### Video Transcript

Find the vector form of the equation of the straight line passing through the point negative five, one, four and the intersection point of the two straight lines 𝑥 plus two over negative two equals 𝑦 plus five over negative two equals 𝑧 plus three over one and 𝑥 plus one over negative three equals 𝑦 minus one over two equals 𝑧 plus three over two.

Okay, so in this example, we have these two straight lines and we know that they intersect at some point. Along with this, we have our given point, with coordinates negative five, one, four. And we want to solve for the equation of the line that passes through both these points. Specifically, we want to solve for the vector form of this line’s equation. A line’s equation written this way is expressed like this. It involves a vector from the origin of a coordinate frame to a known point on the line, a second vector that’s parallel to the line’s axis, and lastly, what’s called a scale factor. Here, we’ve called it 𝑡. This factor ranges across all possible values.

To write the vector form of a line’s equation then, we need to know a vector that’s parallel to that line as well as a point that lies on it. In our problem statement, we’re given a point that lies along our line of interest. And then to solve for a vector parallel to our line, we’ll want to know the point of intersection between the two given lines. And if we know that, we can use that information along with our given point to compute a vector parallel to our line. To solve for this point of intersection, we’ll use the equations of the two lines given. Both of these equations are given in what’s called symmetric form. If we call this first equation, that of line one, and this second equation we’ll say applies to line two, then beginning with line one’s equation, we can write all three of these fractions as being equal to a scale factor we’ll call 𝑡 one.

Using this expression, we can write out the equation of this line in what’s called parametric form. The way it works, there’s one equation for the 𝑥-value, one for the 𝑦-value, and one for the 𝑧-value of the points on this line. And to solve for these equations, we use the equalities in the symmetric form of our line. For example, because 𝑥 plus two over negative two equals 𝑡 one, that means 𝑥 equals negative two times 𝑡 one minus two. In the same way, 𝑦 is equal to negative two times 𝑡 one minus five, while 𝑧 is equal to 𝑡 one minus three. Now that we know the parametric form of line one’s equation, let’s convert line two to a similar form.

For this equation, we can write all three fractions as being equal to a different scale factor we’ll call 𝑡 two. Once again, we can write out separate 𝑥, 𝑦, and 𝑧 equations for this line. 𝑥 is equal to negative three times 𝑡 two minus one. 𝑦 equals two times 𝑡 two plus one, while 𝑧 equals two times 𝑡 two minus three. This is the parametric form of line two’s equation.

Now, returning to our sketch, we know that at the point of intersection between what we’ve called line one and line two, at that point and that point only, we can say that the 𝑥-, 𝑦-, and 𝑧-values of our two lines are equal. That is, at the intersect, these three equations are true. What we’ll do next is use these relationships to solve for 𝑡 one and 𝑡 two. And since that gives us two unknowns to solve for, we’ll use two of these three equations.

Say we pick the first two, the ones representing the 𝑥- and 𝑦-values. Working from the first of these two equations, if we rearrange to solve for 𝑡 one, we find it’s equal to negative three 𝑡 two plus one divided by negative two. What we do next is take that value for 𝑡 one and substitute it in for 𝑡 one in our equation for the 𝑦-value of our intersect. That gives us this relationship here. And note that on the left-hand side, these factors of negative two cancel from numerator and denominator. Combining the positive one and negative five on the left, we get negative three times 𝑡 two minus four equals two times 𝑡 two plus one. If we subtract two times 𝑡 two from both sides and add four to both sides, we get negative five 𝑡 two equals five, which tells us that 𝑡 two equals negative one.

What we do then is take that value and substitute it in for 𝑡 two in our equation for 𝑡 one. 𝑡 one then equals negative three times negative one plus one all divided by negative two. This simplifies to negative two. Now that we know the values for our scale factors, 𝑡 one and 𝑡 two, at this point of intersection, we can choose to substitute, say, the value for 𝑡 one into these equations for 𝑥, 𝑦, and 𝑧. Or, equivalently, we could sub in 𝑡 two to these values here. Whichever way we choose, the result of these substitutions will be the 𝑥-, 𝑦-, and 𝑧-coordinates of our point of intersection.

Let’s say we choose to substitute 𝑡 two into the values on the right. When we do that, the 𝑥-value of this point of intersection turns out to be negative three times negative one minus one or two. The 𝑦-value is two times negative one plus one or negative one. And the 𝑧-value is two times negative one minus three or negative five. We now know the coordinates of the point of intersection of our two lines. And as we mentioned before, we can use these coordinates along with the coordinates of the other known point along our line to solve for the components of a vector parallel to the line, we’ll call it 𝐯. The components of 𝐯 are given by this expression. They work out to seven, negative two, and negative nine.

Now that we know the components of a vector parallel to our line of interest and we also know the coordinates of a point the line passes through, we’ll use negative five, one, four. We can write our line’s equation in vector form like this. What we have is a vector from the origin of a coordinate system to our known point negative five, one, four. And then from that point, we move up and down the line along the direction of our vector 𝐯.

Note that this isn’t the only way to write the vector form of this particular line’s equation. For example, to solve for 𝐯, we could’ve subtracted our points from one another in the opposite order. Even if we had done that, though, we would still get an equation of the same straight line. And so, this answer we have boxed in green here is an acceptable answer as well.