Question

Let L₁ be the line in R³ with equation (x,y,z)=(1,0,2) + t(−1,3,4); t∈R and let L₂ be the line that is parallel to L1 and contains the point (1,−1,3). Let V be the plane that contains both the lines L₁ and L₂.

(a)Find two vectors that are both parallel to the plane V but are not parallel to one another.

(b) FindavectorthatisperpendiculartotheplaneV(c) Find an equation for the plane V.(d) Find an equation for the line L₃ that is perpendicular to the plane V and contains the point (1,−1,4).

Answer 1

(a) Let us compose the parametric equations of the lines  L₁ and L₂ :

${L_1}:\left\{ {\begin{matrix} {x = 1 - t}\\ {y = 3t}\\ {z = 2 + 4t} \end{matrix}} \right.$ , ${L_2}:\left\{ {\begin{matrix} {x = 1 - t}\\ {y = -1+3t}\\ {z = 3+ 4t} \end{matrix}} \right.$

The direction vector of these lines will be the vector $\overrightarrow p = \left( { - 1;3;4} \right)$

Since the lines are parallel to the plane V, the vector $\overrightarrow p$ is also parallel to the plane.

We also know that

${M_1}\left( {1;0;2} \right) \in {L_1}$ , ${M_2}\left( {1; - 1;3} \right) \in {L_2}$

Since the lines are parallel to the plane V, the vector $\overrightarrow {{M_1}{M_2}} = \left( {1 - 1; - 1 - 0;3 - 2} \right) = \left( {0; - 1;1} \right)$ is also parallel to the plane.

Since the coordinates of the vectors $\overrightarrow p$ and $\overrightarrow {{M_1}{M_2}}$ are disproportionate, they are not collinear.

Answer: $\overrightarrow p = \left( { - 1;3;4} \right)$ , $\overrightarrow {{M_1}{M_2}} =\left( {0; - 1;1} \right)$

(b) The vector perpendicular to the plane is found as the vector product of vectors $\overrightarrow p$ and $\overrightarrow {{M_1}{M_2}}$ :

$\overrightarrow n = \left| {\begin{matrix} {\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k }\\ { - 1}&3&4\\ 0&{ - 1}&1 \end{matrix}} \right| = \overrightarrow i \left( {3 + 4} \right) + \overrightarrow j \left( {0 + 1} \right) + \overrightarrow k \left( {1 - 0} \right) = \left( {7;1;1} \right)$

Answer: $\overrightarrow n = \left( {7;1;1} \right)$

(с) Since V contains L₂, ${M_2}\left( {1; - 1;3} \right) \in V$

Let us make an equation of V for a point ${M_2}\left( {1; - 1;3} \right)$ and a normal vector $\overrightarrow n = \left( {7;1;1} \right)$ :

$7(x - 1) + 1(y + 1) + 1(z - 3) = 0 \Rightarrow 7x + y + z - 9 = 0$

Answer: $7x + y + z - 9 = 0$

(d) If  L₃ is perpendicular to the plane V then L₃ is parallel to $n = \left( {7;1;1} \right)$ .

Then

${L_3}:\frac{{x - 1}}{7} = \frac{{y + 1}}{1} = \frac{{z - 4}}{1}$

Answer: ${L_3}:\frac{{x - 1}}{7} = \frac{{y + 1}}{1} = \frac{{z - 4}}{1}$