Video Transcript
How far from the 𝑦𝑧-plane is the
point negative 16, negative 13, 20?
In this question, we are asked to
determine how far a point is away from the 𝑦𝑧-plane. We can recall that this means the
perpendicular distance since it is the shortest distance between the point and the
plane.
There are many ways we can answer
this question, and we will go through two of these. First, we can note that any vector
entirely in the 𝑥-direction is perpendicular to the 𝑦𝑧-plane. Next, we can note that the point
zero, negative 13, 20 lies on the 𝑦𝑧-plane, since its 𝑥-coordinate is zero. The only difference between this
point on the plane and the point off of the plane is the 𝑥-coordinate. We can note that the line between
these points will have a direction vector entirely in the 𝑥-direction. The distance between these two
points will be the magnitude of the 𝑥-coordinate of the point not on the plane,
which is 16.
It is worth noting that this idea
works in general. We can note that the distance
between a point 𝑎, 𝑏, 𝑐 and the 𝑦𝑧-plane will be the magnitude of 𝑎, since the
point zero, 𝑏, 𝑐 lies on the 𝑦𝑧-plane and the line between these points is
perpendicular to the 𝑦𝑧-plane.
This is not the only way we can
answer this question. It can be useful to answer this
question using one of the formulas for the distance between a point and a plane. We can recall that the distance
between the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one and the plane 𝐫 dot 𝑎, 𝑏, 𝑐
equals negative 𝑑 is given by the magnitude of 𝑎𝑥 sub one plus 𝑏𝑦 sub one plus
𝑐𝑧 sub one plus 𝑑 all over the square root of 𝑎 squared plus 𝑏 squared plus 𝑐
squared.
To apply this formula, we need to
find the vector equation of the 𝑦𝑧-plane. We note that the vector one, zero,
zero is perpendicular to the plane and the origin lies on the plane. Since the dot product between the
unit vector in the 𝑥-direction and the position vector of the origin is zero, we
have the vector equation 𝐫 dot one, zero, zero equals zero.
We can now substitute these values
into the formula and evaluate to find the distance between the point and the
plane. We obtain the magnitude of one
times negative 16 plus zero times negative 13 plus zero multiplied by 20 plus zero
all over the square root of one squared plus zero squared plus zero squared. We can then evaluate this
expression. The numerator simplifies to give
the magnitude of negative 16, which is 16, and the denominator evaluates to give
one. Hence, the perpendicular distance
between the point and the plane is 16.
This method is more complicated
than the first method. However, it is more general since
it can be applied to any plane and point provided we can find a vector equation of
the plane.
