Question

Ind an equation of a plane through the point (−5,−1,2))(−5,−1,2)) which is orthogonal to the line x=4t−3,y=−(3+y),z=−(5+3t)x=4t−3,y=−(3+t),z=−(5+3t)

Answer 1

From the line equation you have:

1. 
   `\vec{p}=(4,-1,-3)\parallel \text{ line};`
2. 
   `B(-3,-3,-5)\in \text{ line.}`

About the plane you know that it should pass through the point
`A(-5,-1,2)` and be ortogonal to the given line (also ortogonal to the vector
`\vec{p}`). Then its equation is:

`4\cdot (x-(-5))+(-1)\cdot (y-(-1))+(-3)\cdot (z-2)=0,\\ \\4(x+5)-(y+1)-3(z-2)=0,\\ \\4x+20-y-1-3z+6=0,\\ \\4x-y-3z+25=0.`

Answer:
`4x-y-3z+25=0.`