Video: Finding the Straight-Line Equation Passing through Two Given Points in 3D in Parametric Form

Write the equation of the straight line L passing through the points 𝑃₁ = (4, 1, 5) and 𝑃₂ = (βˆ’2, 1, 3) in parametric form.

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Video Transcript

Write the equation of the straight line 𝐿 passing through the points 𝑃 one equals four, one, five and 𝑃 two equals negative two, one, three in parametric form.

When Cartesian coordinates of a curve or surface are represented as functions of the same variable, usually 𝑑, they’re called parametric equations. Thus, parametric equations in the π‘₯𝑦-plane would be denoted π‘₯ equals 𝑓 of 𝑑, 𝑦 equals 𝑔 of 𝑑, and 𝑧 equals β„Ž of 𝑑. These essentially tell us the π‘₯-, 𝑦-, and 𝑧-coordinates of the curve, given a specific value of 𝑑. And it can be useful to recall the vector equation of a line and use that to find the parametric equation.

For a straight line that passes through a point with a position vector π‘Ÿ nought in a direction vector of π‘Ž, the vector equation is π‘Ÿ equals π‘Ÿ nought plus π‘Ž times 𝑑, where 𝑑 can take values from negative ∞ to ∞. So let’s write the vector equation of our straight line 𝐿. We know it passes through 𝑃 one. And 𝑃 one has a position vector of four, one, five. Remember, that’s just the factor that takes us from the origin to the point 𝑃 one.

But what about the direction vector? Well, the direction vector can be found by moving from 𝑃 one to 𝑃 two, by finding the vector 𝑃 one 𝑃 two. And that’s found by subtracting the vector 𝑂𝑃 one from 𝑂𝑃 two. That’s negative two, one, three minus four, one, five. We can work out 𝑃 one 𝑃 two by subtracting the individual elements. So we get negative six, zero, and negative two. And so in column vector form, we find that the vector equation of our straight line 𝐿 is four, one, five plus negative six, zero, negative two 𝑑.

We’re now going to write π‘Ÿ as a column vector. So we can see what we need to do next. We’re going to write it as π‘₯, 𝑦, 𝑧. We then distribute the second vector by multiplying each individual component by 𝑑. And then, we find the sum of the π‘₯-, 𝑦-, and 𝑧-components. So we have π‘₯, 𝑦, 𝑧 equals four minus six 𝑑, one plus zero 𝑑, and five minus two 𝑑. Of course, we can write the 𝑦-component simply as one.

And now, we should be able to spot what we need to do next. We know that, for these two vectors to be equal, their individual components must themselves be equal. That is, π‘₯ must be equal to four minus six 𝑑. 𝑦 must be equal to one. And 𝑧 is equal to five minus two 𝑑, for values of 𝑑 greater than negative ∞ and less than ∞. And remember, we were expecting our equations to be π‘₯ is equal to some function of 𝑑. 𝑦 is equal to some function of 𝑑. And 𝑧 is equal to some other function of 𝑑. We have indeed achieved this. So we’ve written the equation of the straight line 𝐿 in parametric form.

It’s π‘₯ equals four minus six 𝑑, 𝑦 equals one, and 𝑧 equals five minus two 𝑑 for values of 𝑑 greater than negative ∞ and less than ∞.

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