### Video Transcript

Find, to one decimal place, the
perpendicular distance from point negative three, negative four, zero to the line on
points one, three, one and four, three, two.

Recall that we can find the
perpendicular distance between a point π and a line with direction vector π given
a point π΄ on the line by using the cross product. Between the points π΄, π, and the
point of intersection between the line and its perpendicular to π, we have a right
triangle, with a hypotenuse of length equal to the magnitude of the vector π΄ to
π.

The length of the opposite side of
this right triangle to the angle enclosed by π΄ to π and π is given by the norm of
the hypotenuse, π΄ to π, multiplied by the sine of the angle π enclosed by π΄ to
π and π. Recall that the norm of the cross
product π΄ to π cross π is equal to the norm of π΄ to π multiplied by the norm of
π multiplied by the sine of the angle between the vectors, π.

Rearranging by dividing by the norm
of π, we get the same expression for the perpendicular distance on the right-hand
side: the norm of π΄ to π multiplied by sin π. We can therefore use this
expression on the left-hand side to calculate the perpendicular distance between the
point and the line.

We now need to find both of the
vectors in this expression: π΄ to π and π. The point π is negative three,
negative four, zero. We can let the point π΄ be either
point on the line. So, letβs choose the first, one,
three, one. And letβs call the other point
π΅. We can find the direction vector of
the line π by taking the difference of the position vectors of the points π΅ and
π΄, π to π΅ and π to π΄, respectively. These are given in the question as
four, three, two and one, three, one, respectively. Taking this difference gives us π
equals three, zero, one.

Note that this direction vector is
not unique. And any scalar multiple of π would
also be appropriate, since we divide by the norm of π in the expression. So, itβs only its direction thatβs
matters. The vector π΄ to π is likewise
given by the position vector of π, π to π, minus the position vector of π, π to
π΄. These are given in the question as
negative three, negative four, zero and one, three, one, respectively. Taking this difference gives us π΄
to π equals negative four, negative seven, negative one.

We now have everything we need to
evaluate the perpendicular distance. Letβs clear some space before we
proceed. We have π equals three zero one
and π΄ to π equals negative four, negative seven, negative one. We calculate the cross product, π΄
to π cross π, by taking the determinant of the three-by-three matrix π’, π£, π€,
negative four, negative seven, negative one, three, zero, one. Taking this determinant by
expanding along the top row gives π’ times negative seven minus zero minus π£ times
negative four minus negative three plus π€ times zero minus negative 21. Expanding the parentheses and
simplifying gives us the vector π΄ to π cross π equal to negative seven, one,
21.

The norm of a vector is equal to
the square root of the sum of its components squared. So, taking the norm of π΄ to π
cross π over the norm of π, we get the square root of negative seven squared plus
one squared plus 21 squared all over the square root of three squared plus zero
squared plus one squared. This simplifies to the square root
of 491 over the square root of 10. Performing this calculation gives
us the distance between the point negative three, negative four, zero and the line
on points one, three, one and four, three, two: 7.0 length units to one decimal
place.