Points, straight lines and shapes in 3D

Unit 8 Points, straight lines and shapes in 3D

Key unit competence: Apply the meaning of coordinates in solving problems involving points, lines, planes and shapes in three dimensions.

Key ideas

• Distance $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$; midpoint $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$.
• Line through $A(x_0,y_0,z_0)$ with direction $\vec{u}=(a,b,c)$: vector form $\vec{r}=\vec{a}+t\vec{u}$; parametric $x=x_0+at$, $y=y_0+bt$, $z=z_0+ct$; symmetric $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$.
• Two lines are parallel, intersecting, or skew (not parallel and not meeting).
• Plane with normal $\vec{n}=(a,b,c)$ through $A$: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, i.e. $ax+by+cz+d=0$. A plane through three points uses $\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}$.
• Distance from $P(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$: $\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$. Angle between planes = angle between normals.
• Sphere with centre $(a,b,c)$, radius $r$: $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$.

Formulas to know

Common misconceptions

• "Two lines in space that do not meet are parallel." No: they may be skew.
• "The vector $(a,b,c)$ in $ax+by+cz+d=0$ lies in the plane." No: it is the normal, perpendicular to the plane.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find parametric and symmetric equations of the line through $A(2,-1,3)$ and $B(4,1,0)$.

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   Direction $(2,2,-3)$: $x=2+2t$, $y=-1+2t$, $z=3-3t$; $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-3}{-3}$.

   Common mistake: Using $B$'s coordinates as the direction.

2. Show that the plane $x+2y-2z=9$ is tangent to the sphere $x^2+y^2+z^2=9$.

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   Distance from the centre $O$ to the plane: $\frac{9}{3}=3=r$. Tangent.

   Common mistake: Comparing with $r^2=9$.

3. Find the equation of the plane through $A(1,2,1)$, $B(2,3,0)$, $C(0,1,3)$.

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   $\overrightarrow{AB}=(1,1,-1)$, $\overrightarrow{AC}=(-1,-1,2)$; normal $(1\cdot 2-(-1)(-1),(-1)(-1)-1\cdot 2,-1+1)=(1,-1,0)$. Plane: $x-y+1=0$.

   Common mistake: Forgetting to check that all three points satisfy the equation.

4. Find the point where the line through $(1,1,1)$ with direction $(1,2,3)$ meets the plane $z=0$.

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   $1+3t=0$, $t=-\frac{1}{3}$: $\left(\frac{2}{3},\frac{1}{3},0\right)$.

   Common mistake: Setting $t=0$.

5. Find the distance from $P(1,-2,3)$ to the plane $2x-y+2z+1=0$.

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   $\frac{|2+2+6+1|}{3}=\frac{11}{3}$.

   Common mistake: Dropping the constant term.

6. Find the angle between the line with direction vector $(1,1,0)$ and the plane $z=0$.

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   The direction $(1,1,0)$ is perpendicular to the normal $(0,0,1)$, so the line is parallel to the plane: angle $0^{\circ }$.

   Common mistake: Giving the angle between the direction and the normal ($90^{\circ }$) as the answer.

7. Determine whether the lines $\frac{x-1}{1}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x}{1}=\frac{y-1}{1}=\frac{z}{2}$ intersect.

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   Points $(1+t,2t,-1+t)$ and $(s,1+s,2s)$: $1+t=s$, $2t=1+s$ give $t=2$, $s=3$; then $z$: $1$ vs $6$. They do not meet and are not parallel: skew.

   Common mistake: Using the same parameter for both lines.

8. Find the equation of the sphere with diameter $A(1,0,2)$, $B(3,4,-2)$.

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   Centre $(2,2,0)$, radius $\frac{1}{2}\sqrt{4+16+16}=3$: $(x-2{)}^2+(y-2{)}^2+z^2=9$.

   Common mistake: Using the full length $AB$ as the radius.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the plane containing $(1,2,1)$ and the line $x=t$, $y=1+t$, $z=2t$.	a. $y-x=1$
   2. Which are symmetric equations of the line through $(1,2,3)$ with direction $(2,-1,4)$?	b. $\frac{10}{3}$
   3. Find the plane through $(1,0,0)$, $(0,2,0)$ and $(0,0,3)$.	c. $6x+3y+2z=6$
   4. A sphere has centre $(1,2,-2)$ and touches the plane $2x+y+2z=10$. Find its radius.	d. $\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{4}$

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   1 → a ($y-x=1$) · 2 → d ($\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{4}$) · 3 → c ($6x+3y+2z=6$) · 4 → b ($\frac{10}{3}$)

   • Step 1. $A(0,1,0)$ is on the line, $\vec{d}=(1,1,2)$ and $\overrightarrow{AP}=(1,1,1)$.
     Step 2. $\vec{n}=\vec{d}\times \overrightarrow{AP}=(1-2,2-1,1-1)=(-1,1,0)$.
     Step 3. $-x+y=c$; at $A$: $c=1$. So $y-x=1$.
   • Step 1. Through $(x_0,y_0,z_0)$ with direction $(a,b,c)$: $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$.
     Step 2. Substitute the point $(1,2,3)$ and the direction $(2,-1,4)$.
   • Step 1. Intercept form: $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1$.
     Step 2. Multiply by 6: $6x+3y+2z=6$.
   • Step 1. The radius is the distance from the centre to the plane.
     Step 2. $\frac{|2+2-4-10|}{\sqrt{4+1+4}}=\frac{10}{3}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $4$  ·  $2x-4y+6z=5$  ·  $2$  ·  $(1,-1,0)$
   1. The plane through $(1,1,1)$ containing the line $x=t$, $y=t$, $z=0$ has normal .
   2. The sphere $x^2+y^2+z^2-2x+4y-6z=2$ has radius .
   3. Which plane is parallel to $x-2y+3z=1$? →
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   1: $(1,-1,0)$ · 2: $4$ · 3: $2x-4y+6z=5$

   • Step 1. The line passes through $O$ with direction $(1,1,0)$; $\overrightarrow{OP}=(1,1,1)$.
     Step 2. $(1,1,0)\times (1,1,1)=(1-0,0-1,1-1)=(1,-1,0)$.
   • Step 1. Complete squares: $(x-1{)}^2+(y+2{)}^2+(z-3{)}^2=2+1+4+9$.
     Step 2. $=16$, so $r=4$.
   • Step 1. Parallel planes have parallel normals.
     Step 2. $(2,-4,6)=2(1,-2,3)$, and the constants differ, so the planes are parallel and distinct.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find a direction vector of the line where $x+y+z=1$ meets $x-y=0$.	a. $(1,1,-2)$
   2. Find the plane through $(1,-1,2)$ with normal $(2,3,-1)$.	b. $k=5$
   3. Find $k$ so that $2x+ky+z=1$ and $x-y+3z=4$ are perpendicular.	c. $2x+3y-z+3=0$
   4. Find the midpoint of $A(2,-1,4)$ and $B(6,3,0)$, and the length $AB$.	d. $(4,1,2)$ and $4\sqrt{3}$

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   1 → a ($(1,1,-2)$) · 2 → c ($2x+3y-z+3=0$) · 3 → b ($k=5$) · 4 → d ($(4,1,2)$ and $4\sqrt{3}$)

   • Step 1. The line is perpendicular to both normals, so $\vec{d}=(1,1,1)\times (1,-1,0)$.
     Step 2. $(1\cdot 0-1\cdot (-1),1\cdot 1-1\cdot 0,1\cdot (-1)-1\cdot 1)=(1,1,-2)$.
   • Step 1. $2(x-1)+3(y+1)-(z-2)=0$.
     Step 2. $2x-2+3y+3-z+2=0$, so $2x+3y-z+3=0$.
   • Step 1. Perpendicular planes have perpendicular normals: $(2,k,1)\cdot (1,-1,3)=0$.
     Step 2. $2-k+3=0\Rightarrow k=5$.
   • Step 1. Midpoint: $\left(\frac{2+6}{2},\frac{-1+3}{2},\frac{4+0}{2}\right)=(4,1,2)$.
     Step 2. $\overrightarrow{AB}=(4,4,-4)$, so $AB=\sqrt{48}=4\sqrt{3}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $13$  ·  $k=2$  ·  $k=6$  ·  $(4,-1,7)$
   1. Find where the line $x=2+t$, $y=1-t$, $z=3+2t$ meets the plane $x+y+z=10$. →
   2. Find $k$ so that the line $\vec{r}=(1,2,3)+t(2,k,1)$ is parallel to the plane $x+y-4z=7$. →
   3. Find the distance between $A(1,2,3)$ and $B(4,6,15)$. →
   Show answer

   1: $(4,-1,7)$ · 2: $k=2$ · 3: $13$

   • Step 1. Substitute: $(2+t)+(1-t)+(3+2t)=10\Rightarrow 6+2t=10$.
     Step 2. $t=2$: $(4,-1,7)$.
   • Step 1. The line is parallel to the plane when $\vec{d}\cdot \vec{n}=0$.
     Step 2. $2+k-4=0\Rightarrow k=2$.
   • Step 1. $\overrightarrow{AB}=(3,4,12)$.
     Step 2. $AB=\sqrt{9+16+144}=\sqrt{169}=13$.

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