Introduction to Three Dimensional Geometry: questions with answers

A coordinate plane carries the names of the two axes that lie inside it. The YZ-plane contains the $y$-axis and the $z$-axis, and the third axis, the $x$-axis, stands perpendicular to it. Both other answers name an axis that lies in the plane, which is exactly what reading the plane's name as naming its perpendicular would give.

Both of these axes lie in the ZX-plane, and that pair of letters is what the plane is named after. The other two planes each contain the third axis, which is perpendicular to the plane wanted here.

The point in the middle splits the whole segment into two pieces, so the two distances from it add to the distance between the outer two: $QP + PR = QR$. Equal distances would only say that the middle point is the mid-point, and adding the two smaller squares is the Pythagorean test for a right angle, which no three collinear points satisfy.

Only the square of the LONGEST side can equal the sum of the other two squares. The largest of the three values is 169, belonging to $BC$, so $CA^2 + AB^2 = BC^2$ is the equation to test. Leaving a shorter side standing alone tests something that cannot hold for any triangle, and the test must be carried out on the squares, not on the side lengths.

Lying in a coordinate plane costs exactly one zero: the coordinate belonging to the axis that is not in the plane. For the XY-plane that coordinate is $z$. The two letters in the plane's own name are the ones that stay free, and two zeros would place the point on an axis rather than merely in a plane.

A point of the $y$-axis keeps the coordinate named by the axis and loses the other two, so its $x$- and $z$-coordinates are both zero. The named letter is the one that SURVIVES, not the one that vanishes, and one zero would only place the point in a plane; all three zeros would place it at the origin.

The chapter lists the eight names in the order $XOYZ$, $X'OYZ$, $X'OY'Z$, $XOY'Z$, $XOYZ'$, $X'OYZ'$, $X'OY'Z'$, $XOY'Z'$ for octants I to VIII, so octant VIII is $XOY'Z'$. The name $XOY'Z$ differs only in whether $Z$ is primed, which is the upper and lower halves of the table taken the wrong way round.

Reflecting in the XY-plane changes only the sign of $z$, and Table 11.1 pairs each of octants I to IV with the octant four places later. The point (-4, 9, 2) is in octant II, so its image is in octant VI. Answering II keeps the point where it was, which is what happens when the octants above and below the XY-plane are taken the wrong way round.

The chapter states that $x$, $y$ and $z$ are the perpendicular distances from the YZ-, ZX- and XY-planes respectively, so the $z$-coordinate measures the distance from the XY-plane. Choosing the ZX-plane matches the coordinate to the plane whose name begins with the same letter; choosing the $z$-axis confuses a distance measured ALONG an axis with a distance measured FROM it.

The distances are asked for in the order XY, YZ, ZX, which pairs with $z$, $x$, $y$, so the answer is 9, 1, 4. Reporting 1, 4, 9 gives the same three numbers in the order they were written in the triplet, which answers a different question.

By the distance formula $PQ^2 = 12^2 + 16^2 + 21^2 = 841$, so $PQ = 29$ units. Adding the three gaps instead of squaring them measures the path along the edges of the box, 49, not its diagonal; 841 is $PQ^2$ reported before the square root is taken; and 20 is the distance between the two feet in the XY-plane, found by leaving the gap in $z$ out.

A drop onto a coordinate plane costs one coordinate and leaves two; a drop onto a coordinate axis costs two and leaves one. Treating the axis like a plane keeps one coordinate too many, and keeping only the perpendicular coordinate keeps one too few.

Because the diagonals $AC$ and $BD$ of a parallelogram bisect each other, $A + C = B + D$ and so $D$(5, -1, 4). Pairing $A$ with $B$ as though $AB$ were a diagonal gives (9, 3, -2), and adding all three given vertices gives (9, 5, 0); neither closes the figure.

Whether the squared terms survive depends on the operation, not on the fact that distances have been squared. Equating two squared distances cancels them; adding two squared distances doubles them, which is why the chapter's own answer to $PA^2 + PB^2 = 2k^2$ begins $2x^2 + 2y^2 + 2z^2$.

Each coordinate of the centroid is the average of the three corresponding vertex coordinates, so the three sums -6, 0 and 6 are each divided by 3, giving (-2, 0, 2). Reporting (-6, 0, 6) stops at the sums; (-3, 4, 3) averages only two of the three vertices.