Find parametric equations for the line that intersects $x = 2t+1; y = 2 - t; z = 3 - t$ at a $60°$ degree angle.

Question

Find parametric equations for the line that passes through the point $(-6,-5,-1)$ that intersects the line $x = 2t+1; y = 2 - t; z = 3 - t$ at a $60°$ degree angle. I've never seen a question like this before, and I can do ones where the line is perpendicular or parallel, but I don't know what to do when there is a specific degree involved?

Answer 1

Hints: Required line passes through the point $(-6, -5, -1)$, i.e. we know it would be given by equation: $$\frac{x+6}{l_2}=\frac{y+5}{m_2}=\frac{z+1}{n_2}.$$

The equation of a given line we can rewrite as: $$\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{-1} \space \space \space (l_1=2, \space m_1=-1, \space n_1=-1).$$

Now, we have to utilize that we calculate the angle $\space \varphi \space$ between two lines by formula: $$\cos\varphi=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}},$$

and that we calculate the smallest distance $\space d \space$ between two lines by formula (if $d=0$ the lines intersect each other): $$d=\frac{\pm \begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2\\l_1 &m_1 & n_1\\l_2 & m_2 & n_2 \end{vmatrix}}{\sqrt{\begin{vmatrix}l_1 & m_1\\l_2 & m_2 \end{vmatrix}^2 + \begin{vmatrix}m_1 & n_1\\m_2 & n_2 \end{vmatrix}^2 + \begin{vmatrix}n_1 & l_1\\n_2 & l_2 \end{vmatrix}^2}}.$$

Answer 2

Vector of the given line $L_1$ $= (2 \, -1 \, -1)$

Say, vector of the line $L_2$ at $60^0$ to $L_1 = (a \, \, b \, \, c)$ and equation of the line -

$x = as - 6, y = bs - 5, z = cs - 1$

The dot product of the vectors of two lines is given by,

$\vec{r} \cdot \vec {s} = ||\vec{r}|| \cdot ||\vec{s}|| \cos \theta$

So, $\, (2 \, -1 \, -1) \cdot (a \, \, b \, \, c) = \sqrt{a^2+b^2+c^2} \cdot \sqrt{2^2 + (-1)^2 + (-1)^2} \cos 60^0$

$ 2a - b - c= \frac{1}{2} \sqrt{6(a^2+b^2+c^2)}$

$ 4a - 2b - 2c= \sqrt{6(a^2+b^2+c^2)}$ ...(i)

As the lines $L_1$ and $L_2$ intersect, at the point of intersection -

$2t + 1 = as - 6, 2 - t = bs - 5, 3 - t = cs - 1$

From this, you get $a = 5b - 7c$

Replacing in $(i)$ and solving the quadratic,

you get $14b^2 + 50c^2 - 55bc = 0 = (2b - 5c) (7b - 10c)$

So you get, $a = \frac{11b}{5}, b, c = \frac{2b}{5}$; $a = \frac{b}{10}, b, c = \frac{7b}{10}$ and hence two possible vectors -

$(11 \, \, 5 \, \, 2)$ or $(1 \, \, 10 \, \, 7)$

As the line also passes through point $(-6, -5, -1)$, the parametric equations of the line $L_2$ -

(1) $ \, x = 11s - 6, y = 5s - 5, z = 2s - 1$

(2) $ \, x = s - 6, y = 10s - 5, z = 7s - 1$