Question Video: Finding Distances between Points and Straight Lines | Nagwa Question Video: Finding Distances between Points and Straight Lines | Nagwa

Question Video: Finding Distances between Points and Straight Lines Mathematics • Third Year of Secondary School

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Let 𝐿 be the line through the point (βˆ’6, 8, 9) that makes equal angles with the three coordinate axes. What is the distance between the point (βˆ’4, 5, 3) and 𝐿 to the nearest hundredth?

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Video Transcript

Let 𝐿 be the line through the point negative six, eight, nine that makes equal angles with the three coordinate axes. What is the distance between the point negative four, five, three and 𝐿 to the nearest hundredth?

We’re asked to find the distance between a point negative four, five, three and a line 𝐿 passing through the point negative six, eight, nine that makes equal angles with the three coordinate axes. To do this, we’ll use the formula shown. That is, 𝐷 is the distance from the point to the line is the magnitude of the cross product of a vector 𝚨𝐏 and direction vector 𝐝 over the magnitude of the direction vector 𝐝. And that’s where 𝐎𝐏 is the position vector of the point, which in our case is the vector with components negative four, five, and three. 𝐎𝚨 is the position vector of a point on the line, which for the line 𝐿 has components negative six, eight, and nine. And 𝐝 is the direction vector of the line.

So we’re going to need to find the direction vector of the line. And we’re told that the line makes equal angles with the three coordinate axes. So let’s consider what this tells us. If we consider this first in two dimensions, then a line making equal angles with the π‘₯- and 𝑦-axes must have a slope or gradient equal to one. For every one unit in the π‘₯-direction, we move one unit in the 𝑦-direction. Such a line, therefore, has a direction vector with components one, one.

Now, if we extend the same logic to three dimensions, then the direction vector of the line at equal angles to all three coordinate axes will have components one, one, and one. And so we have our point with position vector negative four, five, three. The point on the line has position vector negative six, eight, nine. And our direction vector has components one, one, and one.

Now, to use the formula to find the distance between the point and the line, we need to find the vector 𝚨𝐏. And that’s given by 𝐎𝐏 minus 𝐎𝚨, that is, negative four, five, three minus negative six, eight, nine. And to subtract a vector from another vector, we subtract the individual components. And this gives us negative four minus negative six is two, five minus eight is negative three, and three minus nine is negative six.

So now making some space, the next element of our formula, we need to find the cross product of the vector 𝚨𝐏 with the direction vector 𝐝. And this is given by the determinant of the three-by-three matrix whose first row is the unit vectors 𝐒, 𝐣, and 𝐀 and whose second and third rows are the two vectors 𝚨𝐏 and the direction vector 𝐝.

Now, expanding along the first row, this is the determinant of the two-by-two matrix with elements negative three, negative six, one, one multiplied by 𝐒 minus the determinant of the two-by-two matrix with elements two, negative six, one, one multiplied by 𝐣 plus the determinant of the two-by-two matrix with elements two, negative three, one, one multiplied by the unit vector 𝐀. Now, recalling that the determinant of a two-by-two matrix with elements π‘Ž, 𝑏, 𝑐, 𝑑 is π‘Žπ‘‘ minus 𝑏𝑐, this gives us negative three minus negative six 𝐒 minus two minus negative six 𝐣 plus two minus negative three 𝐀, that is, three 𝐒 minus eight 𝐣 plus five 𝐀.

And making some space, our next step is to find the magnitude of this cross product. This is given by the square root of the sum of the squares of the components. So that’s the square root of three squared plus negative eight squared plus five squared. And that’s the square root of nine plus 64 plus 25, which is the square root of 98.

The final element of our formula is the magnitude of direction vector 𝐝. And since all three components of vector 𝐝 are equal to one, its magnitude is the square root of one squared plus one squared plus one squared, which is the square root of three. So now we have everything we need to calculate the distance 𝐝. And this is given by the square root of 98 over the square root of three, which is the square root of 98 over three, which in its simplest radical form is seven root six over three. This evaluates to three decimal places to 5.715. And so to the nearest hundredth, the distance from the point negative four, five, three to the line 𝐿, which passes through the point negative six, eight, nine and makes equal angles with the three coordinate axes, is 5.72 units.

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