Question

1. Two given lines are either parallel, skew or intersecting. (6 marks)
2. Determine, if there is one, the point of intersection of the lines given by the equations

x−1/−3=y−8/6=z−3/−2    and    x−8/4=y+3/−5=z−7/2

    b. Give the equations of two lines that meet at the point (–1 5, 2) and which meet at right angles, but do not use that point in either of the equations. Explain your reasoning

Answer 1

Solution (a)

The first line is

$\begin{gathered} \frac{{x - 1}}{{ - 3}} = \frac{{y - 8}}{6} = \frac{{z - 3}}{{ - 2}} = t \\ x = 1 - 3t,\,\,\,y = 8 + 6t,\,\,\,z = 3 - 2t \\ \end{gathered}$

The direction vector of the first line is

${r_1} = \left\langle { - 3,\,6,\, - 2} \right\rangle$

The second line is

The direction vector of the second line is

$\begin{gathered} \frac{{x - 8}}{4} = \frac{{y + 3}}{{ - 5}} = \frac{{z - 7}}{2} = s \\ x = 8 + 4s,\,\,\,y = - 3 - 5s,\,\,\,z = 7 + 2s \\ \end{gathered}$

The direction vector of the second line is ${r_2} = \left\langle {4,\, - 5,\,2} \right\rangle$

Since we can see that the direction vectors of both the lines are different, then for sure they are not parallel.

 

Now there are two possibilities, they intersect, or both the lines are skewed.

In case both intersect, then for some values of “$s$ ” and “$t$ ”, we must have 

$\begin{gathered} 1 - 3t = 8 + 4s & & \Rightarrow & 3t + 4s = - 7 & & ... & \left( i \right) \\ 8 + 6t = - 3 - 5s & & \Rightarrow & 6t + 5s = - 11 & & ... & \left( {ii} \right) \\ 3 - 2t = 7 + 2s & & \Rightarrow & 2t + 2s = - 4 & & ... & \left( {iii} \right) \\ \end{gathered}$

Solving (i) and (ii) 

$\begin{gathered} 6t - 6t + 5s - 8s = - 11 + 14 \\ - 3s = 3 & & \Rightarrow & s = - 1 \\ \end{gathered}$

Replace this value of $s$ in (i) 

$\begin{gathered} 3t + 4\left( { - 1} \right) = - 7 \\ 3t = - 3 & & \Rightarrow & t = - 1 \\ \end{gathered}$

Now verifying values of $s$ and $t$ in (iii)

$2\left( { - 1} \right) + 2\left( { - 1} \right) = - 4$ (TRUE)

Hence for $s=t=-1$ , the two lines intersect.

 

The point of intersection will be

$\begin{gathered} x = 1 - 3\left( { - 1} \right),\,\,\,y = 8 + 6\left( { - 1} \right),\,\,\,z = 3 - 2\left( { - 1} \right) \\ \left( {x,y,z} \right) = \left( {4,\,2,\,5} \right) \\ \end{gathered}$

Solution (b)

We know that the general form of the equation of a line for some value of “$t$ ” is 

$(x, y, z) = (x_0, y_0, z_0) +t (v_1, v_2, v_3)$

Where $(x_0, y_0, z_0)$ is any point on the line and $(v_1, v_2, v_3)$ is the direction vector of the line.

Given that the two lines must pass through the point $(-1, 5, 2)$ and perpendicular to each other.

Then the two lines for some values of "$s$ " and "$t$ " will be

$(x, y, z) = (-1, 5, 2) +t (v_1, v_2, v_3)$ ... (Line 1)

$(x, y, z) = (-1, 5, 2) +s (u_1, u_2, u_3)$ ... (Line 2)

Here $(u_1, u_2, u_3)$ is the direction vecator of the second line.

Let us consider,

$(v_1, v_2, v_3)=(1, 1, 1)$ and $(u_1, u_2, u_3)=(a, b, c)$

Then for the two lines being perpendicular, we must have the dot product of the two direction vectors equal to zero.

That is

$(1, 1, 1) . (a, b, c) = 0$

$a+b+c=0$

so we can choose any values of $a, b$ and $c$ such that $a+b+c=0$

Let us choose $a=1, b=-1, c=0$

Therefore, the equations of the two lines will be

$(x, y, z) = (-1, 5, 2) +t (1, 1, 1)$ ... (Line 1)

$(x, y, z) = (-1, 5, 2) +s (1, -1, 0)$ ... (Line 2)

Since it is required that the lines must not use the point $(-1, 5, 2)$ , we find other points on the two lines respectively.

For $t=1$ , from the line 1, we have

$(x, y, z) = (-1, 5, 2) +1 (1, 1, 1)=(0, 6, 3)$

And for $s=-1$ , from the line 2, we have

$(x, y, z) = (-1, 5, 2) +(-1) (1, -1, 0)=(-2, 6, 2)$

Hence the two required lines, perpendicular to each other and not using the point $(-1, 5, 2)$ are

$(x, y, z) = (0, 6, 3) +t (1, 1, 1)$ ... (Line 1)

$(x, y, z) = (-2, 6, 2) +s (1, -1, 0)$ ... (Line 2)