Find the distance from the point (1, 1, 1) to the line given as the intersection of planes x+2y−
2z = 5 and 6x − 3y + 2z = 8.
To find the distance from a point to a line in 3D, we can use the formula derived from vector projection. Here are the steps to find the distance:
1. Find the direction vector of the line:
To find the direction vector of the line, we can take the cross product of the normal vectors of the two intersecting planes. Let's call the normal vectors of the planes "n1" and "n2".
For the first plane: x + 2y - 2z = 5
The normal vector n1 is [1, 2, -2].
For the second plane: 6x - 3y + 2z = 8
The normal vector n2 is [6, -3, 2].
Now, take the cross product of n1 and n2 to get the direction vector d of the line.
d = n1 x n2
2. Find a point on the line:
To find a point on the line, we can find the intersection point of the two planes. To do this, we need to solve the system of equations formed by the two planes.
Equations of the planes:
x + 2y - 2z = 5 ---- (1)
6x - 3y + 2z = 8 ---- (2)
Solve equations (1) and (2) simultaneously to find the coordinates of the intersection point.
3. Calculate the distance:
Now that we have the direction vector of the line (d) and a point on the line (let's call it P), we can calculate the distance between P and the given point (1, 1, 1) using the following formula:
Distance = |(P - 1,1,1) · d| / ||d||
In this formula, "." represents the dot product, and "|| ||" represents the magnitude or length of a vector.
By following these steps, you would be able to find the distance between the given point (1, 1, 1) and the line given as the intersection of the two planes x + 2y - 2z = 5 and 6x - 3y + 2z = 8.