Question

the foot of perpendicular from origin to the line x+2y+3z+4=0=2x+3y+4z+5 is​

Answer 1

Given:

Line x+2y+3z+4=0=2x+3y+4z+5

To Find:

The foot of perpendicular from origin to the line.

Solution:

To get the equation of line of intersection of these 2 planes,

• x + 2y + 3z + 4 = 0;
• 2x + 3y + 4z + 5 = 0

Let z be a free variable.

• z = t ; foor of the perpendicular on the line.
• x + 2y = -3t - 4 - ( 1 )
• 2x + 3y = -4t - 5 - (2)

Solving for x and y in terms of t ,

• 2x + 4y = -6t - 8
• 2x + 3y = -4t - 5

• y = -2t -3
• x = -3t - 4 - 2 ( -2t - 3 )
• x = -3t - 4 +4t + 6
• x = t + 2

Therefore parametrized form of line,

• ( t + 2 )i + (-2t - 3 )j + tk
• Equation of line = 2i -3j + t ( i -2j +k)

If ( t + 2 )i + (-2t - 3 )j + tk is the foot of the perpendicular,

• Dot product vector from origin to the foot and the parallel vector to line = 0

• (( t + 2 )i + (-2t - 3 )j + tk ).( i - 2j + k) = 0

• t + 2 + 4t + 6 + t = 0
• 6t = -8
• t = -4/3

Foot of perpendicular from origin to the line x+2y+3z+4=0=2x+3y+4z+5 is​ ( 2/3 , -1/3 , -4/3).