Question bank — Lines and planes in 3D — vector equations

The lines $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ and $\mathbf{r}=\mathbf{a}+t(2\mathbf{d})$ are different lines. ::: False. Doubling the direction only changes how fast the dial $t$ moves you, not which points you reach. The set of points is identical, so it's the same line. Two lines with parallel direction vectors must be the same line. ::: False. Parallel directions make them parallel lines, but they can be shifted apart (like two rails). They coincide only if they also share at least one point. A single linear equation like $2x-y+3z=5$ describes a line in 3D. ::: False. In 3D one linear equation pins down one degree of freedom, leaving a 2D flat sheet — a plane. A line in 3D needs two such equations (or one vector equation). The vector $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ passes through the origin for some $t$ only if the origin lies on the line. ::: True. Reaching the origin means $\mathbf{a}+t\mathbf{d}=0$ has a solution $t$; that solution exists exactly when the origin is a point of the line — it isn't guaranteed. In $ax+by+cz=d$, scaling the whole equation by $2$ gives a different plane. ::: False. $2ax+2by+2cz=2d$ has the same solution set; the normal just doubled in length, still pointing the same way. Same plane. The parametric plane $\mathbf{r}=\mathbf{a}+s\mathbf{u}+t\mathbf{v}$ requires $\mathbf{u}$ and $\mathbf{v}$ to be perpendicular. ::: False. They only need to be non-parallel so they span two genuinely different directions. Skewed (non-perpendicular) spanning vectors still sweep out the whole plane. The distance from a point to a plane can be negative. ::: False. Distance is a length, so the formula uses absolute value $|\mathbf{p}\cdot \mathbf{n}-d|$ with $d=\mathbf{a}\cdot \mathbf{n}$. The unsigned quantity $\mathbf{p}\cdot \mathbf{n}-d$ can be negative — it only tells you which side of the plane you're on. If a line's direction vector is $\mathbf{d}=(2,4,0)$, the line is horizontal (constant height $z$). ::: True. The $z$-component of $\mathbf{d}$ is $0$, so walking along the line never changes $z$. Every point sits at whatever fixed height $\mathbf{a}$ started at. Every point in a plane satisfies $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$, and only plane points do. ::: True. In-plane displacements $\mathbf{r}-\mathbf{a}$ are perpendicular to $\mathbf{n}$ (dot $=0$); any point off the plane gives a displacement with a nonzero component along $\mathbf{n}$, so the dot product isn't zero.

"Plane $3x+y-2z=7$, so its normal is $(3,1,-2,7)$." ::: The normal is only $(3,1,-2)$ — the coefficients of $x,y,z$. The $7$ is the constant $d=\mathbf{a}\cdot \mathbf{n}$; it slides the plane along the normal but carries no direction, so it can't be part of the normal vector. "Line through $(1,0,2)$ and $(3,4,2)$: symmetric form $\frac{x-1}{2}=\frac{y}{4}=\frac{z-2}{0}$." ::: You may never divide by $0$. Since $\mathbf{d}$ has $c=0$, write $z=2$ separately and use symmetric form only for $x$ and $y$: $\frac{x-1}{2}=\frac{y}{4},z=2$. "To get the plane's normal from spanning vectors $\mathbf{u},\mathbf{v}$, compute $\mathbf{u}\cdot \mathbf{v}$." ::: The dot product returns a number, not a direction. The normal must be a vector perpendicular to both $\mathbf{u}$ and $\mathbf{v}$, which is exactly what $\mathbf{u}\times \mathbf{v}$ (the cross product) produces — see the perpendicular arrow in Figure s02. "Line and plane are parallel because the line's direction equals the plane's normal." ::: That's backwards. If the direction equals the normal, the line pierces the plane at a right angle. A line is parallel to a plane when its direction is perpendicular to the normal (dot product zero). "Point–to–line distance $=\frac{(\mathbf{p}-\mathbf{a})\times \mathbf{d}}{|\mathbf{d}|}$." ::: The numerator must be a magnitude: $\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}$. A cross product is a vector; distance is a single non-negative number, so you take its length. "The point $A=(2,-1,3)$ lies on plane $x+2y+2z=6$? Check: $2+2(-1)+2(3)=2-2+6=8\ne 6$." ::: Arithmetic slip: $2-2+6=6$, not $8$. It equals $d=6$, so $A$ does lie on the plane — as it must, since the plane was built through $A$.

Why does the line equation separate a point $\mathbf{a}$ from a direction $\mathbf{d}$ instead of using slope like $y=mx+c$? ::: Slope divides by a run and breaks for vertical lines; splitting position from direction avoids division entirely and works in any dimension, including 3D where "slope" isn't even well-defined. Why does the normal form $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ capture a whole 2D plane with just one equation? ::: One dot-product-equals-zero condition removes exactly one degree of freedom (the "how far along $\mathbf{n}$" direction), leaving the two in-plane freedoms intact — precisely a plane. Why do we use the cross product, not the dot product, to build a plane's normal? ::: We need a vector perpendicular to two given directions at once. The cross product is defined to output exactly that orthogonal vector (Figure s02); the dot product outputs a scalar and can't point anywhere. Why does dividing $|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|$ by $|\mathbf{d}|$ give the perpendicular distance to a line? ::: The cross product's length is the parallelogram area $|\mathbf{p}-\mathbf{a}||\mathbf{d}|\sin \theta$ (Figure s01). Dividing out $|\mathbf{d}|$ leaves $|\mathbf{p}-\mathbf{a}|\sin \theta$, which is exactly the perpendicular height from $P$ to the line. Why does the point–to–plane formula divide by $|\mathbf{n}|$? ::: You must measure how far the displacement projects along a unit normal; dividing $\mathbf{n}$ by its own length turns it into $\hat{\mathbf{n}}$, so $|(\mathbf{p}-\mathbf{a})\cdot \hat{\mathbf{n}}|$ is a true length, not scaled by $\mathbf{n}$'s size. Why is a line through two points written $(1-t)\mathbf{a}+t\mathbf{b}$ so convenient? ::: The weights $(1-t)$ and $t$ always sum to $1$, so $t=0$ lands on $A$, $t=1$ lands on $B$, and $0\le t\le 1$ traces exactly the segment between them. Why can two different $(\mathbf{a},\mathbf{d})$ pairs describe the same line? ::: A line is a set of points, not a formula. Any point on it works as $\mathbf{a}$, and any scalar multiple of the direction works as $\mathbf{d}$, so infinitely many descriptions give one geometric object.

What happens to the vector equation of a line if $\mathbf{d}=0$? ::: It collapses: every $t$ gives the single point $\mathbf{a}$. A zero direction has no "way to walk," so you get a point, not a line — that's why the definition demands $\mathbf{d}\ne 0$. What if you try to build a plane $\mathbf{r}=\mathbf{a}+s\mathbf{u}+t\mathbf{v}$ with $\mathbf{u}$ parallel to $\mathbf{v}$? ::: Both spanning arrows point the same way, so combinations only sweep out a line, not a plane. Their cross product $\mathbf{u}\times \mathbf{v}=0$ signals the failure — no normal exists. Can three points fail to determine a plane? ::: Yes — if the three points are collinear (all on one line), infinitely many planes contain that line. Then $\mathbf{u}=B-A$ and $\mathbf{v}=C-A$ are parallel and $\mathbf{u}\times \mathbf{v}=0$. What does the point–to–line distance formula give when $P$ lies on the line? ::: $\mathbf{p}-\mathbf{a}$ is then parallel to $\mathbf{d}$, so $\sin \theta =0$ and the cross product is $0$. Distance $=0$ — correctly, the point is on the line. What does $\mathbf{p}\cdot \mathbf{n}-d=0$ mean in the plane-distance formula (with $d=\mathbf{a}\cdot \mathbf{n}$)? ::: The point $P$ satisfies the plane equation exactly, so it lies on the plane and its distance is $0$. The sign of that expression (when nonzero) tells you which side you're on. If a plane passes through the origin, what is its constant $d$? ::: $d=\mathbf{a}\cdot \mathbf{n}$ with $\mathbf{a}=0$ gives $d=0$. So the equation is $ax+by+cz=0$ — a plane through the origin always has zero constant term. A line lies entirely within a plane — what must be true of the line's direction $\mathbf{d}$ and the plane's normal $\mathbf{n}$? ::: $\mathbf{d}\cdot \mathbf{n}=0$ (direction perpendicular to normal, so the line doesn't rise out of the plane) and one point of the line satisfies the plane equation. Both conditions are needed; the first alone only makes them parallel.