### 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 letβs begin by recalling weβve
got a line that passes through the points π΄ and π΅. The shortest distance from a third
point, πΆ, 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, letβs define
π΄ as being the point with coordinates one, three, one and π΅ as being the point
with coordinates four, three, two. πΆ has coordinates negative three,
negative four, zero. But weβre going to begin by finding
the equation of the line that passes through π΄ and π΅.

Weβre going to use the vector
equation of a straight line. Given that the line contains a
point with a position vector of π and the line itself has a direction vector of π,
then the equation of the line π« is π plus π‘ times π.

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 π«
equals one, three, one plus π‘ times the direction vector. And thatβs the vector ππ.

Now of course, to find the
direction vector ππ, we subtract the position vector of π from the position
vector of π. Thatβs ππ minus ππ, 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 ππ is
three, zero, one. We now have the vector of the line
passing through points π and π. Itβs one, three, one plus π‘ times
three, zero, one.

Weβve cleared some space for the
next step. The next thing weβre going to do is
to find the point on the line where the perpendicular from πΆ meets the line as
point π·. Then we can say that the position
vector of π β thatβs vector ππ β is one plus three π‘, three plus zero π‘, and
one plus π‘. Essentially, weβve broken down the
individual components of our line into π₯-, π¦-, and π§-coordinates. And so that simplifies to one plus
three π‘, three, and one plus π‘.

And that means we can now find the
vector that takes us from πΆ to π·. Itβs the difference between the
vector ππ and ππ. Thatβs of course one plus three π‘,
three, one plus π‘ minus negative three, negative four, zero. And this simplifies to four plus
three π‘, seven, and one plus π‘.

And now, at this stage, we have a
few options. We know that the vectors joining π΄
to π΅ and πΆ to π· are perpendicular. And so we could use the dot product
and say that this means that the dot product of the vector ππ and ππ is equal to
zero.

Alternatively, we could use the
distance formula. We know that the magnitude of ππ
is equal to the square root of the sum of the squares of its individual
components. So itβs the square root of four
plus three π‘ squared plus seven squared plus one plus π‘ squared.

We distribute the parentheses
inside this root. And when we simplify, we find that
the magnitude of ππ is the square root of 10π‘ squared plus 26π‘ 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 ππ.

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

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

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

Now that we know the value of π‘
that minimizes the expression the square root of 10π‘ squared plus 26π‘ per 66,
letβs 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,
thatβs 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.