Introduction to Three Dimensional Geometry

A point on the x-axis has coordinates of the form **$(x,0,0)$. So its y-coordinate and z-coordinate are both 0**.

A point in the XZ-plane has its y-coordinate equal to 0. So the answer is 0.

Use the signs of $(x,y,z)$ for each point:

- $(1,2,3)$ → $(+,+,+)$ → I
- $(4,-2,3)$ → $(+,-,+)$ → IV
- $(4,-2,-5)$ → $(+,-,-)$ → VIII
- $(4,2,-5)$ → $(+,+,-)$ → V
- $(-4,2,-5)$ → $(-,+,-)$ → VI
- $(-4,2,5)$ → $(-,+,+)$ → II
- $(-3,-1,6)$ → $(-,-,+)$ → III
- $(-2,-4,-7)$ → $(-,-,-)$ → VII

So the octants are I, IV, VIII, V, VI, II, III, VII. Since the printed answer list in the source is by the same order of points, the mapping above is the required result.

From the chapter:

1. The $x$-axis and $y$-axis together determine the XY-plane.
2. Points in the XY-plane have coordinates of the form **$(x,y,0)$.
3. The coordinate planes divide space into eight** octants.

Use the distance formula:
$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

(i) $(2,3,5)$ and $(4,3,1)$
$$\sqrt{(4-2)^2+(3-3)^2+(1-5)^2} =\sqrt{4+0+16}=\sqrt{20}=2\sqrt5$$
So the distance is **$2\sqrt5$.

(ii)** $(-3,7,2)$ and $(2,4,-1)$
$$\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2} =\sqrt{25+9+9}=\sqrt{43}$$
So the distance is **$\sqrt{43}$.

(iii)** $(-1,3,-4)$ and $(1,-3,4)$
$$\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2} =\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}$$
So the distance is **$2\sqrt{26}$.

(iv)** $(2,-1,3)$ and $(-2,1,3)$
$$\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2} =\sqrt{16+4+0}=\sqrt{20}=2\sqrt5$$
So the distance is **$2\sqrt5$**.

The computed answers are not among the short placeholder options shown here, so the correct distances are the values above.

Let the points be $P(-2,3,5)$, $Q(1,2,3)$ and $R(7,0,-1)$.

Find the distances:

$$PQ=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2} =\sqrt{9+1+4}=\sqrt{14}$$

$$QR=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2} =\sqrt{36+4+16}=\sqrt{56}=2\sqrt{14}$$

$$PR=\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2} =\sqrt{81+9+36}=\sqrt{126}=3\sqrt{14}$$

Now,
$$PQ+QR=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=PR$$

Since the sum of two sides equals the third side, the three points are collinear.

The chapter verifies three statements:

(i) Points $(0,7,-10)$, $(1,6,-6)$ and $(4,9,-6)$ form an isosceles triangle.

(ii) Points $(0,7,10)$, $(-1,6,6)$ and $(-4,9,6)$ form a right-angled triangle.

(iii) Points $(-1,2,1)$, $(1,-2,5)$, $(4,-7,8)$ and $(2,-3,4)$ are the vertices of a parallelogram.

So the verification is exactly as stated in the textbook examples/exercise.