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)

[20]

Answer 1

For two mismatched lines to be parallel on the plane, it is necessary and sufficient that the direction vectors of the given lines be collinear, or the normal vectors of the given lines be colinear, or the direction vector of one line is perpendicular to the normal vector of the other line.On the plane is based on the collinearity of vectors or the condition of perpendicularity of two vectors. If

$l 1 ​ ​ =(l 1x ​ ,l 1y ​ ) and \vec{l_2}=({l_{2x},l_{2y})} l 2 ​ ​ =(l 2x ​ ,l 2y ​ )$

$l 1 ​ ​ =(l 1x ​ ,l 1y ​ ) and \vec{l_2}=({l_{2x},l_{2y})} l 2 ​ ​ =(l 2x ​ ,l 2y ​ )$

8.2

Plane equation

$ax+by+cz+d=0\\ \cfrac{a}{-2}=\cfrac{b}{4}=\cfrac{c}{-5}\\ a=-2, b=4, c=-6.\\ 2(-2)+4\cdot 4-5(-3)+d=0\\ d=-27\\ ax+by+cz+d=0−2 a ​=4 b ​=−5 c ​a=−2,b=4,c=−6.2(−2)+4⋅4−5(−3)+d=0d=−27$

the plane that passes through the point (2, 4, −3) and is parallel to

the plane $−2x + 4y − 5z + 6 = 0$

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

8.3

The line that passes through the point (2, 5, 3) and is perpendicular to the plane

2x − 3y + 4z + 7 = 0.

Normal vector of plane and equation of line are

$\vec{n}=(2,-3,4)\\ \cfrac{x-2}{2}=\cfrac{y-5}{-3}=\cfrac{z-3}{4}$

​

Parametric equation of line

$t=\cfrac{x-2}{2},t=\cfrac{y-5}{-3},t=\cfrac{z-3}{4}\\ x=2t+2,y=-3t+5,z=4t+3$ .

8.4

Equation for line passing through the points (4, −2, 5) and (0, 2, 4)

$\cfrac{x-4}{-4}=\cfrac{y+2}{4}=\cfrac{z-5}{-1}\\$

​

Equation of the plane passing through the point(−2, 3, 4) and is perpendicular to the line

$-4(x+2)+4(y-3)-(z-4)=0\\.$

$−4x+4y−z−16=0.$