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Please Help. Any kind of help is appreciated. 

 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

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