Question Video: Finding Distances between Points and Straight Lines Mathematics

Let 𝐿 be the line through point (7, 5, 5) in the direction of vector <2, 4, βˆ’9>. Find the distance between 𝐿 and the point (2, 6, 6), to the nearest hundredth.


Video Transcript

Let 𝐿 be the line through point seven, five, five in the direction of vector two, four, negative nine. Find the distance between 𝐿 and the point two, six, six to the nearest hundredth.

Okay, so in this example, we have this line 𝐿. And we’re told that this line passes through the point seven, five, five and that it’s parallel to a vector we’ll call 𝐬, where the components of 𝐬 are given as two, four, negative nine. Knowing this, as well as the coordinates of a point in space we’ll call 𝑃, we want to solve for the perpendicular distance between this given point and our line. And we’ll call this distance 𝑑.

To start doing this, we can recall that, in general, if we’re given a point in space 𝑃 𝑠, a point on a line 𝑃 𝐿, and a vector parallel to that line 𝐬, then the perpendicular distance between that line and the point in space 𝑑 is given by this expression. We see it involves the vector 𝐬, the one parallel to the line of interest, as well as a vector called 𝐏 𝐬 𝐏 𝐋. That is, this is a vector that goes from our point in space to the point on our line.

If we were to sketch in this vector, for the information given in this scenario, the vector would look like this. We can call this vector 𝐏𝐋. It goes from our point to our line. And as far as the components of 𝐏𝐋, it’s equal to the vector form of the difference between the point on our line and the point in space. Performing this subtraction, we get five, negative one, negative one.

Knowing this, our next step according to our equation for 𝑑 is to cross this vector with a vector parallel to our line. That is, we want to calculate 𝐏𝐋 cross 𝐬, where 𝐬 is this given vector. That cross product equals the determinant of this three-by-three matrix.

Note that in the first row, we have the 𝐒, 𝐣, and 𝐀 hat unit vectors. And then in the second and third rows, we have the corresponding components of the vectors 𝐏𝐋 and 𝐬, respectively. Starting with the 𝐒 hat component, that has a magnitude equal to the determinant of this two-by-two matrix. Negative one times negative nine minus negative one times four comes out to 13. For the 𝐣-component, we have negative the determinant of this two-by-two matrix, where five times negative nine minus negative one times two equals negative 43. Lastly, the 𝐀-component, five times four minus negative one times two is equal to 22. This then is our cross product. And we can write this as 13, 43, 22 in vector notation.

Now that we’ve calculated 𝐏𝐋 cross 𝐬, we’re ready to take the magnitude of this vector and divide it by the magnitude of 𝐬. In our numerator, the magnitude of 𝐏𝐋 cross 𝐬 equals the square root of 13 squared plus 43 squared plus 22 squared, while in the denominator, the magnitude of 𝐬 equals the square root of two squared plus four squared plus negative nine squared. Entering this entire expression in our calculator and rounding to the nearest hundredth, we get 4.98.

Since we’ve calculated a distance, this value will be in some type of units. Since we don’t know what those are specifically, we’ll report our answer as 4.98 length units.

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