Answer
See the explanation
Work Step by Step
To find the distance from a point to a line in space, you can use the formula:
$d = \frac{|(P - Q) × V|} {|V|}$
where:
- P is the given point,
- Q is a point on the line,
- V is the direction vector of the line,
- × denotes the cross product, and
- | | represents the magnitude or length of a vector.
Let's consider an example to illustrate this. Suppose we have a line defined by the equation:
x = 2 + t
y = 1 - t
z = 3t
where t is a parameter. We want to find the distance from the point P(4, -1, 2) to this line.
First, we need to find a point Q on the line. We can choose any value for t and substitute it into the equations to find the corresponding coordinates. Let's set t = 0:
Q(2, 1, 0)
Next, we find the direction vector V of the line. We can take the coefficients of t in the equations:
V = (1, -1, 3)
Now, we can substitute the values into the distance formula:
$d =\frac{ |(P - Q) × V| }{|V|}$
$ = \frac{|(4 - 2, -1 - 1, 2 - 0) × (1, -1, 3)|} {|(1, -1, 3)|}$
Simplifying further:
$d =\frac{ |(2, -2, 2) × (1, -1, 3)| }{|(1, -1, 3)|}$
$= \frac{|(-4, -4, 0)|}{\sqrt{(1^2 + (-1)^2 + 3^2)}}$
$=\frac{8}{ \sqrt{11}}$
Therefore, the distance from the point P(4, -1, 2) to the given line is \frac{8}{\sqrt{11}}.
Now, let's move on to finding the distance from a point to a plane in space.
To find the distance from a point to a plane, you can use the formula:
$d =\frac {|(P - A) · n|}{ |n|}$
where:
- P is the given point,
- A is a point on the plane,
- n is the normal vector of the plane,
- · denotes the dot product, and
- | | represents the magnitude or length of a vector.
Consider an example where we have a plane defined by the equation:
2x + 3y - z = 5
and we want to find the distance from the point P(1, -2, 4) to this plane.
First, we need to find a point A on the plane. We can choose any values for x and y and substitute them into the equation to find the corresponding z-coordinate. Let's set x = 0 and y = 0:
A(0, 0, -5)
Next, we find the normal vector n of the plane. We can take the coefficients of x, y, and z in the equation:
n = (2, 3, -1)
Now, we can substitute the values into the distance formula:
$d = \frac{|(P - A) · n| }{|n|}$
$ = \frac{|(1 - 0, -2 - 0, 4 - (-5)) · (2, 3, -1)|}{|(2, 3, -1)|}$
Simplifying further:
$d = \frac{|(1, -2, 9) · (2, 3, -1)|}{\sqrt{(2^2 + 3^2 + (-1)^2)}}$
$= \frac{|(2 - 6 - 9)|}{\sqrt{14}}$
$= \frac{|-13|}{\sqrt{14}}$
$ = \frac{13}{\sqrt{14}}$
Therefore, the distance from the point P(1, -2, 4) to the given plane is $\frac{13 }{\sqrt{14}}$.
