Answer to Question #112306 in Analytic Geometry for Caylin

Question

1) Let L1 and L2 be lines defined by :

x = w0 + su, s e R(element of R)
y= w1 + tv, t e R(element of R) respectively

Show that L1 and L2 are parallel if and only if u = kv for some k e R. (element of R)

2) Find the plane that passes through the point (2,4,-3) and is parallel to the plane - 2x + 4y - 5z + 6 = 0

3) Find the line that passes through the point (2,5,3) and is perpendicular to the plane 2x - 3y + 4z + 7 = 0

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).

Answer 1

1.) Let L1 and L2 be lines defined by :

$x = w\omicron + su, s\in \real$

$y= w1 + tv, t\in R\phantom{i} respectively$

Show that L1 and L2 are parallel if and only if u = kv for some $k \in R$

Let L₁ and L₂ be lines defined by

$x = w\omicron + su, s\in \real$ ....................................1

and

$y= w1 + tv, t\in R$ ,..................................2

respectively.

Show that L₁ and L₂ are parallel if and only if $u = kv$ for some $k\in R$ .

PROPOSITION

L1 AND L2 ARE PARALLEL

we show that:

$u=kv$

for L1 and L2 to be parallel .

their directional vector should be scalar multiple of each other.

directional vector of L1 is given by $su$

directional vector of L2 is given by $tv$

then if L1 and L2 are parallel , $su$ and $tv$ shall be scalar multiple of each other

that is:

$su=m*tv$ , where m is scalar

substituting $k=m*t/s$ to the equation $u=kv$ then k is a scalar since t,s and m are all scalar,

therefore, L1 and L2 are parallel since $u=kv$ and k is the scalar multiple, u and v are directional vectors of L1 and L2.

2.) Find the plane that passes through the point $(2,4,−3)$ and is parallel to the plane $- 2x + 4y - 5z + 6 = 0$

the equation of the plane parallel to $- 2x + 4y - 5z + 6 = 0$ is given be

$$- 2x + 4y - 5z + c = 0$$

since it passes through the point $(2,4,-3)$ . to obtain c we substitute the point to the equation

$$- 2x + 4y - 5z + c = 0\\ -2(2)+4(4)-5(-3)+c=0\\-4+16+15+c=0\\\implies c=-27$$

hence the equation of the plane parallel to $- 2x + 4y - 5z + 6 = 0$ is

$$- 2x + 4y - 5z -27 = 0$$

3.) Find the line that passes through the point$(2,5,3)$ and is perpendicular to the plane $2x - 3y + 4z + 7 = 0$

the equation perpendicular to $2x - 3y + 4z + 7 = 0$ is given by

$$\dfrac{x-x\omicron}{2}=\dfrac{y-y\omicron}{-3}=\dfrac{z-z\omicron}{4}$$

and since the line has pass through $(2,5,3)$ then the equation of the line perpendicular to $2x - 3y + 4z + 7 = 0$ will be

$$\dfrac{x-2}{2}=\dfrac{y-5}{-3}=\dfrac{z-3}{4}$$

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).

• solution

The line passing through $(4,-2,5)$ and $(0,2,4)$ has the directional ratio of

$$<4-0,-2-2,5-4>\\=<4,-4,1>$$

hence the equation of the plane passing through $(x\omicron,y\omicron,z\omicron)$ and the directional ratio of the line perpendicular to the plane

$<a,b,c>$

is given by

$$a(x-x\omicron)+b(y-y\omicron)+c(z-z\omicron)=0$$

then:

the required equation of the plane passing through $(-2,3,4)$ and the directional ratio of the line perpendicular to the plane $<4,-4,1>$ is given by:

$$4(x+2)-4(y-3)+1(z-4)=0\\\implies 4x+8-4y+12+z-4=0\\putting\phantom{i}like\phantom{i}terms\phantom{i}together:$$ $$4x-4y+z+16=0$$