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Question Video: Finding Distances between Points and Straight Lines Mathematics

Determine, to the nearest hundredth, the distance between the point (7, −5, −4) and the straight line passing through the point (0, −2, 2) with direction ratios (−9, 7, −5).

04:54

Video Transcript

Determine, to the nearest hundredth, the distance between the point seven, negative five, negative four and the straight line passing through the point zero, negative two, two with direction ratios negative nine, seven, negative five.

We’re given the direction ratios of a line, which are the components of the direction vector 𝐝 of the line. Remember a direction vector is parallel to and in the same direction as the line. We’re told that the line passes through the point zero, negative two, two. And if we call this point 𝐴, then vector from the origin to 𝐴, that’s 𝐎𝐀, has components zero, negative two, two. We’re given a point — let’s call it 𝑃 — with coordinates seven, negative five, negative four so that the vector 𝐎𝐏 from the origin to 𝑃 then has components seven, negative five, negative four. And we want to find the distance from the point 𝑃 to the line. So let’s call this distance uppercase 𝐷.

And to find this distance, we can use the formula shown, which tells us that the distance 𝐷 is the magnitude of the cross product of vector 𝐀𝐏 with the direction vector 𝐝 divided by the magnitude of direction vector 𝐝. Now we know that the direction vector 𝐝 has components negative nine, seven, negative five. And we can find the vector 𝐀𝐏 from our two vectors 𝐎𝐀 and 𝐎𝐏. And so we have 𝐀𝐏 is 𝐎𝐏 minus 𝐎𝐀, that is, seven, negative five, negative four, minus zero, negative two, two. And subtracting like-for-like components, this gives us seven minus zero, negative five minus negative two, and negative four minus two. And evaluating this gives us the vector with components seven, negative three, negative six.

So now making some space, the next thing we need to find for our distance is the cross product of 𝐀𝐏 with the direction vector 𝐝. And we can find this cross product by evaluating the determinant of the three-by-three matrix whose first row is the unit vectors 𝐢, 𝐣, and 𝐤 and whose second and third rows are our two vectors 𝐀𝐏 and 𝐝. Expanding along the first row of our three-by-three matrix, we have the determinant of the two-by-two matrix with elements negative three, negative six, seven, negative five multiplied by 𝐢 minus the determinant of the two-by-two matrix with elements seven, negative six, negative nine, negative five multiplied by 𝐣 plus the determinant of the two-by-two matrix with elements seven, negative three, negative nine, and seven multiplied by the unit vector 𝐤.

And now recalling that the determinant of a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. Our first two-by-two determinant is negative three times negative five minus negative six times seven. That’s 15 plus 42 𝐢, which is 57𝐢. And our second determinant is seven times negative five minus negative six times negative nine, which we multiply by 𝐣. That’s negative 35 minus 54 𝐣, which is negative 89𝐣. And finally, our third determinant is seven times seven minus negative three times negative nine. And this is multiplied by 𝐤. And that’s 49 minus 27 𝐤, which is 22𝐤. So our cross product of 𝐀𝐏 with direction vector 𝐝 is 57𝐢 minus negative 89𝐣 plus 22𝐤. And since minus negative 89 is just 89, our cross product is 57𝐢 plus 89𝐣 plus 22𝐤. That is the vector with components 57, 89, and 22.

So now making a little space, we next need to find the magnitude of this vector and that’s given by the positive square root of 57 squared plus 89 squared plus 22 squared, that is, the square root of the sum of the squares of the components. And this evaluates to the square root of 11654. And we leave this in this form for the moment where we next need to find the magnitude of the direction vector 𝐝. And since 𝐝 has components negative nine, seven, and negative five, its magnitude is the positive square root of negative nine squared plus seven squared plus negative five squared. And that’s the square root of 81 plus 49 plus 25 which is the square root of 155.

So now using our two magnitudes, that’s 𝐀𝐏 cross 𝐝 and direction vector 𝐝, into our formula for the distance, that’s uppercase 𝐷, we have the square root of 11654 divided by the square root of 155. And we can collect our two radicals into one. This evaluates to 8.6710 and so on, which is 8.67 to the nearest hundredth. Hence to the nearest hundredth, the distance between the given point and the straight line is 8.67 units.

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