Question

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).

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.

The condition of parallelism of lines $l_1$ and $l_2$ on the plane is based on the collinearity of vectors or the condition of perpendicularity of two vectors. If $\vec{l_1}=({l_{1x},l_{1y})}$ and $\vec{l_2}=({l_{2x},l_{2y})}$ are directing vectors of lines and$\vec{n_{l1}}=({n_{l1x},n_{l1y})},\vec{n_{l2}}=({n_{l2x},n_{l2y})}$ are normal vectors of lines

$l_1$  and $l_2$, then the above necessary and sufficient condition:

${\vec l_1}=k\cdot\vec{l_2} \Leftrightarrow \left\{ l_{1x}=k\cdot l_{2x}; l_{1y}=k\cdot l_{2y}\right.$ or ${\vec n_{l}}_1=k\cdot\vec{n_{l}}_2$ or $\vec{l}_1\cdot\vec{n}_{l2}=0$ .

8.2

Plane eqation

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

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

$-4(x+2)+4(y-3)-(z-4)=0\\ -4x+4y-z-16=0.$