Find the line passing through point (0.2.4) and parallel to two planes x + 2Z = 1 and y-3z = 2 I J K I J K calculated by matrix n= 1 0 2 n2=3 6 -3 0 1 -3 2 -1 -1 How can I get N2~

Find the line passing through point (0.2.4) and parallel to two planes x + 2Z = 1 and y-3z = 2 I J K I J K calculated by matrix n= 1 0 2 n2=3 6 -3 0 1 -3 2 -1 -1 How can I get N2~

When x = 0, z = 1 / 2, y = 7 / 2, so the point (0,7 / 2,1 / 2) is on the intersection of two planes x + 2Z = 1 and y-3z = 2; when y = 0, z = - 2 / 3, x = 7 / 3, so the point (7 / 3,0, - 2 / 3) is also on the intersection of two planes x + 2Z = 1 and y-3z = 2

Find the distance from the point P (2,3, - 1) to the straight line (expressed by two equations 2x-2y + Z + 3 = 0 and 3x-2y + 2Z + 17 = O)

The linear equation is
x-2y=11
x+z+14=0
The square of the distance is the minimum of the following function
f(x)=(2-x)^2 + (3-y)^2 + (z+1)^2 = (2-x)^2 + (8.5-x/2)^2 + (13+x)^2
= (x^2-4x+4) + (8.5^2 -8.5x +(x^2)/4) + (13^2 + 26x +x^2)
=2.25x^2 +13.5x +(4 + 8.5^2 +169)
When x = - 13.5 / (2 * 2.25) = - 3
So the square of the distance is 2.25 * 9 + 13.5 * (- 3) + (4 + 8.5 ^ 2 + 169)
Just work out the formula

2x+y-z=-3,3x+2y+z=2,x-5y+2z=17
Come on, I can't thank you enough!

2X + Y-Z = - 3 (1) 3x + 2Y + Z = 2 (2) x-5y + 2Z = 17 (3) (1) + (2) to get 5x + 3Y = - 1 (4) (1) × 2 + (3) to get 5x-3y = 11 (5) (4) + (5) to get 10x = 10  x = 1 to substitute x = 1 into (4) to get y = - 2 to substitute x = 1, y = - 2 into (1) to get 2-2-z = - 3  z = 3 the solution of the equations is x = 1y = - 2Z = 3

The distance between X + 2y-5 = 0 and 2x + 4Y + 5 = 0 is

That is, to find the distance between the straight line 2x + 4y-10 = 0 and 2x + 4Y + 5 = 0 is (integer is easier to calculate than fraction)
|-10-5|/√(2²+4²)
=15/√20
=15√5/10
=3√5/2