Lines and Planes Practice

These questions follow Equations of Lines, Equations of Planes, and Intersections of Lines and Planes — building the equations, converting between forms, and settling where things meet. Audit every equation you build by substituting a point you know is on it.

Lines

1. Determine vector and parametric equations of the line through $(3,2,-1)$ and $(0,2,1)$, and represent the same line as the intersection of two planes.
2. Consider the system $2x+y=5$ and $4x+2y=7$ in two-space. How many solution points are there, and what does the answer look like geometrically?
3. Show that the lines $\vec{r}=(1,2,0)+t(1,1,1)$ and $\vec{r}=(2,4,6)+s(1,-1,2)$ are skew.

Answer 1

Direction: $(0,2,1)-(3,2,-1)=(-3,0,2)$. Vector equation: $\vec{r}=(3,2,-1)+t(-3,0,2)$. Parametric: $x=3-3t$, $y=2$, $z=-1+2t$. For the two planes: $y=2$ is one of them already ($y$ never moves). Eliminate $t$ between $x$ and $z$: $t=\frac{3-x}{3}=\frac{z+1}{2}$, so $2(3-x)=3(z+1)$, which tidies to $2x+3z-3=0$. The line is the crease where $y-2=0$ meets $2x+3z-3=0$. Audit with $(3,2,-1)$: $6-3-3=0$. ✓

Answer 2

Doubling the first equation gives $4x+2y=10$, but the second insists $4x+2y=7$. No pair $(x,y)$ can satisfy both — no solutions. Geometrically: two parallel, distinct lines (same slope, different intercepts) that never meet. A system’s algebra and its picture always tell the same story.

Answer 3

The directions $(1,1,1)$ and $(1,-1,2)$ are not scalar multiples — not parallel. Do they intersect? Match components: $x$: $1+t=2+s$ and $y$: $2+t=4-s$. Adding the two conditions ($t-s=1$ and $t+s=2$) gives $t=\frac{3}{2}$, $s=\frac{1}{2}$. Now the $z$ test: $t=\frac{3}{2}$ versus $6+2s=7$. Since $\frac{3}{2}\ne 7$, no common point exists. Not parallel and not intersecting: skew — the highways at different heights.

Planes

4. Determine the scalar equation of the plane through $(3,2,5)$, $(0,-2,2)$, and $(1,3,1)$.
5. Represent the plane $\vec{r}=(2,1,0)+s(1,-1,3)+t(2,0,-5)$ with a scalar equation.

Answer 4

Two vectors in the plane, first point to the others: $\vec{u}=(-3,-4,-3)$ and $\vec{v}=(-2,1,-4)$. Normal: $\vec{n}=\vec{u}\times \vec{v}=((-4)(-4)-(-3)(1),(-3)(-2)-(-3)(-4),(-3)(1)-(-4)(-2))=(19,-6,-11)$ Using $(3,2,5)$: $19(3)-6(2)-11(5)=57-12-55=-10$, so the plane is $19x-6y-11z+10=0$. Audit with the other two points: $0+12-22+10=0$ ✓ and $19-18-11+10=0$ ✓. Three points, one plane, fully checked.

Answer 5

The normal is the cross product of the two direction vectors: $(1,-1,3)\times (2,0,-5)=((-1)(-5)-3(0),3(2)-1(-5),1(0)-(-1)(2))=(5,11,2)$ Through $(2,1,0)$: $5(2)+11(1)+2(0)=21$, so $5x+11y+2z-21=0$. Audit: the normal dotted with each direction vector gives $5-11+6=0$ and $10+0-10=0$ — perpendicular to both, as a normal must be.

Where they meet

6. Determine the point where the line $\vec{r}=(1,0,2)+t(2,1,-1)$ meets the plane $x-y+2z=9$.
7. Determine the intersection of the planes $x+y+z=6$, $x-y+z=2$, and $2x+y-z=1$, and describe the configuration.
8. Determine the distance from the point $(3,1,2)$ to the plane $2x-y+2z-5=0$.

Answer 6

Substitute the parametric coordinates $x=1+2t$, $y=t$, $z=2-t$ into the plane: $(1+2t)-t+2(2-t)=5-t=9\Rightarrow t=-4$ The point: $(1-8,-4,2+4)=(-7,-4,6)$. Audit in the plane: $-7+4+12=9$. ✓ One value of $t$, one crossing — the line pierces the plane.

Answer 7

Subtract the second equation from the first: $2y=4$, so $y=2$. Then $x+z=4$ (first equation) and, from the third, $2x-z=-1$. Adding: $3x=3$, so $x=1$ and $z=3$. Unique solution $(1,2,3)$ — audit in all three: $1+2+3=6$ ✓, $1-2+3=2$ ✓, $2+2-3=1$ ✓. Three planes meeting in a single point, like the corner of a room.

Answer 8

The distance is measured along the normal: $\begin{array}{rl}d& =\frac{|2(3)-1(1)+2(2)-5|}{\sqrt{2^2+(-1{)}^2+2^2}}\\ & =\frac{|6-1+4-5|}{3}=\frac{4}{3}\end{array}$ The numerator is the plane’s equation evaluated at the point — zero would mean the point lies on the plane, and the farther from zero, the farther from the plane. Dividing by $|\vec{n}|=3$ converts that raw imbalance into metres of perpendicular clearance: the certification step of The Flight Path.