Video: Finding the Perpendicular Distance from a Given Point to the Line on Two Given Points

Find, to one decimal place, the perpendicular distance from point (βˆ’3, βˆ’4, 0) to the line on points (1, 3, 1) and (4, 3, 2).

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Video Transcript

Find, to one decimal place, the perpendicular distance from the point negative three, negative four, zero to the line on points one, three, one and four, three, two.

And let’s begin by recalling we’ve got a line that passes through the points 𝐴 and 𝐡. The shortest distance from a third point, 𝐢, to the line is the perpendicular distance from that point. And actually, there are a number of ways that we can calculate this distance.

Before we do though, let’s define 𝐴 as being the point with coordinates one, three, one and 𝐡 as being the point with coordinates four, three, two. 𝐢 has coordinates negative three, negative four, zero. But we’re going to begin by finding the equation of the line that passes through 𝐴 and 𝐡.

We’re going to use the vector equation of a straight line. Given that the line contains a point with a position vector of 𝐚 and the line itself has a direction vector of 𝐝, then the equation of the line 𝐫 is 𝐚 plus 𝑑 times 𝐝.

Now in our case, we know that a point on our line has coordinates one, three, one. So the position vector is one, three, one. And the equation of our line is 𝐫 equals one, three, one plus 𝑑 times the direction vector. And that’s the vector 𝐀𝐁.

Now of course, to find the direction vector 𝐀𝐁, we subtract the position vector of 𝐎 from the position vector of 𝐁. That’s 𝐎𝐁 minus πŽπ€, which is four, three, two minus one, three, one. We can, of course, subtract the individual components of each vector. And we find that the vector 𝐀𝐁 is three, zero, one. We now have the vector of the line passing through points 𝐀 and 𝐁. It’s one, three, one plus 𝑑 times three, zero, one.

We’ve cleared some space for the next step. The next thing we’re going to do is to find the point on the line where the perpendicular from 𝐢 meets the line as point 𝐷. Then we can say that the position vector of 𝐝 β€” that’s vector πŽπƒ β€” is one plus three 𝑑, three plus zero 𝑑, and one plus 𝑑. Essentially, we’ve broken down the individual components of our line into π‘₯-, 𝑦-, and 𝑧-coordinates. And so that simplifies to one plus three 𝑑, three, and one plus 𝑑.

And that means we can now find the vector that takes us from 𝐢 to 𝐷. It’s the difference between the vector πŽπƒ and πŽπ‚. That’s of course one plus three 𝑑, three, one plus 𝑑 minus negative three, negative four, zero. And this simplifies to four plus three 𝑑, seven, and one plus 𝑑.

And now, at this stage, we have a few options. We know that the vectors joining 𝐴 to 𝐡 and 𝐢 to 𝐷 are perpendicular. And so we could use the dot product and say that this means that the dot product of the vector 𝐀𝐁 and 𝐂𝐃 is equal to zero.

Alternatively, we could use the distance formula. We know that the magnitude of 𝐂𝐃 is equal to the square root of the sum of the squares of its individual components. So it’s the square root of four plus three 𝑑 squared plus seven squared plus one plus 𝑑 squared.

We distribute the parentheses inside this root. And when we simplify, we find that the magnitude of 𝐂𝐃 is the square root of 10𝑑 squared plus 26𝑑 plus 66. Now, of course, we know that the perpendicular distance from the point to our line is the shortest possible distance. And so what we could actually do is look to minimize the value of the magnitude of 𝐂𝐃.

When we’re looking to minimize or find the minimum, we differentiate with respect to 𝑑. Of course, the smaller the number, the smaller the square root. So actually, we’re only looking to differentiate 10𝑑 squared plus 26𝑑 plus 66. We’ll differentiate term by term. The derivative of 10𝑑 squared with respect to 𝑑 is two times 10𝑑, which is 20𝑑. The derivative of 26𝑑 is 26, and the derivative of 66 is equal to zero.

To find the critical point, we’ll set this equal to zero and solve for 𝑑. And when we do, subtracting 26 from both sides and dividing by 20, we find 𝑑 is equal to negative 26 over 20, which is negative 1.3. So how do we know that this is a minimum?

Well, usually we’d look to perform a second- or first-derivative test. But actually, we can look at the expression 10𝑑 squared plus 26𝑑 plus 66. Imagine we were trying to draw the graph of 𝑦 equals 10𝑑 squared plus 26𝑑 plus 66. It’s a quadratic equation, so we get a parabola. The leading coefficient here, the coefficient of 𝑑 squared, is positive. So we get the U-shaped parabola. And that means the only turning point is at the minimum. It’s the absolute minimum value.

Now that we know the value of 𝑑 that minimizes the expression the square root of 10𝑑 squared plus 26𝑑 per 66, let’s substitute it in. We get the square root of 10 times negative 1.3 squared plus 26 times negative 1.3 plus 66, which is 7.007 and so on. Correct to one decimal place, that’s 7.0.

And so the perpendicular distance from the point negative three, negative four, zero to the line on points one, three, one and four, three, two is 7.0 length units.

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