Question Video: Finding the Cartesian Form of the Equation of a Straight Line Mathematics

Find the Cartesian form of the equation of the straight line passing through the origin and the intersection point of the two straight lines 𝐿₁: π‘Ÿ = 〈1, 1, βˆ’2βŒͺ + π‘‘βŒ©1, 4, 3βŒͺ and 𝐿₂: π‘₯ = 3, (𝑦 βˆ’ 5)/βˆ’4 = (𝑧 βˆ’ 3)/βˆ’1.

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Video Transcript

Find the Cartesian form of the equation of the straight line passing through the origin and the intersection point of the two straight lines 𝐿 one π‘Ÿ equals one, one, negative two plus 𝑑 times one, four, three and 𝐿 two π‘₯ equals three 𝑦 minus five over negative four equals 𝑧 minus three over negative one.

Alright, so here’s the idea. We have these two straight lines 𝐿 one and 𝐿 two. And we know that there’s a point where they intersect. What we want to do is solve for the equation of the line that passes through this intersection point as well as the origin. Our first step in doing this will be to figure out where 𝐿 one and 𝐿 two meet. The equation for 𝐿 two is given in what’s called symmetric form. Interpreting this form, we can say that while π‘₯ is always equal to positive three, 𝑦 minus five over negative four and 𝑧 minus three over negative one are both equal to a scale factor we can call 𝑑 two.

This means we can write the equation of 𝐿 two in what’s called parametric form. Written this way, π‘₯ is equal to three as always, while 𝑦, we can say, is equal to negative four times 𝑑 two plus five. This is so because 𝑦 minus five over negative four is equal to 𝑑 two. In the same way, because 𝑧 minus three over negative one equals 𝑑 two, we can say that 𝑧 equals negative 𝑑 two plus three. Now that we have our line 𝐿 two written in parametric form, we’re going to do the same thing with our line 𝐿 one. This line, we see, is given to us in what’s called vector form. If we write this equation in parametric form, we have that π‘₯ is equal to one times 𝑑 or 𝑑 plus one. Then the 𝑦-component is equal to four times 𝑑 plus one and the 𝑧-component, three times 𝑑 minus two.

Now here’s why we’ve written our two lines equations in parametric form. At the point of intersection that we want to solve for between 𝐿 one and 𝐿 two, we can say that the π‘₯-values of line two and line one are equal and the same thing for the 𝑦- and the 𝑧-values. And note that this is only at the single point of intersection of these lines. At the single point of intersection then, we can say that these three equations hold true. And what we’ll do now is use these equations to solve for the parameters 𝑑 and 𝑑 two.

Since there are two unknowns we want to solve for, we’ll pick two equations to use to find them out. Let’s choose the equations for the π‘₯- and 𝑦-values of our point. Our π‘₯-equation, that three is equal to 𝑑 plus one, implies that 𝑑 equals three minus one or two. If we then substitute this value for 𝑑 into 𝑑 in our 𝑦-equation, we find that negative four times 𝑑 two plus five equals four times two, that’s our value for 𝑑, plus one.

Rearranging this equation a bit, this means that negative four 𝑑 two is equal to positive four. And then if we divide both sides of this equation by negative four, we find that 𝑑 two equals negative one. Now that we know the values of 𝑑 and 𝑑 two, we can substitute them into these three equations everywhere they appear, and the values we’ll solve for will be the π‘₯-, 𝑦-, and 𝑧-coordinates of our point of intersection of line one and line two. Doing this in our first equation confirms that π‘₯ equals three where these lines meet.

In our second equation, it tells us that the 𝑦-value at this point is positive nine. And from our third equation, we find that 𝑧 equals four. The point of intersection of our lines, then, is three, nine, four. And now that we know this, let’s recall a bit about the Cartesian form of the equation of a straight line. A line like this can be written in general as π‘₯ minus π‘₯ one over 𝑙 being equal to 𝑦 minus 𝑦 one over π‘š being equal to 𝑧 minus 𝑧 one over 𝑛. Here, π‘₯ one, 𝑦 one, and 𝑧 one are coordinates of a point that this line passes through. And 𝑙, π‘š, and 𝑛 are the components of a vector parallel to the line.

So when it comes to the equation of the straight line we want to solve for, if we know a point that line passes through as well as a vector parallel to it, we can write that line’s equation in Cartesian form. We do already know a point that our line passes through. Actually, we know two, the origin and the point we solve for, three, nine, four. We do not yet know the components of a vector parallel to our line. But note that we could solve for such a vector by connecting up our two known points along the line. If we say that this vector has components 𝑙, π‘š, and 𝑛, then those values are given by the differences between the various coordinates of our two points. We find this vector simply has components three, nine, four.

We now have everything we need to write out the Cartesian form of the equation of this straight line. If we choose to let the origin correspond to the point π‘₯ one, 𝑦 one, 𝑧 one that our line passes through, then we can say that the equation of our line is π‘₯ divided by three equals 𝑦 divided by nine equals 𝑧 divided by four. Note that our final answer would look differently if we had picked another point that lies along this line. Either way, though, the equation would still be describing the same line in space. The Cartesian form we’ll report as our final answer then is π‘₯ over three equals 𝑦 over nine equals 𝑧 over four.

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