Question Video: Finding the Distance between Three Parallel Lines

Find to the nearest hundredth the distance between the parallel lines π‘₯ = 6 + 𝑑, 𝑦 = 8 + 2𝑑, 𝑧 = 7 + 3𝑑 and (π‘₯ βˆ’ 2)/3 = (𝑦 βˆ’ 3)/6 = (𝑧 βˆ’ 1)/9.

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Video Transcript

Find to the nearest hundredth the distance between the parallel lines π‘₯ equals six plus 𝑑, 𝑦 equals eight plus two 𝑑, 𝑧 equals seven plus three 𝑑 and π‘₯ minus two over three equals 𝑦 minus three over six equals 𝑧 minus one over nine.

Okay, so here we have these two parallel lines, but say that this is line one and this is line two. We want to know the distance between the lines, meaning the shortest possible distance or the perpendicular distance between them. We’ll call this 𝑑. To solve for this distance, there are three things we’ll need to know. We’ll need to know the coordinates of a point on line one. We’ll call that 𝑃 one. We’ll also need to know the coordinates of a point on line two. We’ll call that 𝑃 two. And lastly, we’ll need to know the components of a vector that runs parallel to both these lines. We’ll call this vector 𝐬. If we can solve for these three bits of information, then we can use this relationship to calculate the distance between the lines.

In this equation, the vector 𝐬 is indeed a vector that’s parallel to both lines, and the vector 𝐏 one 𝐏 two would look like this on our sketch. It goes from point one to point two. Let’s now start solving for the vector 𝐏 one 𝐏 two by figuring out a point on line one and one on line two. Starting with line one, we see that this is given to us in what’s called parametric form. We have separate equations for the π‘₯-, 𝑦-, and 𝑧-coordinates of every point on line one. It’s possible to combine together these three equations into what’s called the vector form of a line. And we would do it by saying that π‘₯, 𝑦, and 𝑧 are components of a vector which is equal to another vector that starts at the origin of a coordinate frame and goes to the point six, eight, seven and then moves up and down the line in this direction multiplied by the scale factor 𝑑.

We get two important pieces of information from this vector form of line one’s equation. First, we can say that the line passes through the point six, eight, seven. We’ve solved then for our point 𝑃 one. Along with this, we know that this vector with components one, two, three is parallel to line one. That means it’s parallel to line two as well. And therefore, these can be our vector 𝐬 with components one, two, three. Knowing all this about line one, we can now clear some space and start to work with our given equation for line two.

Now that we know 𝑃 one as well as 𝐬, all we need to solve for is a point somewhere on line two. The equation of this line is written in what’s called symmetric form. This means that our line is expressed using a series of equalities. These equalities all hold true because these three fractions are all equal to the same scale factor we can call 𝑑 two. So then π‘₯ minus two over three equals 𝑑 two as does 𝑦 minus three over six as does 𝑧 minus one over nine. All this lets us rewrite the equation of line two in parametric form. For example, since π‘₯ minus two over three equals 𝑑 two, it must be true that π‘₯ equals three times 𝑑 two plus two. In a similar way, 𝑦 equals six times 𝑑 two plus three, while 𝑧 equals nine times 𝑑 two plus one. We’ve rewritten line two in parametric form because this way it’s easier to see a point that this line passes through.

If we again consider combining these three equations into one vector equation, on the left we would have a vector with components π‘₯, 𝑦, 𝑧, while on the right we multiply our scale factor 𝑑 two by a vector parallel to the line with components three, six, nine. And to this is added a vector that goes from the origin of a coordinate system to a point on the line. In other words, the point with coordinates two, three, one lies along line two.

Now that we know all this, we can move ahead with calculating our vector 𝐏 one 𝐏 two. It will be the vector of point 𝑃 one subtracted from point 𝑃 two. And substituting in our values for 𝑃 one and 𝑃 two, we find this is a vector with components negative four, negative five, negative six. At this point, we’re ready to calculate this cross product of 𝐏 one 𝐏 two and the vector 𝐬. It’s equal to the determinant of this matrix. Here in the top row, we have the 𝐒 hat, 𝐣 hat, and 𝐀 hat unit vectors and then below that the corresponding π‘₯-, 𝑦-, and 𝑧-components of 𝐏 one 𝐏 two and the π‘₯-, 𝑦-, and 𝑧-components of 𝐬.

The 𝐒 hat component of this vector equals the determinant of this two-by-two matrix. Negative five times three minus negative six times two comes out to negative three. The negative 𝐣 hat component equals the determinant of this matrix. Negative four times three minus negative six times one equals negative six. And lastly, the 𝐀-component of our cross product equals the determinant of this two-by-two matrix. Negative four times two minus negative five times one comes out to negative three altogether. Then this is our cross product, and we can write this instead in vector form with components negative three, six, negative three.

Alright, now that we know 𝐏 one 𝐏 two cross 𝐬, we’re ready to solve for 𝑑 by calculating the magnitude of this cross product and dividing it by the magnitude of 𝐬. Writing all that out, the magnitude of 𝐏 one 𝐏 two cross 𝐬 equals the square root of negative three squared plus six squared plus negative three squared while the magnitude of the vector 𝐬 equals the square root of one squared plus two squared plus three squared. Entering this fraction on our calculator, rounded to the nearest hundredth, our answer is 1.96. That’s the shortest distance between these two parallel lines.

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