Video: Finding the Parametric Equations of a Straight Line in Three Dimensions given Its Equation

Find the parametric equations of the straight line (3π‘₯ βˆ’ 7)/βˆ’9 = (8𝑦 βˆ’ 3)/4 = (βˆ’8 βˆ’ 6𝑧)/βˆ’9.

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Video Transcript

Find the parametric equations of the straight line three π‘₯ minus seven over negative nine equals eight 𝑦 minus three over four which equals negative eight minus six 𝑧 over negative nine.

We’ve been given an equation in Cartesian form. Now, in general, if a line passes through a point 𝑝 nought which has coordinates π‘₯ nought, 𝑦 nought, 𝑧 nought and is parallel to the vector π‘Žπ‘π‘. That its parametric equations are π‘₯ equals π‘₯ nought plus π‘Žπ‘‘, 𝑦 equals 𝑦 nought plus 𝑏𝑑, and 𝑧 equals 𝑧 nought plus 𝑐𝑑. In Cartesian form, we can compare this and say that the equation would be π‘₯ minus π‘₯ nought over π‘Ž equals 𝑦 minus 𝑦 nought over 𝑏 which equals 𝑧 minus 𝑧 nought over 𝑐. So, to write parametric equations, we’re going to begin by comparing our Cartesian equation to the general form.

We begin by noticing that the coefficient of π‘₯, 𝑦, and 𝑧 throughout is simply positive one. To achieve this in the first part of our equation, we’re going to divide both the numerator and denominator of the fraction by three. Now, remember, as long as we do the same to both the numerator and the denominator, we’re just creating an equivalent fraction. So, it doesn’t change the size. Three π‘₯ divided by three is π‘₯. Negative seven divided by three is negative seven-thirds and negative nine divided by three is negative three.

Now, we have the first part of the equation in the right form. That is, the coefficient of π‘₯ is one. We’re going to repeat this process for the second part of the equation. This time, we want the coefficient of 𝑦 two to be one. So, we’re going to divide both the numerator and the denominator of our fraction by eight. Eight 𝑦 divided by eight is 𝑦. Negative three divided by eight is negative three-eighths. And four divided by eight is one-half.

We repeat this process one more time. This time, we want the coefficient of 𝑧 to be positive one. So, we’re going to have to divide the numerator and denominator of this fraction by negative six. Eight divided by negative six is four-thirds. Negative six 𝑧 divided by negative six is 𝑧. And negative nine divided by negative six is three over two. Well, we write this last bit ever so slightly so it compares quite nicely to our original equation. And of course, we know that these three expressions are equal.

Let’s compare π‘₯ minus seven over three over negative three to π‘₯ minus π‘₯ nought over π‘Ž. We can let π‘₯ nought be equal to seven over three here and π‘Ž be equal to negative three. Using the parametric form π‘₯ equals π‘₯ nought plus π‘Žπ‘‘, we can say that the parametric equation for π‘₯ is π‘₯ equals seven-thirds minus three 𝑑. Let’s repeat this process for the second part. 𝑦 nought here is going to be equal to three-eighths, and 𝑏 is equal to one-half. And so, using the parametric form 𝑦 equals 𝑦 nought plus 𝑏𝑑, we find 𝑦 equals three-eighths plus one-half 𝑑.

We’ll repeat this process for the final part. This time, we could say that 𝑧 nought is equal to negative four-thirds and 𝑐 is equal to three over two. Once again, we compare this to the parametric form, when we get 𝑧 equals negative four-thirds plus three over two 𝑑. And so, by comparing our equation to the general form and manipulating it somewhat, we found the parametric equations of our straight line are π‘₯ equals seven-thirds minus three 𝑑, 𝑦 equals three-eighths plus a half 𝑑, and 𝑧 equals negative four-thirds plus three over two 𝑑.

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