Answer to Question #114075 in Analytic Geometry for Azwinndini

Question

QUESTION 8 8.1 Let L1 and L2 be lines defined by x = w0 + su, s ∈ R and y = w1 + tv, t ∈ R, respectively. Show that L1and L2 are parallel if and only if u = kv for some k ∈ R. (8) 8.2 Find the plane that passes through the point (2, 4, −3) and is parallel to the plane −2x + 4y − 5z + 6 = 0. (4) 8.3 Find the line that passes through the point (2, 5, 3) and is perpendicular to the plane 2x − 3y + 4z + 7 = 0. (4) 8.4 Find an equation of the plane passing through the point(−2, 3, 4) and is perpendicular to the line passing through the points (4, −2, 5) and (0, 2, 4). (4)

Answer 1

8.8 Let $L_1$ be defined by $\vec{x} = \vec{w_0} + s\vec{u}, s \in \R$ and $L_2$ defined by $\vec{y} = \vec{w_1} + t\vec{v}, t \in \R.$ We may rewrite these equations in a parametric form

$\begin{cases}x_1=w_{0,1}+su_1,\\x_2=w_{0,2}+su_2,\end{cases} \begin{cases}y_1=w_{1,1}+tv_1,\\y_2=w_{1,2}+tv_2\end{cases} .$

We should show that $\exists k\in \R: \, \vec{u}=k\vec{v}$, that is,

$u_1 = kv_1, u_2=kv_2.$

Let us determine the slope of the lines.

$k_1 = \dfrac{x_2-w_{0,2}}{x_1-w_{0,1}} = \dfrac{su_2}{su_1} = \dfrac{u_2}{u_1},$

$k_2 = \dfrac{y_2-w_{1,2}}{y_1-w_{1,1}} = \dfrac{tv_2}{tv_1} = \dfrac{v_2}{v_1}.$

If the two lines are parallel, then $k_1=k_2$ , therefore

$\dfrac{u_2}{u_1} = \dfrac{v_2}{v_1}.$

This formula means that vectors $\vec{u}$ and $\vec{v}$ are proportional, therefore $\exists k\in \R: \, \vec{u}=k\vec{v}.$

8.2 Let the equation of plane be $A x + B y + C z + D = 0$. Two planes are parallel if

$\dfrac{A_1}{A_2} = \dfrac{B_1}{B_2} = \dfrac{C_1}{C_2} = k.$

So we get

$\dfrac{A}{-2} = \dfrac{B}{4} = \dfrac{C}{-5} = k.$

Let $k=1$ , because rescale the equation to get $k( A x + B y + C z + D )= 0.$ So $A=-2, \; B=4,\; C=-5.$

The plane passes through the point $(2, 4, −3)$ , so we obtain

$-2\cdot2+4\cdot4 -5\cdot (-3)+D = 0,$

therefore $D=-27.$

The equation of the plane is $-2x+4y-5z-27=0.$

8.3 Let the equation of the line be $\vec{r}=\vec{r_0} + t\vec{a},$

where $\vec{r_0} = (2,5,3)$ is radius vector of point $(2,5,3).$ The plane defined by $Ax+By+Cz+D=0$ and line are perpendicular if $\vec{a}$ is parallel to $\vec{v}=(A,B,C),$

so we get

$\dfrac{a_x}{A}=\dfrac{a_y}{B}=\dfrac{a_z}{C}$ ,

therefore

$\dfrac{a_x}{2}=\dfrac{a_y}{-3}=\dfrac{a_z}{4}$ .

We may choose $a_x = 2,\; a_y=-3, \; a_z = 4.$ So the equation of the line may be rewritten in parametric form as

$\begin{cases} x = x_0 + a_xt,\\ y=y_0 + a_yt, \\ z=z_0 + a_zt \end{cases}$ and $\begin{cases} x = 2 +2t,\\ y=5 -3t, \\ z=3 + 4t .\end{cases}$

8.4 Let the equation of plane be $A x + B y + C z + D = 0$ .

First we determine the equation of the line passing through two given points:

$\dfrac{x-x_1}{x_2-x_1} = \dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}.$

Therefore

$\dfrac{x-4}{0-4} = \dfrac{y+2}{2+2}=\dfrac{z-5}{4-5}, \;\; \dfrac{x-4}{-4} = \dfrac{y+2}{4}=\dfrac{z-5}{-1}, \;\;$ so the line is parallel to vector $(-4,4,-1).$

As the plane and line are perpendicular. we get

$\dfrac{A}{-4}=\dfrac{B}{4} = \dfrac{C}{-1}.$

We may choose $A=-4, \; B=4,\; C=-1.$ We know that plane passes through $(-2,3,4),$ so

$-2A+3B+4C+D=0, \;\; -2\cdot(-4) + 3\cdot4 +4\cdot(-1) + D = 0,$

therefore $D=-16.$

So the equation of the plane is $-4x+4y-z-16=0.$