Points, Directed Lines, Dot Products, Cross Products, and Equations of Lines and Planes

Points, Directed Lines, Dot Products, and Cross Products

In $\mathbb{R}^3$R3 with axes labeled $x$x, $y$y, and $z$z, the point $(a, b, c)$(a,b,c) is plotted by showing dashed red lines projecting its coordinates $a$a, $b$b, and $c$c onto the respective axes to form a rectangular box, and the position vector $\vec{u}$u is drawn in blue from the origin to the point $(a, b, c)$(a,b,c), illustrating how points in three-dimensional space are located.

In $\mathbb{R}^n$Rn each point has $n$n coordinates:

$a = (a_1, a_2, a_3, \ldots, a_n)$a=(a1​,a2​,a3​,…,an​) $b = (b_1, b_2, b_3, \ldots, b_n)$b=(b1​,b2​,b3​,…,bn​)

Directed line

Given two points in $\mathbb{R}^n$Rn

$P = (a_1, a_2, a_3, \ldots, a_n)$P=(a1​,a2​,a3​,…,an​) $Q = (b_1, b_2, b_3, \ldots, b_n)$Q=(b1​,b2​,b3​,…,bn​)

we can define a directed line

$\vec{PQ} = \langle b_1 - a_1,\ b_2 - a_2,\ \ldots,\ b_n - a_n \rangle$PQ​=⟨b1​−a1​, b2​−a2​, …, bn​−an​⟩

Note $\vec{PQ} \neq \vec{QP}$PQ​=QP​ but $\vec{PQ} = -\vec{QP}$PQ​=−QP​

$|\vec{PQ}| = \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2 + \cdots + (b_n - a_n)^2}$∣PQ​∣=(b1​−a1​)2+(b2​−a2​)2+⋯+(bn​−an​)2​

is the length of $\vec{PQ}$PQ​.

Example in $\mathbb{R}^2$R2

Suppose $P$P at $(1,2)$(1,2) and $Q$Q at $(2,2)$(2,2)

$|\vec{PQ}| = \sqrt{(c-a)^2 + (d-b)^2}$∣PQ​∣=(c−a)2+(d−b)2​

is its length.

$|\vec{PQ}| = \sqrt{(2-1)^2 + (2-2)^2} = 1$∣PQ​∣=(2−1)2+(2−2)2​=1

Vectors

Given a directed line $\vec{PQ}$PQ​, we define a vector $\vec{v}$v by setting

$\vec{v} = \langle b_1 - a_1,\ b_2 - a_2,\ \ldots,\ b_n - a_n \rangle$v=⟨b1​−a1​, b2​−a2​, …, bn​−an​⟩

$\vec{u} = \langle a, b \rangle = a\vec{i} + b\vec{j}$u=⟨a,b⟩=ai+bj​  $\vec{v} = \langle a, b, c \rangle = a\vec{i} + b\vec{j} + c\vec{k}$v=⟨a,b,c⟩=ai+bj​+ck

$\vec{i} = \langle 1, 0, 0 \rangle$i=⟨1,0,0⟩ $\vec{j} = \langle 0, 1, 0 \rangle$j​=⟨0,1,0⟩ $\vec{k} = \langle 0, 0, 1 \rangle$k=⟨0,0,1⟩

Defining characteristics of vectors

1. Magnitude (the length of the directed line)

2. Direction (the direction of the directed line)

Operations on Vectors

$\vec{u} = \langle u_1, u_2, \dots, u_n \rangle$u=⟨u1​,u2​,…,un​⟩ $\vec{v} = \langle v_1, v_2, \dots, v_n \rangle$v=⟨v1​,v2​,…,vn​⟩

(1) Addition

Geometric way of adding vectors

Adding vectors:

$= \langle u_1 + v_1,\, u_2 + v_2,\, \dots,\, u_n + v_n \rangle$=⟨u1​+v1​,u2​+v2​,…,un​+vn​⟩

(2) Scaling

$k\,v = \langle v_1 + v_1,\, v_2 + v_2,\, \dots,\, v_n \rangle = \langle k\,v_1,\, k\,v_2,\, \dots,\, k\,v_n \rangle$kv=⟨v1​+v1​,v2​+v2​,…,vn​⟩=⟨kv1​,kv2​,…,kvn​⟩

• If $k > 1$k>1, then $k$k gets stretch.

• If $0 < k < 1$0<k<1, then $k$k shrinks $v$v.

• If $k < 0$k<0, the direction of $v$v also reverses.

(3) Subtraction

$\vec{u} - \vec{v} = \vec{u} + (-\vec{v})$u−v=u+(−v)

Examples

The zero vector has each component equal to zero.

$\vec{v} = \langle 0, 0, \dots, 0 \rangle$v=⟨0,0,…,0⟩

Rmk $|\vec{v}| = 0$∣v∣=0

Given a nonzero vector $\vec{v}$v, its unit vector is given by

$\vec{u} = \frac{\vec{v}}{|\vec{v}|}$u=∣v∣v​

Rmk We use a unit vector for its direction.

Dot Product (Another operation on vectors)

Law of Cosines

FACTS

1. $\sin^2 A + \cos^2 A = 1$sin2A+cos2A=1

2. $\cos(\pi - \theta) = \cos(\theta - \pi) = -\cos\theta$cos(π−θ)=cos(θ−π)=−cosθ

3. $\sin(\pi - \theta) = \sin\theta$sin(π−θ)=sinθ

4. $(A \pm B)^2 = A^2 + B^2 \pm 2AB$(A±B)2=A2+B2±2AB

   $\begin{aligned} c^2 &= (a - b\cos\theta)^2 + b^2 \sin^2\theta \\ &= a^2 - 2ab\cos\theta + b^2\cos^2\theta + b^2\sin^2\theta \\ &= a^2 + b^2 - 2ab\cos\theta\end{aligned}$c2​=(a−bcosθ)2+b2sin2θ=a2−2abcosθ+b2cos2θ+b2sin2θ=a2+b2−2abcosθ​

Dot Product

$\vec{u} = \langle u_1, u_2, u_3 \rangle$u=⟨u1​,u2​,u3​⟩ $\vec{v} = \langle v_1, v_2, v_3 \rangle$v=⟨v1​,v2​,v3​⟩ $\vec{w} = \vec{u} - \vec{v} = \langle u_1 - v_1, \; u_2 - v_2, \; u_3 - v_3 \rangle$w=u−v=⟨u1​−v1​,u2​−v2​,u3​−v3​⟩

$\Downarrow$⇓

$\begin{aligned} |w| &= \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 + (u_3 - v_3)^2} \\ &= u_1^2 + v_1^2 + u_2^2 + v_2^2 + u_3^2 + v_3^2 - 2u_1 v_1 - 2u_2 v_2 - 2u_3 v_3\end{aligned}$∣w∣​=(u1​−v1​)2+(u2​−v2​)2+(u3​−v3​)2​=u12​+v12​+u22​+v22​+u32​+v32​−2u1​v1​−2u2​v2​−2u3​v3​​ $\begin{aligned} |w|^2 &= |\vec{u}|^2 + |\vec{v}|^2 - 2|\vec{u}||\vec{v}|\cos\theta \\ &= u_1^2 + u_2^2 + u_3^2 + v_1^2 + v_2^2 + v_3^2 - 2|\vec{u}||\vec{v}|\cos\theta\end{aligned}$∣w∣2​=∣u∣2+∣v∣2−2∣u∣∣v∣cosθ=u12​+u22​+u32​+v12​+v22​+v32​−2∣u∣∣v∣cosθ​ $-2|\vec{u}||\vec{v}|\cos\theta = -2u_1 v_1 - 2u_2 v_2 - 2u_3 v_3$−2∣u∣∣v∣cosθ=−2u1​v1​−2u2​v2​−2u3​v3​ $|\vec{u}||\vec{v}|\cos\theta = u_1 v_1 + u_2 v_2 + u_3 v_3$∣u∣∣v∣cosθ=u1​v1​+u2​v2​+u3​v3​

The dot product of $\vec{u}$u and $\vec{v}$v is defined as follows:

algebraic $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 + \dots + u_n v_n$u⋅v=u1​v1​+u2​v2​+u3​v3​+⋯+un​vn​

geometric $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$u⋅v=∣u∣∣v∣cosθ

$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$cosθ=∣u∣∣v∣u⋅v​

Note: $\vec{u} \cdot \vec{v}$u⋅v is not a vector, but a scalar.

Projection

Scalar Projection

Let

$\vec{a} = \vec{PS}$a=PS $\vec{b} = \vec{PR}$b=PR

Then $|\vec{PQ}| = |\vec{b}|\cos\theta$∣PQ​∣=∣b∣cosθ is called the scalar projection of $\vec{b}$b onto $\vec{a}$a or the component of $\vec{b}$b along $\vec{a}$a and is denoted by

$\operatorname{comp}_{\vec{a}} \vec{b}$compa​b $|\vec{a}||\vec{b}|$∣a∣∣b∣ $|\vec{b}|\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$∣b∣cosθ=∣a∣a⋅b​ $\boxed{\operatorname{comp}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}}$compa​b=∣a∣a⋅b​​

$\longleftarrow$⟵ Scalar Projection

Vector Projection

$\operatorname{Proj}_{\vec{a}} \vec{b} = \operatorname{comp}_{\vec{a}} \vec{b} \;\frac{\vec{a}}{|\vec{a}|}$Proja​b=compa​b∣a∣a​ $= \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \frac{\vec{a}}{|\vec{a}|}$=∣a∣a⋅b​∣a∣a​ $\boxed{\operatorname{Proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \;\vec{a}}$Proja​b=∣a∣2a⋅b​a​

$\longleftarrow$⟵ Vector Projection

Cross Product

Given two vectors in $\mathbb{R}^3$R3

$A = \langle a_1, a_2, a_3 \rangle$A=⟨a1​,a2​,a3​⟩ $B = \langle b_1, b_2, b_3 \rangle$B=⟨b1​,b2​,b3​⟩

their cross product is given by

$A \times B = \langle a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1 \rangle$A×B=⟨a2​b3​−a3​b2​,a3​b1​−a1​b3​,a1​b2​−a2​b1​⟩

In linear algebra

$\begin{aligned} ax + by &= c_1 \\ cx + dy &= c_2 \end{aligned} \quad \longrightarrow \quad \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$ax+bycx+dy​=c1​=c2​​⟶(ac​bd​)(xy​)=(c1​c2​​)

The determinant of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$(ac​bd​)

$= \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc = \det\begin{pmatrix} a & b \\ c & d \end{pmatrix}$=​ac​bd​​=ad−bc=det(ac​bd​) $A \times B = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$A×B=​ia1​b1​​j​a2​b2​​ka3​b3​​​ $= \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \vec{i} - \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \vec{j} + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \vec{k}$=​a2​b2​​a3​b3​​​i−​a1​b1​​a3​b3​​​j​+​a1​b1​​a2​b2​​​k $= \left\langle \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}, -\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}, \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \right\rangle$=⟨​a2​b2​​a3​b3​​​,−​a1​b1​​a3​b3​​​,​a1​b1​​a2​b2​​​⟩ $= \langle a_2 b_3 - a_3 b_2,\ a_3 b_1 - a_1 b_3,\ a_1 b_2 - a_2 b_1 \rangle$=⟨a2​b3​−a3​b2​, a3​b1​−a1​b3​, a1​b2​−a2​b1​⟩ $A \times B \neq B \times A = -(A \times B)$A×B=B×A=−(A×B)

So the cross product is anti-commutative.

Direction of $A \times B$A×B is determined by the right hand rule.

$B \times A = -(A \times B)$B×A=−(A×B)

Magnitude of $A \times B$A×B is the area of the parallelogram determined by the vectors $A$A and $B$B.

$|A \times B| = |A|\,|B|\sin\theta$∣A×B∣=∣A∣∣B∣sinθ $\vec{a} \cdot \vec{b} = |a|\,|b|\cos\theta$a⋅b=∣a∣∣b∣cosθ $A = \langle 1, 2, 3 \rangle$A=⟨1,2,3⟩ $B = \langle 0, -1, 2 \rangle$B=⟨0,−1,2⟩ $A \cdot B = (1)(0) + (2)(-1) + (3)(2) = 4$A⋅B=(1)(0)+(2)(−1)+(3)(2)=4 $|A| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$∣A∣=12+22+32​=14​ $|B| = \sqrt{0^2 + (-1)^2 + (2)^2} = \sqrt{5}$∣B∣=02+(−1)2+(2)2​=5​ $\cos\theta = \frac{A \cdot B}{|A||B|} = \frac{4}{\sqrt{70}} = \frac{4\sqrt{70}}{70} = \frac{2\sqrt{70}}{35}$cosθ=∣A∣∣B∣A⋅B​=70​4​=70470​​=35270​​ $\operatorname{Proj}_A B = \frac{A \cdot B}{|A|^2} A = \frac{4}{14} \langle 1, 2, 3 \rangle = \frac{2}{7} \langle 1, 2, 3 \rangle$ProjA​B=∣A∣2A⋅B​A=144​⟨1,2,3⟩=72​⟨1,2,3⟩ $\operatorname{Proj}_B A = \frac{B \cdot A}{|B|^2} B = \frac{4}{5} \langle 0, -1, 2 \rangle$ProjB​A=∣B∣2B⋅A​B=54​⟨0,−1,2⟩ $A \times B = \langle 4 - (-3),\, (0)(3) - (0)(2),\, (-1)(1) - (0)(2) \rangle$A×B=⟨4−(−3),(0)(3)−(0)(2),(−1)(1)−(0)(2)⟩ $= \langle 7, -2, -1 \rangle$=⟨7,−2,−1⟩ $A \cdot (A \times B) = 7 - 4 - 3 = 0 \quad \text{Yay! :)}$A⋅(A×B)=7−4−3=0Yay! :) $B \cdot (A \times B) = 0 + 2 - 2 = 0$B⋅(A×B)=0+2−2=0

§ 12.5: Equations of Lines and Planes

Lines

$\boxed{\vec{r} = \vec{r}_0 + t\,\vec{v}} \qquad \text{Vector equation of } L$r=r0​+tv​Vector equation of L

Because

$\vec{r} = \langle x, y, z \rangle$r=⟨x,y,z⟩ $\vec{r}_0 = \langle x_0, y_0, z_0 \rangle$r0​=⟨x0​,y0​,z0​⟩ $t\vec{v} = \langle ta, tb, tc \rangle$tv=⟨ta,tb,tc⟩

we have

$\boxed{ \begin{aligned} x &= x_0 + ta \\ y &= y_0 + tb \\ z &= z_0 + tc \end{aligned} } \qquad \text{Parametric equations of } L$xyz​=x0​+ta=y0​+tb=z0​+tc​​Parametric equations of L

$\vec{v} = \langle a, b, c \rangle$v=⟨a,b,c⟩

$t$t is called a parameter.

$-\infty < t < \infty$−∞<t<∞, $t \in (-\infty, \infty)$t∈(−∞,∞)

$\boxed{\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}} \qquad \text{Symmetric equations of } L$ax−x0​​=by−y0​​=cz−z0​​​Symmetric equations of L

Line Segments

$\vec{r} = \vec{r}_0 + t(\vec{r}_1 - \vec{r}_0), \quad t \in [0,1]$r=r0​+t(r1​−r0​),t∈[0,1] $\vec{r} = \vec{r}_0 + t\,\vec{r}_1 - t\,\vec{r}_0$r=r0​+tr1​−tr0​ $\boxed{\vec{r} = \vec{r}_0\,(1-t) + t\,\vec{r}_1\,, \quad t \in [0,1]}$r=r0​(1−t)+tr1​,t∈[0,1]​

$\vec{v} = \langle a, b, c \rangle$v=⟨a,b,c⟩

A plane is determined by a point $(x_0, y_0, z_0)$(x0​,y0​,z0​) on the plane and a vector $\vec{n}$n which is orthogonal to the plane and which we call the normal vector.

Hence $\vec{n}$n and $\vec{r} - \vec{r}_0$r−r0​ are orthogonal. So

$\boxed{\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0}$n⋅(r−r0​)=0​

Vector equation of a plane.

$\therefore \quad \vec{n} \cdot \vec{r} = \vec{n} \cdot \vec{r}_0$∴n⋅r=n⋅r0​

$ax + by + cz = ax_0 + by_0 + cz_0 = -d$ax+by+cz=ax0​+by0​+cz0​=−d $\boxed{ax + by + cz + d = 0}$ax+by+cz+d=0​

Standard equation of a plane.

$\boxed{a(x - x_0) + b(y - y_0) + c(z - z_0) = 0}$a(x−x0​)+b(y−y0​)+c(z−z0​)=0​

Scalar equation of a plane through the point $(x_0, y_0, z_0)$(x0​,y0​,z0​) with $\vec{n} = \langle a, b, c \rangle$n=⟨a,b,c⟩.

Examples (Lines)

Find the vector, parametric, and symmetric equations of the line

2. The line through the point $(4, 2, -3)$(4,2,−3) and parallel to the vector $2\mathbf{i} - \mathbf{j} + 6\mathbf{k}$2i−j+6k.

Recall $\vec{r} = \vec{r_0} + t\,\mathbf{v}$r=r0​​+tv (vector)

$= \langle 4, 2, -3 \rangle + t\,\langle 2, -1, 6 \rangle$=⟨4,2,−3⟩+t⟨2,−1,6⟩ $\left. \begin{aligned} x &= x_0 + ta &= 4 + 2t \\ y &= y_0 + tb &= 2 - t \\ z &= z_0 + tc &= -3 + 6t \end{aligned} \right\} \quad \text{Parametric}$xyz​=x0​+ta=y0​+tb=z0​+tc​=4+2t=2−t=−3+6t​⎭⎬⎫​Parametric $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ax−x0​​=by−y0​​=cz−z0​​ $\boxed{\frac{x - 4}{2} = \frac{y - 2}{-1} = \frac{z + 3}{6}}$2x−4​=−1y−2​=6z+3​​

4. The line through the point $(6, 0, -2)$(6,0,−2) and parallel to the line

$\begin{aligned} x &= 4 - 3t \\ y &= -1 + 4t \\ z &= 6 + 5t\end{aligned}$xyz​=4−3t=−1+4t=6+5t​

Vector

$\vec{r} = \vec{r_0} + \vec{v}\,t = \langle 6, 0, -2 \rangle + t\,\langle -3, 4, 5 \rangle$r=r0​​+vt=⟨6,0,−2⟩+t⟨−3,4,5⟩

Parametric

$\begin{aligned} x &= 6 - 3t \\ y &= 0 + 4t \\ z &= -2 + 5t\end{aligned}$xyz​=6−3t=0+4t=−2+5t​

Symmetric

$\frac{x - 6}{-3} = \frac{y - 0}{4} = \frac{z + 2}{5}$−3x−6​=4y−0​=5z+2​

b. A line through the points $A = (1, 6, -2)$A=(1,6,−2) and $B = (-5, 2, 5)$B=(−5,2,5).

$\vec{v} = \vec{AB} = \langle -6, -4, 7 \rangle$v=AB=⟨−6,−4,7⟩ $\vec{r}_0 = \langle 1, 6, -2 \rangle$r0​=⟨1,6,−2⟩ $\boxed{\vec{r} = \langle 1, 6, -2 \rangle + t\langle -6, -4, 7 \rangle}$r=⟨1,6,−2⟩+t⟨−6,−4,7⟩​

vector

$\vec{AB} = \langle -5-1,\; 2-6,\; 5-(-2) \rangle$AB=⟨−5−1,2−6,5−(−2)⟩

Parametric

$\begin{aligned} x &= 1 - 6t \\ y &= 6 - 4t \\ z &= -2 + 7t\end{aligned}$xyz​=1−6t=6−4t=−2+7t​

Symmetric

$\frac{x-1}{-6} = \frac{y-6}{-4} = \frac{z+2}{7}$−6x−1​=−4y−6​=7z+2​

Examples (Planes + Lines)

Find an equation of the line.

12. The line of intersection of the planes

$x + 2y + 3z = 1$x+2y+3z=1 $x - y + z = 1$x−y+z=1

$\vec{n}_1 = \langle 1, 2, 3 \rangle$n1​=⟨1,2,3⟩

$\vec{n}_2 = \langle 1, -1, 1 \rangle$n2​=⟨1,−1,1⟩

$\vec{v} = \vec{n}_1 \times \vec{n}_2 = \langle 5, 2, -3 \rangle$v=n1​×n2​=⟨5,2,−3⟩

Let $z = 0$z=0

$\begin{aligned} x + 2y &= 1 \\ -x + y &= -1\end{aligned}$x+2y−x+y​=1=−1​

$\therefore$∴ $3y = 0$3y=0

therefore $y = 0$y=0

$\therefore$∴ $x = 1$x=1

because

$\therefore$∴ The $(1, 0, 0)$(1,0,0) is on the line we seek.

$\therefore \quad \boxed{\vec{r} = \langle 1, 0, 0 \rangle + t\,\langle 5, 2, -3 \rangle}$∴r=⟨1,0,0⟩+t⟨5,2,−3⟩​

Exercises on Planes

Find an equation of the plane

24. The plane through the point $(-3, 4, 2)$(−3,4,2) and has normal vector $\langle 6, 1, -1 \rangle$⟨6,1,−1⟩.

Recall the scalar equation of a plane with normal vector $\mathbf{n} = \langle a, b, c \rangle$n=⟨a,b,c⟩ and a point $(x_0, y_0, z_0)$(x0​,y0​,z0​) is given by

$\boxed{a(x - x_0) + b(y - y_0) + c(z - z_0) = 0}$a(x−x0​)+b(y−y0​)+c(z−z0​)=0​

$\therefore$∴

$\boxed{6(x + 3) + y - 4 - (z - 2) = 0}$6(x+3)+y−4−(z−2)=0​ $6x + 18 + y - 4 - z + 2 = 0$6x+18+y−4−z+2=0 $6x + y - z + 14 + 2 = 0$6x+y−z+14+2=0 $\boxed{6x + y - z = -16}$6x+y−z=−16​

26. The plane through the origin and perpendicular to the line

$x = 1 - 8t$x=1−8t $y = -t$y=−t $z = 4$z=4

$(x_0, y_0, z_0) = (0, 0, 0)$(x0​,y0​,z0​)=(0,0,0)

$\vec{n} = \langle -8, -1, 2 \rangle$n=⟨−8,−1,2⟩

$\therefore$∴ $\boxed{-8x - 7y + 2z = 0}$−8x−7y+2z=0​

28. The plane through $(9, -4, -5)$(9,−4,−5) and parallel to

$-2x + 3y + z = 2.$−2x+3y+z=2.

Since we have $-2x + 3y + z = 0$−2x+3y+z=0 and this plane is parallel to the one we seek, we can take $\mathbf{n}_2 = \langle -2, 3, 1 \rangle$n2​=⟨−2,3,1⟩.

$\therefore \quad -2(x - 9) + 3(y + 4) + (z + 5) = 0$∴−2(x−9)+3(y+4)+(z+5)=0 $-2x + 3y + z + 35 = 0$−2x+3y+z+35=0 $-2x + 3y + z = -35$−2x+3y+z=−35 $2x - 3y - z = 35$2x−3y−z=35 $\boxed{2x - 3y - z - 35 = 0}$2x−3y−z−35=0​

34.* The plane through the points $A(3,0,1)$A(3,0,1), $B(-2,-2,3)$B(−2,−2,3) and $C(7,1,-4)$C(7,1,−4).

$\vec{u} = \overrightarrow{AB} = \langle -5, -2, 2 \rangle$u=AB=⟨−5,−2,2⟩ $\vec{v} = \overrightarrow{AC} = \langle 4, 1, -5 \rangle$v=AC=⟨4,1,−5⟩ $\vec{n} = \vec{u} \times \vec{v} = \langle 8, -17, 3 \rangle$n=u×v=⟨8,−17,3⟩ $\boxed{8(x-3) - 17(y-0) + 3(z-1) = 0}$8(x−3)−17(y−0)+3(z−1)=0​ $8x - 24 - 17y + 3z - 3 = 0$8x−24−17y+3z−3=0 $8x + 17y + 3z - 27 = 0$8x+17y+3z−27=0

. The plane passes through the line of intersection of the plane

$x - z = 1 \quad \text{and} \quad y + 2z = 3$x−z=1andy+2z=3

and is perpendicular to the plane

$x + y - 2z = 1$x+y−2z=1 $\vec{n} = \vec{v} \times \vec{n}_3$n=v×n3​ $\begin{aligned} n_1 &= \langle 1, 0, -1 \rangle \\ n_2 &= \langle 0, 1, 2 \rangle \\ \vec{v} &= \langle 1, -2, 1 \rangle \\ \vec{n}_3 &= \langle 1, 1, -2 \rangle\end{aligned}$n1​n2​vn3​​=⟨1,0,−1⟩=⟨0,1,2⟩=⟨1,−2,1⟩=⟨1,1,−2⟩​

Then $\vec{n} = \langle 3, 3, 3 \rangle$n=⟨3,3,3⟩.

From

$\begin{aligned} x - z &= 1 \\ y + 2z &= 3\end{aligned}$x−zy+2z​=1=3​

set $z = 0$z=0. Then

$\begin{aligned} x &= 1 \\ y &= 3\end{aligned}$xy​=1=3​

The point $(1, 3, 0)$(1,3,0) is on the plane we are seeking.

$3(x-1) + 3(y-3) + 3(z-0) = 0$3(x−1)+3(y−3)+3(z−0)=0 $x - 1 + y - 3 + z = 0$x−1+y−3+z=0 $\boxed{x + y + z = 4}$x+y+z=4​

Angle between two planes

Two parallel planes don’t intersect and their normal vectors are parallel too.

$\theta = \cos^{-1}\left(\frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|}\right)$θ=cos−1(∣n1​∣∣n2​∣n1​⋅n2​​)

The angle between two intersecting planes is the acute angle between their normal vectors.

Distances

$\vec{b} = \langle x - x_0, y - y_0, z - z_0 \rangle$b=⟨x−x0​,y−y0​,z−z0​⟩ $\vec{n} = \langle a, b, c \rangle$n=⟨a,b,c⟩ $|\vec{n}| = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}}$∣n∣=a2+b2+c2​∣ax+by+cz+d∣​

This is the distance between the point $(x, y, z)$(x,y,z) and the plane $ax + by + cz + d = 0$ax+by+cz+d=0.

Exercise

22. Find the distance between the point $(-6, 3, 5)$(−6,3,5) and the plane $x - 2y - 4z = 8$x−2y−4z=8.

$n = \langle 1, -2, -4 \rangle$n=⟨1,−2,−4⟩ $(x, y, z) = (-6, 3, 5)$(x,y,z)=(−6,3,5) $D = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}}$D=a2+b2+c2​∣ax+by+cz+d∣​ $= \frac{|-6 - 6 - 20 - 8|}{\sqrt{1 + 4 + 16}} = \frac{40}{\sqrt{21}}$=1+4+16​∣−6−6−20−8∣​=21​40​