Coordinates in Space and the Distance Formula, Worked Solutions

Ques­tion 1: Nam­ing octants from signs#

The prob­lem#

Name the octant in which each point lies: $(-2, 3, 4)$; $(4, -2, -5)$; $(-3, -1, 6)$; $(2, 5, -1)$.

Under­stand­ing the prob­lem#

The three coor­di­nate planes divide space into eight octants, num­bered I to VIII. Which octant a point is in depends only on the signs of its three coor­di­nates, not their sizes.

The idea#

Write the sign pat­tern $(\pm, \pm, \pm)$ of each point and match it with the octant table in the les­son:

Step-by-step solu­tion#

Step 1. $(-2, 3, 4)$ has signs $(-, +, +)$. In the table this is col­umn II.

Step 2. $(4, -2, -5)$ has signs $(+, -, -)$: col­umn VIII.

Step 3. $(-3, -1, 6)$ has signs $(-, -, +)$: col­umn III.

Step 4. $(2, 5, -1)$ has signs $(+, +, -)$: col­umn V.

Check­ing the answer#

A quick mem­ory aid: octants I to IV have $z > 0$ (above the $XY$-plane) and V to VIII have $z < 0$. Points 1 and 3 have pos­i­tive $z$ and got octants II and III; points 2 and 4 have neg­a­tive $z$ and got VIII and V. Con­sis­tent.

Answer#

II; VIII; III; V.

Ques­tion 2: Points on an axis and in a coor­di­nate plane#

The prob­lem#

What are the coor­di­nates of a gen­eral point on the $y$-axis? Of a gen­eral point in the $ZX$-plane?

Under­stand­ing the prob­lem#

The coor­di­nates $(x, y, z)$ are the signed dis­tances of a point from the $YZ$-, $ZX$- and $XY$-planes. You must say which coor­di­nates are forced to be zero.

The idea#

A point on the $y$-axis lies in both the $XY$-plane ($z = 0$) and the $YZ$-plane ($x = 0$). The $ZX$-plane is the plane $y = 0$.

Step-by-step solu­tion#

Step 1. On the $y$-axis the point lies in the $YZ$-plane, so $x = 0$, and in the $XY$-plane, so $z = 0$. Only $y$ is free.

$$(0,\ y,\ 0)$$

Step 2. In the $ZX$-plane the dis­tance from that plane, which is $y$, is zero. The other two are free.

$$(x,\ 0,\ z)$$

Check­ing the answer#

$(0, 5, 0)$ is clearly on the $y$-axis; $(2, 0, -3)$ has no $y$-com­po­nent, so it lies in the $ZX$-plane.

Answer#

On the $y$-axis: $(0, y, 0)$. In the $ZX$-plane: $(x, 0, z)$.

Ques­tion 3: Using the dis­tance for­mula#

The prob­lem#

Find the dis­tance between (a) $(2, -1, 3)$ and $(-4, 1, 0)$; (b) $(a, b, c)$ and $(-a, -b, -c)$.

Under­stand­ing the prob­lem#

Both parts are direct uses of the dis­tance for­mula. In (b) the answer will be an expres­sion in $a$, $b$, $c$.

The idea#

Sub­tract coor­di­nates in order, square each dif­fer­ence, add, and take the square root. Squar­ing removes any minus signs, so the order of sub­trac­tion does not mat­ter.

Step-by-step solu­tion#

Part (a)

Step 1. Dif­fer­ences: $x$: $-4 - 2 = -6$; $y$: $1 - (-1) = 2$; $z$: $0 - 3 = -3$.

Step 2. Square and add.

$$(-6)^2 + 2^2 + (-3)^2 = 36 + 4 + 9 = 49$$

Step 3. Take the square root.

$$d = \sqrt{49} = 7$$

Part (b)

Step 1. Dif­fer­ences: $-a - a = -2a$, $-b - b = -2b$, $-c - c = -2c$.

Step 2. Square and add.

$$(-2a)^2 + (-2b)^2 + (-2c)^2 = 4a^2 + 4b^2 + 4c^2 = 4(a^2 + b^2 + c^2)$$

Step 3. Take the square root; $\sqrt{4} = 2$ comes out­side.

$$d = 2\sqrt{a^2 + b^2 + c^2}$$

Check­ing the answer#

In (b) the two points are reflec­tions of each other in the ori­gin, so the dis­tance should be twice the dis­tance of $(a, b, c)$ from $O$, which is $\sqrt{a^2 + b^2 + c^2}$. It is.

Answer#

(a) $7$ units; (b) $2\sqrt{a^2 + b^2 + c^2}$ units.

Ques­tion 4: Prov­ing three points collinear#

The prob­lem#

Show that $(-1, 4, 2)$, $(1, 3, 0)$ and $(5, 1, -4)$ are collinear.

Under­stand­ing the prob­lem#

Collinear means all three points lie on one straight line. You must prove it, not just state it.

The idea#

As in Exam­ple 4, three points are collinear exactly when the two shorter dis­tances add up to the longest one; then the mid­dle point lies on the seg­ment join­ing the other two. So find all three dis­tances.

Step-by-step solu­tion#

Step 1. Name the points $P(-1, 4, 2)$, $Q(1, 3, 0)$, $R(5, 1, -4)$.

Step 2. Find $PQ$. Dif­fer­ences $2, -1, -2$.

$$PQ = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

Step 3. Find $QR$. Dif­fer­ences $4, -2, -4$.

$$QR = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$$

Step 4. Find $PR$. Dif­fer­ences $6, -3, -6$.

$$PR = \sqrt{36 + 9 + 36} = \sqrt{81} = 9$$

Step 5. Com­pare: the two shorter dis­tances add up to the longest.

$$PQ + QR = 3 + 6 = 9 = PR$$

If $Q$ were off the line $PR$, the tri­an­gle $PQR$ would give $PQ + QR > PR$. Equal­ity means $Q$ lies on seg­ment $PR$, so the points are collinear.

Check­ing the answer#

The dif­fer­ences from $P$ to $Q$ are $(2, -1, -2)$ and from $Q$ to $R$ are $(4, -2, -4)$, exactly twice as big and in the same direc­tion. That also shows the three points are in a line.

Answer#

$PQ = 3$, $QR = 6$, $PR = 9$ and $PQ + QR = PR$, so the three points are collinear.

Ques­tion 5: Prov­ing a tri­an­gle equi­lat­eral#

The prob­lem#

Show that $(1, 2, 3)$, $(3, 2, 5)$ and $(1, 4, 5)$ are the ver­tices of an equi­lat­eral tri­an­gle.

Under­stand­ing the prob­lem#

An equi­lat­eral tri­an­gle has all three sides equal. You must com­pute the three side lengths and show they are the same.

The idea#

Apply the dis­tance for­mula to each pair of points.

Step-by-step solu­tion#

Step 1. Name the points $A(1, 2, 3)$, $B(3, 2, 5)$, $C(1, 4, 5)$.

Step 2. $AB$: dif­fer­ences $2, 0, 2$.

$$AB = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2}$$

Step 3. $BC$: dif­fer­ences $-2, 2, 0$.

$$BC = \sqrt{4 + 4 + 0} = \sqrt{8} = 2\sqrt{2}$$

Step 4. $CA$: dif­fer­ences $0, -2, -2$.

$$CA = \sqrt{0 + 4 + 4} = \sqrt{8} = 2\sqrt{2}$$

Step 5. All three sides are $2\sqrt{2}$, and the points are not collinear (the longest side is not the sum of the other two), so $ABC$ is an equi­lat­eral tri­an­gle.

Check­ing the answer#

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \approx 2.83$ for all three.

Answer#

$AB = BC = CA = 2\sqrt{2}$, so the tri­an­gle is equi­lat­eral.

Ques­tion 6: Prov­ing four points form a par­al­lel­o­gram#

The prob­lem#

Show that $(0, 0, 0)$, $(2, 1, 1)$, $(3, 3, 2)$ and $(1, 2, 1)$ form a par­al­lel­o­gram.

Under­stand­ing the prob­lem#

Take the points in the given order as $O(0, 0, 0)$, $A(2, 1, 1)$, $B(3, 3, 2)$, $C(1, 2, 1)$, so the quadri­lat­eral is $OABC$ with diag­o­nals $OB$ and $AC$. In space, equal oppo­site sides alone do not guar­an­tee a flat par­al­lel­o­gram, so use a test that does.

The idea#

A quadri­lat­eral is a par­al­lel­o­gram when its diag­o­nals bisect each other, that is, when the two diag­o­nals have the same mid­point. The mid­point of a seg­ment has coor­di­nates equal to the aver­ages of the end coor­di­nates. Then con­firm with side lengths from the dis­tance for­mula.

Step-by-step solu­tion#

Step 1. Mid­point of diag­o­nal $OB$: aver­age the coor­di­nates of $O(0,0,0)$ and $B(3,3,2)$.

$$\displaystyle \left(\frac{0 + 3}{2},\ \frac{0 + 3}{2},\ \frac{0 + 2}{2}\right) = \left(\tfrac{3}{2},\ \tfrac{3}{2},\ 1\right)$$

Step 2. Mid­point of diag­o­nal $AC$: aver­age $A(2,1,1)$ and $C(1,2,1)$.

$$\displaystyle \left(\frac{2 + 1}{2},\ \frac{1 + 2}{2},\ \frac{1 + 1}{2}\right) = \left(\tfrac{3}{2},\ \tfrac{3}{2},\ 1\right)$$

Step 3. The two diag­o­nals have the same mid­point, so they bisect each other and $OABC$ is a par­al­lel­o­gram.

Step 4. Con­firm with the sides.

$$\begin{aligned} OA &= \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}, & AB &= \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6},\\ BC &= \sqrt{(-2)^2 + (-1)^2 + (-1)^2} = \sqrt{6}, & CO &= \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}. \end{aligned}$$

Oppo­site sides are equal, as a par­al­lel­o­gram needs. In fact all four sides are equal, so it is a rhom­bus.

Step 5. It is not a square, because the diag­o­nals are unequal:

$$OB = \sqrt{9 + 9 + 4} = \sqrt{22}, \qquad AC = \sqrt{1 + 1 + 0} = \sqrt{2}$$

Check­ing the answer#

The move from $O$ to $A$ is $(2, 1, 1)$ and from $C$ to $B$ is also $(2, 1, 1)$: the oppo­site sides $OA$ and $CB$ are par­al­lel and equal, which again con­firms a par­al­lel­o­gram.

Answer#

The diag­o­nals $OB$ and $AC$ bisect each other at $\displaystyle \left(\tfrac{3}{2}, \tfrac{3}{2}, 1\right)$ and oppo­site sides are equal (each $\sqrt{6}$), so the points form a par­al­lel­o­gram (in fact a rhom­bus).

Ques­tion 7: A point on the $x$-axis equidis­tant from two points#

The prob­lem#

Find the point on the $x$-axis that is equidis­tant from $(1, 2, 3)$ and $(3, 5, -2)$.

Under­stand­ing the prob­lem#

Any point on the $x$-axis has the form $(x, 0, 0)$, with one unknown, $x$. "Equidis­tant" means its dis­tance to each of the two given points is the same.

The idea#

Set the two dis­tances equal. Square both sides first to remove the square roots; the $x^2$ terms will then can­cel, leav­ing a lin­ear equa­tion.

Step-by-step solu­tion#

Step 1. Let the point be $P(x, 0, 0)$. Squared dis­tance to $(1, 2, 3)$:

$$(x - 1)^2 + (0 - 2)^2 + (0 - 3)^2 = (x - 1)^2 + 13$$

Step 2. Squared dis­tance to $(3, 5, -2)$:

$$(x - 3)^2 + (0 - 5)^2 + (0 + 2)^2 = (x - 3)^2 + 29$$

Step 3. Set them equal and expand.

$$x^2 - 2x + 1 + 13 = x^2 - 6x + 9 + 29$$

Step 4. Can­cel $x^2$ and sim­plify.

$$-2x + 14 = -6x + 38 \quad\Longrightarrow\quad 4x = 24 \quad\Longrightarrow\quad x = 6$$

Check­ing the answer#

From $(6, 0, 0)$: to $(1, 2, 3)$ the squared dis­tance is $25 + 4 + 9 = 38$; to $(3, 5, -2)$ it is $9 + 25 + 4 = 38$. Equal.

Answer#

$(6, 0, 0)$.

Ques­tion 8: The locus of points equidis­tant from two points#

The prob­lem#

Find the equa­tion of the set of points equidis­tant from $(1, 2, 3)$ and $(3, 2, -1)$.

Under­stand­ing the prob­lem#

You need a con­di­tion on a gen­eral point $P(x, y, z)$ that says its dis­tances to $A(1, 2, 3)$ and $B(3, 2, -1)$ are equal. As in Exam­ple 5, the answer will be the equa­tion of a plane (the per­pen­dic­u­lar bisec­tor plane of $AB$).

The idea#

Write $PA^2 = PB^2$, expand, and can­cel the squared terms that appear on both sides.

Step-by-step solu­tion#

Step 1. Write the con­di­tion.

$$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$$

Step 2. The $(y - 2)^2$ terms are the same on both sides; can­cel them. Expand the rest.

$$x^2 - 2x + 1 + z^2 - 6z + 9 = x^2 - 6x + 9 + z^2 + 2z + 1$$

Step 3. Can­cel $x^2$ and $z^2$, and note that both sides have con­stant $10$.

$$-2x - 6z = -6x + 2z$$

Step 4. Col­lect terms on the left.

$$4x - 8z = 0 \quad\Longrightarrow\quad x - 2z = 0$$

Check­ing the answer#

The mid­point of $AB$ is $(2, 2, 1)$, which must be on the locus: $2 - 2(1) = 0$. Cor­rect.

Answer#

$x - 2z = 0$.

Ques­tion 9: Points at a fixed dis­tance from the ori­gin#

The prob­lem#

Find the locus of points whose dis­tance from the ori­gin is $4$.

Under­stand­ing the prob­lem#

A locus is the set of all points sat­is­fy­ing a con­di­tion, described by an equa­tion. Here the con­di­tion is $OP = 4$ for a gen­eral point $P(x, y, z)$.

The idea#

Use $OP = \sqrt{x^2 + y^2 + z^2}$ and square to remove the root.

Step-by-step solu­tion#

Step 1. Write the con­di­tion.

$$\sqrt{x^2 + y^2 + z^2} = 4$$

Step 2. Square both sides (both are non-neg­a­tive, so noth­ing is lost).

$$x^2 + y^2 + z^2 = 16$$

Step 3. Describe it: every point exactly $4$ units from $O$ in every direc­tion, which is a sphere with cen­tre $O$ and radius $4$.

Check­ing the answer#

$(4, 0, 0)$, $(0, 0, -4)$ and $(2, 2, 2\sqrt{2})$ all sat­isfy the equa­tion, and each is $4$ units from $O$.

Answer#

The sphere $x^2 + y^2 + z^2 = 16$, with cen­tre at the ori­gin and radius $4$.

Ques­tion 10: The longest rod in a box#

The prob­lem#

A box sits in the first octant with one cor­ner at the ori­gin and edges $3$, $4$ and $12$ along the axes. How long is the longest rod that fits inside?

Under­stand­ing the prob­lem#

The box has cor­ners at the ori­gin and at $(3, 4, 12)$, the cor­ner far­thest from $O$. The longest straight rod that fits runs from one cor­ner to the diag­o­nally oppo­site cor­ner, through the inside of the box.

The idea#

Find the dis­tance from $(0, 0, 0)$ to $(3, 4, 12)$ with the for­mula $OP = \sqrt{x^2 + y^2 + z^2}$.

Step-by-step solu­tion#

Step 1. Iden­tify the oppo­site cor­ners: $O(0, 0, 0)$ and $P(3, 4, 12)$.

Step 2. Apply the for­mula.

$$OP = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$$

Check­ing the answer#

Pythago­ras twice: the floor diag­o­nal is $\sqrt{9 + 16} = 5$; then with height $12$, $\sqrt{25 + 144} = 13$. Same result.

Answer#

The longest rod is $13$ units long.