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 𝑥- and 𝑦-axes must have a
slope or gradient equal to one. For every one unit in the
𝑥-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 𝐢, 𝐣,
and 𝐤 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 𝐢 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 𝐤. Now, recalling that the determinant
of a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐, this gives
us negative three minus negative six 𝐢 minus two minus negative six 𝐣 plus two
minus negative three 𝐤, that is, three 𝐢 minus eight 𝐣 plus five 𝐤.
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.
