Corrected exercises: geometric vectors and lines in space (MATH 133)

a) Always the END minus the START: $\overrightarrow{AB} = B - A = (5 - 1, 2 - 1) = (4, 1)$, $\overrightarrow{AC} = (2 - 1, 4 - 1) = (1, 3)$ and $\overrightarrow{BC} = (2 - 5, 4 - 2) = (-3, 2)$. Then $\overrightarrow{AB} + \overrightarrow{BC} = (4 - 3, 1 + 2) = (1, 3) = \overrightarrow{AC}$. This is the whole content of vector addition: going from $A$ to $B$ and then from $B$ to $C$ is the same trip as going from $A$ to $C$. Writing $A - B$ instead of $B - A$ gives $(-4, -1)$, the right length pointing the wrong way, and every later answer inherits the sign.

b) Componentwise: $\mathbf{u} + \mathbf{v} = (5, 4)$, $\mathbf{u} - \mathbf{v} = (4 - 1, 1 - 3) = (3, -2)$, and $2\mathbf{u} - 3\mathbf{v} = (8, 2) - (3, 9) = (5, -7)$. Scalar multiplication stretches a move, and a negative scalar also reverses it: $-3\mathbf{v}$ is three copies of $\mathbf{v}$ walked backwards.

c) In $ABDC$ the sides are $AB$, $BD$, $DC$, $CA$, so the diagonal from $A$ is $AD$. The parallelogram rule says exactly this: starting at $A$, the moves $\mathbf{u}$ and $\mathbf{v}$ combine into the diagonal $\overrightarrow{AD} = \mathbf{u} + \mathbf{v}$. Hence $D = A + \mathbf{u} + \mathbf{v} = (1 + 5, 1 + 4) = (6, 5)$. Check with opposite sides: $\overrightarrow{CD} = (6 - 2, 5 - 4) = (4, 1) = \overrightarrow{AB}$. Two opposite sides equal AS VECTORS, same length and same direction, is the definition that makes the proof one line.

d) The fourth vertex is opposite one of the three given points, and there are three choices. Opposite $A$: $D = B + C - A = (6, 5)$, found above. Opposite $B$, the parallelogram $ABCD$ in that order: $D_2 = A + C - B = (-2, 3)$, and indeed $\overrightarrow{D_2C} = (4, 1) = \overrightarrow{AB}$. Opposite $C$: $D_3 = A + B - C = (4, -1)$, and $\overrightarrow{D_3B} = (1, 3) = \overrightarrow{AC}$. The trap is to answer c) with $(-2, 3)$: the ORDER of the letters decides which point is opposite which, and a question that says $ABDC$ and one that says $ABCD$ have different answers.

e) $\mathbf{u} - \mathbf{v} = \overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{CA} = \overrightarrow{CB}$, and $B - C = (3, -2)$ agrees with b). So the parallelogram built on $\mathbf{u}$ and $\mathbf{v}$ carries BOTH operations: the sum is the diagonal leaving the common starting point, the difference is the other diagonal, from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$. Reversing it, from $B$ to $C$, gives $\mathbf{v} - \mathbf{u}$.

a) $\|\mathbf{u}\| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. Dividing by the norm keeps the direction and makes the length $1$: $\frac{1}{3}\mathbf{u} = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$, and its norm is $\sqrt{\frac{4 + 1 + 4}{9}} = 1$, the check worth writing. A length of $12$ is four times $3$, and opposite means a negative scalar: $-4\mathbf{u} = (-8, 4, -8)$, of norm $|-4| \times 3 = 12$. The rule is $\|c\mathbf{v}\| = |c|\,\|\mathbf{v}\|$, with the absolute value: a length is never negative.

b) The distance between two points is the norm of the vector joining them: $\overrightarrow{P_1P_2} = (2, 3, -6)$ and $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$. Subtract first, square second. Squaring the coordinates of each point separately and subtracting the results, a common slip, gives a number with no geometric meaning.

c) Square both sides, since both are non-negative: $k^2 + 4 + (k - 1)^2 = 9$, so $2k^2 - 2k + 5 = 9$, that is $k^2 - k - 2 = 0 = (k - 2)(k + 1)$. Both roots work: $(2, 2, 1)$ has norm $\sqrt{4 + 4 + 1} = 3$ and $(-1, 2, -2)$ has norm $\sqrt{1 + 4 + 4} = 3$. Geometrically the vectors $(k, 2, k - 1)$ have their tips on a line, and the sphere of radius $3$ cuts that line twice; stopping at one root loses half the answer.

d) First pair: $\frac{-3}{2} = \frac{6}{-4} = \frac{-9}{6} = -\frac{3}{2}$, the same scalar three times, so $(-3, 6, -9) = -\frac{3}{2}(2, -4, 6)$: parallel, pointing in opposite directions. Second pair: $\frac{1}{2}$, $\frac{-2}{-4} = \frac{1}{2}$, but $\frac{4}{6} = \frac{2}{3}$. Two components agreeing prove NOTHING; the third one breaks it, so the vectors are not parallel. This is the direction test that decides between parallel and non-parallel lines in exercises 4 and 5, and it always needs all three components.

e) $M = \frac{1}{2}(A + B) = \left(4, \frac{1}{2}, 2\right)$. For $P$, the ratio $2 : 1$ cuts $AB$ into $2 + 1 = 3$ equal parts and $P$ sits after two of them, so $P = A + \frac{2}{3}\overrightarrow{AB} = (1, 2, -1) + \frac{2}{3}(6, -3, 6) = (1, 2, -1) + (4, -2, 4) = (5, 0, 3)$. Equivalently $P = \frac{1}{3}A + \frac{2}{3}B$: the weight $\frac{2}{3}$ goes on $B$, because $P$ is CLOSER to $B$. Check: $\|\overrightarrow{AP}\| = \|(4, -2, 4)\| = 6$ and $\|\overrightarrow{PB}\| = \|(2, -1, 2)\| = 3$, a ratio of $2$. The two classic errors are $\frac{1}{2}$ of the way, which is $M$, and $A + 2\overrightarrow{AB} = (13, -4, 11)$, a point beyond $B$.

a) Vector form: $(x, y, z) = (2, -1, 3) + t(4, 0, -2)$. Parametric: $x = 2 + 4t$, $y = -1$, $z = 3 - 2t$. For the symmetric form, the $y$ equation does not contain $t$ at all, so it cannot be solved for $t$: it stays as it is. The answer is $\frac{x - 2}{4} = \frac{z - 3}{-2}$ together with $y = -1$. Writing $\frac{y + 1}{0}$ is not an equation, it is a division by zero, and it costs the mark on every paper. The direction may also be halved to $(2, 0, -1)$, giving $\frac{x - 2}{2} = \frac{z - 3}{-1}$: a different-looking and equally correct answer.

b) Direction: $\overrightarrow{AB} = (3, -2, 2)$, so $(x, y, z) = (1, 2, 3) + t(3, -2, 2)$. Starting from $B$ with the reversed direction, $(x, y, z) = (4, 0, 5) + s(-3, 2, -2)$ is another equation. Same line? The directions are parallel, $(-3, 2, -2) = -1 \times (3, -2, 2)$, and the point $(4, 0, 5)$ of the second is on the first, at $t = 1$. Two tests, direction then point, and never a comparison of the numbers written in the two equations, which have nothing in common here.

c) Every numerator must read coordinate MINUS number, and every coefficient of $x$, $y$, $z$ must be $1$. Rewrite: $\frac{1 - y}{2} = \frac{y - 1}{-2}$ and $\frac{2z - 4}{6} = \frac{2(z - 2)}{6} = \frac{z - 2}{3}$. The form is now $\frac{x + 2}{3} = \frac{y - 1}{-2} = \frac{z - 2}{3}$, which reads point $(-2, 1, 2)$ and direction $(3, -2, 3)$. Check with a second point: the common value $1$ gives $(1, -1, 5)$, and indeed $\frac{3}{3} = \frac{2}{2} = \frac{6}{6} = 1$. Reading $(-2, -1, 4)$ as the point or $(3, 2, 6)$ as the direction are the two answers this conversion is designed to catch.

d) $Q$ is on the line if ONE value of $t$ gives all three coordinates. $1 + 3t = 7$ gives $t = 2$; $2 - 2t = -2$ gives $t = 2$; $3 + 2t = 7$ gives $t = 2$. Same $t$ three times: $Q$ is on the line. For $R$, the first two equations again give $t = 2$, but $3 + 2t = 8$ demands $t = \frac{5}{2}$. No single $t$ works, so $R$ is NOT on the line, although it differs from $Q$ by one unit in one coordinate. Checking two coordinates out of three is the error; the third is the one that decides.

e) The student used the coordinates of $A$ as if they were a direction. A direction must be a difference of two points of the line. The proposed line does pass through $B$ ($t = 0$), but $(1, 2, 3)$ is not parallel to $(3, -2, 2)$ since $\frac{3}{1} \ne \frac{-2}{2}$. To be sure it misses $A$: $4 + t = 1$ forces $t = -3$, and then $y = 2t = -6 \ne 2$. It is a different line through $B$, and every question built on it afterwards is answered about the wrong object.

a) Rename the parameter of $L_2$ as $s$ and equate coordinates: $1 + t = 4 + s$, $2 - t = 2 + 2s$, $2t = 3 - s$. The first gives $t = 3 + s$; substituted in the second, $2 - 3 - s = 2 + 2s$, so $-3 = 3s$ and $s = -1$, then $t = 2$. Now the THIRD equation, which was not used: $2t = 4$ and $3 - s = 4$. It holds, so the lines meet. The point, computed from BOTH lines as a check: $L_1$ at $t = 2$ gives $(3, 0, 4)$, and $L_2$ at $s = -1$ gives $(4 - 1, 2 - 2, 3 + 1) = (3, 0, 4)$. Two lines, two parameter values, one point.

b) With the same letter, the equation $1 + t = 4 + t$ asks for a value of the parameter at which both lines give the same point, that is, whether two particles moving along the lines with these parametrizations are at the same place at the same TIME. They are not, but that says nothing about the paths. The lines do meet, at $(3, 0, 4)$, reached at $t = 2$ on $L_1$ and $s = -1$ on $L_2$. The letter in an equation of a line is a dummy variable: renaming it is always the first step, and skipping it turns a correct problem into a wrong conclusion worth zero.

c) Equate: $1 + t = 2s$, $2 - t = s$, $2t = s$. Adding the first two: $3 = 3s$, so $s = 1$ and $t = 1$. The first two equations are solved, and a student who stops here announces the point $(2, 1, 2)$. But the third reads $2t = 2$ against $s = 1$: $2 \ne 1$. No pair $(t, s)$ satisfies all three, so the lines do NOT meet. Since $(2, 1, 1)$ is not a multiple of $(1, -1, 2)$, as $\frac{2}{1} \ne \frac{1}{-1}$, they are not parallel either: $L_1$ and $L_3$ are skew. At $t = s = 1$ they pass over each other, at $(2, 1, 2)$ and $(2, 1, 1)$, one unit apart vertically.

d) Only the $z$ equation changes: $2t = c + s$. The first two still force $t = s = 1$, so the lines meet exactly when $2 = c + 1$, that is $c = 1$. The point is $L_1$ at $t = 1$, $(2, 1, 2)$, and $L_4$ at $s = 1$ gives $(2, 1, 1 + 1) = (2, 1, 2)$. Moving the second line by one unit in $z$ was the whole difference between skew and intersecting, which is why the third equation can never be skipped.

e) In the plane, equating two lines gives two equations in the two unknowns $t$ and $s$; when the directions are not parallel the system has exactly one solution, so the lines meet. In space there are three equations for the same two unknowns. Two of them fix $t$ and $s$, and the third is then a CONDITION that holds or fails. For two lines chosen at random it fails, so skew lines are the typical case in $\mathbb{R}^3$, not an exotic one. This is exactly why the third equation decides.

a) Direction test: $(-4, 2, -6) = -2(2, -1, 3)$, the same factor $-2$ on all three components, so the directions are parallel. Point test: is $(5, -2, 5)$ on $L_1$? $1 + 2t = 5$ gives $t = 2$, then $y = -2$ and $z = -1 + 6 = 5$. Yes. Parallel directions AND a common point: $L_1$ and $L_2$ are the same line, written with another point and a direction reversed and doubled. Nothing in the two equations looks alike, which is exactly why equations are never compared term by term.

b) Direction: $(4, -2, 6) = 2(2, -1, 3)$, parallel again. Point: $1 + 2t = 3$ forces $t = 1$, and then $y = -1$, not $1$. So $(3, 1, 2)$ is NOT on $L_1$, and two parallel lines sharing no point are parallel and distinct. The directions here are proportional to those of a), and the conclusion is the opposite one: proportional directions decide only that the lines are parallel or equal, never which.

c) Direction: $\frac{1}{2} \ne \frac{1}{-1}$, not parallel, so the lines either meet once or are skew. System: $1 + 2t = 4 + s$, $-t = -3 + s$, $-1 + 3t = 1 + s$. The second gives $s = 3 - t$; in the first, $1 + 2t = 7 - t$, so $t = 2$ and $s = 1$. The third equation decides: $-1 + 6 = 5$ on the left, $1 + 1 = 2$ on the right. $5 \ne 2$, no solution, and non-parallel lines with no common point are skew.

d) From the second equation $-t = 1 - 2s$, $t = 2s - 1$, and the first becomes $1 + 4s - 2 = 3 + 4s$, that is $-1 = 3$: no solution, immediately. The system alone says only that the lines have no common point, and two lines with no common point can be PARALLEL or SKEW. Answering skew in b) because the system failed is the classic error. The order that avoids it is always the same: direction test first, then the system or the point test.

e) Parallel to $L_1$ means same direction: $L_5 : (x, y, z) = t(2, -1, 3)$. Is the origin on $L_1$? $1 + 2t = 0$ gives $t = -\frac{1}{2}$, but $y = -t = 0$ gives $t = 0$: no. So $L_5$ is parallel to $L_1$ and distinct from it. Since $L_2$ is $L_1$ by a), the same answer holds for $L_2$, with no new computation. Parallel to a line never means equal to it until a point has been tested.

a) $\overrightarrow{AB} = \overrightarrow{DC}$ reads $B - A = C - D$, which rearranges to $A + C = B + D$. Dividing by $2$: $\frac{1}{2}(A + C) = \frac{1}{2}(B + D)$, that is, the midpoint of $AC$ is the midpoint of $BD$. Every step is reversible, so the equivalence goes both ways: the diagonals of a parallelogram bisect each other, and a quadrilateral whose diagonals bisect each other is a parallelogram. The whole proof is one rearrangement, and no figure is used, which is exactly what makes it a proof.

b) $P = \frac{1}{2}(A + B)$ and $Q = \frac{1}{2}(B + C)$, so $\overrightarrow{PQ} = Q - P = \frac{1}{2}(C - A) = \frac{1}{2}\overrightarrow{AC}$. In the same way $S = \frac{1}{2}(D + A)$ and $R = \frac{1}{2}(C + D)$ give $\overrightarrow{SR} = \frac{1}{2}(C - A)$. So $\overrightarrow{PQ} = \overrightarrow{SR}$, the definition of the parallelogram $PQRS$. Nothing assumed that $A$, $B$, $C$, $D$ are in one plane: the computation lives in $\mathbb{R}^3$ from start to finish. This is Varignon's theorem, and it is a result a vector proof gives for free where a proof by drawing would not even know how to start.

c) $P = (2, 0, 1)$, $Q = (5, 2, 1)$, $R = (3, 3, 2)$, $S = (0, 1, 2)$. Then $\overrightarrow{PQ} = (3, 2, 0)$ and $\overrightarrow{SR} = (3, 2, 0)$, equal, and both are $\frac{1}{2}\overrightarrow{AC} = \frac{1}{2}(6, 4, 0)$. The other pair agrees too: $\overrightarrow{PS} = \overrightarrow{QR} = (-2, 1, 1) = \frac{1}{2}\overrightarrow{BD}$. These four points $A$, $B$, $C$, $D$ are not coplanar (the $3 \times 3$ determinant of $\overrightarrow{AB}$, $\overrightarrow{AC}$, $\overrightarrow{AD}$ is $88$, not $0$), and the midpoints still form a flat parallelogram.

d) By a), the centre is the common midpoint of the diagonals: $\frac{1}{2}(P + R) = \frac{1}{2}(5, 3, 3) = \left(\frac{5}{2}, \frac{3}{2}, \frac{3}{2}\right)$, and $\frac{1}{2}(Q + S) = \frac{1}{2}(5, 3, 3)$, the same point. In general $\frac{1}{2}(P + R) = \frac{1}{2}\left(\frac{A + B}{2} + \frac{C + D}{2}\right) = \frac{1}{4}(A + B + C + D)$, and here $\frac{1}{4}(10, 6, 6)$ agrees.

e) Parallel vectors can have different lengths, and a quadrilateral with one pair of parallel sides of different lengths is a trapezoid. With $A(0, 0)$, $B(4, 0)$, $C(3, 2)$, $D(1, 2)$: $\overrightarrow{AB} = (4, 0)$ and $\overrightarrow{DC} = (2, 0)$ are parallel, $\overrightarrow{AB} = 2\overrightarrow{DC}$, but not equal, and $ABCD$ is not a parallelogram. The missing line is the equality of the vectors, which the proof in b) actually delivers, same direction AND same length. On a paper it costs the conclusion mark.