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Aug 15, 2022

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Hello Guest!

Notice, these lines are in 3D ($$R^3$$)
So, I think the most efficient method is to solve these using vectors. That is, let us write the vector equations of the

two lines:

This is the vector equation of a line: $$r_1(t)=r_0+td_1$$

Where:

$$r_0$$ is the position vector of a point on the line (for example: $$r_0=\begin{pmatrix} 7\\ -3 \\ 1 \end{pmatrix}$$ )

$$t$$ is a scalar ( t could be any real number).

$$d_1$$ is the direction vector of the line. (That is, the vector that is passing through any two points in the direction of the line.)

So: $$d_1=\begin{pmatrix} 7\\ -3 \\ 1 \end{pmatrix} -\begin{pmatrix} 5\\ 2 \\ 2 \end{pmatrix}=\begin{pmatrix} 2\\ -5 \\ -1 \end{pmatrix}$$

Now the equation of line $$l_1:$$ $$r_1(t)=\begin{pmatrix} 7\\ -3 \\ 1 \end{pmatrix}+t\begin{pmatrix} 2\\ -5 \\ -1 \end{pmatrix}$$

Similarly, for $$l_2:$$  $$r_2(\lambda)=\begin{pmatrix} 8\\ -1 \\ -1 \end{pmatrix}+\lambda\begin{pmatrix} 1\\ -4 \\ 0 \\ \end{pmatrix}$$ where $$\lambda$$ is a real scalar.

Now if the lines intersect, then: $$r_1(t)=r_2(\lambda) \\ \text{For some unique values of } \hspace{0.1cm} t \text{ and } \lambda$$

$$\begin{pmatrix} 7+2t\\ -3-5t \\ 1-t \end{pmatrix}=\begin{pmatrix} 8+\lambda\\ -1-4\lambda \\ -1 \end{pmatrix}$$

So we get a system of three equations:

$$7+2t=8+\lambda \\ -3-5t=-1-4\lambda \\ 1-t=-1 \\$$

From the third equation: $$t=2$$

Then, substituting in the first equation: $$11=8+\lambda \implies \lambda=3$$

Now, we must check that the second equation is satisfied:  (Because, the lines might be parallel in two components, but the third component is different!)

$$-3-5(2) = -13 \text{ Left-hand side} \\ -1-4(3) = -13 \text{ Right-hand side} \\ \text{Therefore, the lines do intersect when} \space \space \lambda=3 \text{ and } t=2$$

Now we substitute $$\lambda=3 \text{ or } t=2$$ in the respective equation of a line.
I.e. : $$r_1(2)=\begin{pmatrix} 7+2(2)\\ -3+2(-5) \\ 1+2(-1) \end{pmatrix}=\begin{pmatrix} 11\\ -13 \\ -1 \end{pmatrix}$$. Moreover, $$r_2(3)=\begin{pmatrix} 8+3(1)\\ -1+3(-4)\\ -1+3(0) \end{pmatrix}=\begin{pmatrix} 11 \\ -13 \\ -1 \end{pmatrix}$$

so they give the same point; hence the two lines intersect at the point: $$(11,-13,-1)$$

I hope this helps!

Aug 21, 2022