
Find its length.

Show solution

$\overrightarrow{AB}=(6, 4, -3)$ , so $AB=\sqrt{36+16+9}=\sqrt{61}\approx7.81$ m.

Answer. $7.81$ .

Exercise 8 Short answer

A drone starts at the origin $O$ and flies in a straight line; after $t$ seconds it is at $M(2t, 2t, t)$ (in metres). The no-fly zone is the region of space where $x+2y+2z\ge30$ . After how many seconds does the drone start to enter the no-fly zone?

What you need and how to start

What you need: the drone reaches the boundary of the zone when the coordinates of $M$ satisfy $x+2y+2z=30$ .

Steps:

Substitute the coordinates of $M$ into $x+2y+2z$ .
Solve the equation for $t$ .

Show solution

$2t+4t+2t=8t=30\iff t=3.75$ . For $t<3.75$ we have $8t<30$ , so before that moment the drone is outside the zone.

Answer. $3.75$ .

Exercise 9 Written answer

Let $A(3, 1, -2)$ and the plane $(P)\colon x+2y-2z=0$ . Write the equations of the line $d$ through $A$ perpendicular to $(P)$ , find the orthogonal projection $H$ of $A$ on $(P)$ , and find $d(A, (P))$ .

What you need and how to start

What you need: $d$ takes $\vec n_P$ as its direction vector; $H=d\cap(P)$ .

Steps:

Write $d\colon x=3+t$ , $y=1+2t$ , $z=-2-2t$ .
Substitute into $(P)$ to find $t$ , hence $H$ ; then find $AH$ .

Show solution

Substituting into $(P)$ : $(3+t)+2(1+2t)-2(-2-2t)=0\iff 9+9t=0\iff t=-1$ , so $H(2, -1, 0)$ . Then $AH=\sqrt{1+4+4}=3$ , which agrees with $d(A, (P))=\dfrac{|3+2+4|}{3}=3$ .

Answer. $d\colon x=3+t$ , $y=1+2t$ , $z=-2-2t$ ; $H(2, -1, 0)$ ; $d(A, (P))=3$ .

Exercise 10 Short answer (challenge)

The sphere $(S)$ has centre $I(1, 2, -1)$ and touches the plane $(P)\colon 2x-y+2z+11=0$ at the point $H(a, b, c)$ . Find $a+b+c$ .

What you need and how to start

What you need: the radius is $R=d(I, (P))$ ; the point of contact is the orthogonal projection of $I$ on $(P)$ .

Steps:

Find $R$ .
Write the line through $I$ perpendicular to $(P)$ and find where it meets $(P)$ .

Show solution

$R=\dfrac{|2-2-2+11|}{3}=3$ . The line $x=1+2s$ , $y=2-s$ , $z=-1+2s$ meets $(P)$ when $2(1+2s)-(2-s)+2(-1+2s)+11=0\iff 9s+9=0\iff s=-1$ , so $H(-1, 3, -3)$ and $a+b+c=-1$ .

Answer. $-1$ .

Vectors and coordinates

Equations of planes

Equations of lines and angle formulas

Spheres and real-life coordinate models

Key terms

Exercises

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $(S)$ has centre $I(1, -2, 0)$ .	True	False
b) $(S)$ has radius $9$ .	True	False
c) The distance from the centre $I$ to $(P)$ is $1$ .	True	False
d) $(P)$ cuts $(S)$ in a circle of radius $2\sqrt2$ .	True	False