Vectors — scalar product, angle-finding, and skew/parallel/intersecting lines

$A=(1,0,2)$, $B=(3,4,1)$, $C=(2,1,5)$. Find the angle $ABC$, to 1 decimal place.

• $39.6°$, from $\overrightarrow{BA}\cdot\overrightarrow{BC} = 18$, $|\overrightarrow{BA}|=\sqrt{21}$, $|\overrightarrow{BC}|=\sqrt{26}$, giving $\cos\theta \approx 0.7704$

  Correct. $\overrightarrow{BA}=(-2,-4,1)$ and $\overrightarrow{BC}=(-1,-3,4)$, both correctly formed pointing away from vertex B. The scalar product is positive, so the angle is acute — and stays acute, since this is what the real geometry gives; there is no rule that vertex angles in this topic must come out obtuse.

• B$61.8°$, using $\overrightarrow{AC}\cdot\overrightarrow{BC}$ instead of $\overrightarrow{BA}\cdot\overrightarrow{BC}$

  $\overrightarrow{AC}$ starts at A, not at B — this pair does not share a common vertex, so its scalar product is a real number but not the cosine of any angle in this triangle. This is precisely the wrong-pairing error real scripts are documented making on this exact question type.

• C$140.4°$, using $\overrightarrow{AB}\cdot\overrightarrow{BC}$ instead of $\overrightarrow{BA}\cdot\overrightarrow{BC}$

  $\overrightarrow{AB}$ points INTO B, not out of it — using it unreversed flips the sign of the scalar product and turns a genuinely acute angle into a wrong obtuse-looking one. Both vectors in the formula must point AWAY from the vertex the angle sits at.

• DThe scalar product is the vector $(2, 12, 4)$, so no single angle can be found without more information

  This treats the scalar product as if it were a componentwise (vector) product rather than a single number — the documented real confusion between what a "scalar" product actually returns and what a vector product would. $\overrightarrow{BA}\cdot\overrightarrow{BC}$ is one number, $18$, not three unadded components.

Traps tested: Wrong vectors selected for the angle calculation · Scalar vs vector product confusion

For what value of $t$ are the vectors $(2, t, -3)$ and $(4, -1, t)$ perpendicular?

• A$t = -2$

  This is what a sign slip on the $-4t$ term produces: treating $8-t-3t$ as $8+t+3t=8+4t$ and setting THAT to zero gives $t=-2$. The two negative terms in the scalar product are real and must both survive to the final equation.

• $t = 2$

  Correct. Perpendicular means the scalar product is zero: $(2)(4)+(t)(-1)+(-3)(t) = 8 - t - 3t = 8-4t = 0$, so $t=2$. Check: $(2,2,-3)\cdot(4,-1,2) = 8-2-6=0$. ✓

• C$t = 8$

  This comes from dropping the third component entirely — computing only $(2)(4)+(t)(-1)=8-t=0$, as though these were two-dimensional vectors. All three components of a 3D scalar product must be included; dropping one silently turns the calculation into a different, smaller problem.

• D$t = -4$

  This combines two separate slips — a sign error on one term and mismatching which vector's $t$ belongs where. It is a plausible-looking number, but $(2,-4,-3)\cdot(4,-1,-4) = 8+4+12=24 \neq 0$ — not actually perpendicular, which a quick check catches in seconds.

Traps tested: Sign slip combining scalar product terms · Component dropped from 3d scalar product

Which pair of direction vectors below are scalar multiples of each other (and therefore represent parallel lines)?

• $(2,-4,6)$ and $(-1,2,-3)$

  Correct. $(-1,2,-3) = -\tfrac{1}{2}(2,-4,6)$ — check every component: $-\tfrac{1}{2}(2)=-1$ ✓, $-\tfrac{1}{2}(-4)=2$ ✓, $-\tfrac{1}{2}(6)=-3$ ✓. A single consistent $k=-\tfrac{1}{2}$ works throughout, even though the two vectors look nothing alike at a glance — which is exactly the point of checking properly rather than eyeballing it.

• B$(1,2,3)$ and $(2,4,5)$

  Close, but not quite: $k=2$ fits the first two components ($2=2\times1$, $4=2\times2$) but the third would need to be $2\times3=6$, and it is $5$. Checking only the first two components and stopping there is exactly how this one gets missed.

• C$(3,-1,2)$ and $(6,-2,5)$

  Also close: $k=2$ fits the first two components ($6=2\times3$, $-2=2\times-1$) but the third would need $2\times2=4$, not $5$. The same near-miss pattern as option B, with different numbers — the third component is what genuinely decides it every time.

• D$(4,0,-2)$ and $(-2,1,1)$

  $k=-0.5$ fits the first component ($-2 = -0.5\times4$), but the second would then need $-0.5\times0=0$, and it is $1$. Not a scalar multiple, and therefore not parallel.

Traps tested: Non parallel shown via non identical not non scalar multiple

$l: \mathbf{r}=(1,2,3)+\lambda(1,0,1)$ and $m: \mathbf{r}=(2,2,4)+\mu(0,1,1)$. What is the relationship between $l$ and $m$?

• AThey are parallel, since both direction vectors contain a zero component

  Having a zero in the same-looking position is a coincidence of how these two vectors happen to be written, not a test for parallel direction. The actual test is whether one vector is a scalar multiple of the other — and $(0,1,1)$ is not $k(1,0,1)$ for any $k$, since that would force the second component to stay zero.

• BThey are skew, since their direction vectors are not parallel and two non-parallel lines in 3D cannot meet

  The first half is right and the conclusion drawn from it is not: not-parallel rules out ONE of the two ways lines can fail to be skew, not both. Non-parallel lines in three dimensions can still intersect — this pair does, at (2,2,4) — and checking the third equation is exactly the step that tells the two cases apart.

• They intersect, at $(2,2,4)$

  Correct. The direction vectors $(1,0,1)$ and $(0,1,1)$ are not scalar multiples (no $k$ can turn a nonzero first component into a zero one). Equating gives $x$: $1+\lambda=2 \Rightarrow \lambda=1$; $y$: $2=2+\mu \Rightarrow \mu=0$; checking the $z$-equation: $3+\lambda = 4+\mu \Rightarrow 3+1=4+0 \Rightarrow 4=4$ — consistent. Substituting $\lambda=1$ into $l$ gives $(2,2,4)$.

• DNot enough information — two unknowns cannot be determined from a system with a third equation left over

  Two unknowns need exactly two equations to solve, which the $x$- and $y$-equations already supply here; the third equation is not an obstacle, it is the deliberate check for whether the solution found also satisfies the geometry in all three dimensions at once.

Traps tested: Parallel judged by inspection not ratio · Not parallel assumed to mean skew · Miscounts equations needed

Two lines have direction vectors $(1,1,0)$ and $(-3,-3,1)$. Their scalar product gives $\cos\theta = -6/\sqrt{38} \approx -0.9732$, i.e. $\theta \approx 166.7°$. What is the angle BETWEEN the two lines, to 1 decimal place?

• A$90°$, since a negative scalar product means the lines are perpendicular

  A negative scalar product means the RAW angle computed is obtuse — nothing more specific than that. Perpendicularity is the one special case where the scalar product is exactly zero, not merely negative.

• B$166.7°$, reported exactly as the formula gives it

  Correct for a genuine named-vertex angle like ∠ABC — but a between-two-lines question has no vertex to anchor the sign of θ to, so the raw obtuse value here is an artefact of which direction each vector happened to be written in, not a property of the actual lines.

• CCannot be found, since the direction vectors point in nearly opposite directions

  Reversing a direction vector never changes which line it describes — $(-3,-3,1)$ and $(3,3,-1)$ define the same line. "Nearly opposite" is a fact about how the two vectors were chosen to be written, not a fact that prevents the angle between the lines from being found.

• $13.3°$, since a line has no built-in direction and the angle between two lines is conventionally the acute value, $180° - 166.7°$

  Correct. Reversing either direction vector describes the same line, so the raw $166.7°$ (which depends on which way the two vectors happened to be written) is not itself a property of the lines — only its acute supplement is. This is the opposite adjustment from a real, named vertex angle like ∠ABC, which is reported exactly as computed, obtuse or not.

Traps tested: Negative scalar product assumed to mean perpendicular · Between lines angle not converted to acute · Direction vector orientation treated as fixed

A student shows two lines do not intersect, then argues: "the direction vectors $(1,0,1)$ and $(0,1,1)$ are not perpendicular, since their scalar product is $1$, not $0$ — so the lines are not skew." What is wrong with this argument?

• AThe scalar product calculation itself is wrong — it should be $0$

  $(1)(0)+(0)(1)+(1)(1) = 1$ is correct arithmetic. The error in the argument is not a wrong number; it is checking the wrong PROPERTY entirely.

• Perpendicularity is irrelevant to skewness — the missing condition is that the direction vectors must be shown NOT to be scalar multiples of each other

  Correct, and this is a directly documented real error, not a hypothetical one: an examiner report records candidates who, having shown two lines did not intersect, went on to check perpendicularity instead of checking for parallel direction vectors. Skew lines can be perpendicular, or nowhere near it — the angle between them has no bearing on whether they are skew.

• CThe lines could still be parallel, so more information is needed regardless of the perpendicularity check

  True that parallel needs checking too — but that alone doesn't diagnose what's wrong with THIS argument specifically: even after confirming the lines are not parallel by the correct test, checking perpendicularity would still contribute nothing to a skew conclusion, because it was never one of the two conditions skewness requires.

• DNon-perpendicular lines are always intersecting, so this actually proves the lines are NOT skew

  Non-perpendicular says nothing about whether lines intersect — most pairs of intersecting lines meet at some angle other than 90°, and most pairs of skew lines are not perpendicular either. The claim inverts a property that was never linked to intersection in the first place.

Traps tested: Arithmetic blamed for a conceptual error · Identifies a real gap but not the one in this argument · Skew shown via not perpendicular instead of not parallel