Direction vector of straight line A unit vector collinear with the direction vector of the line 3x + 4Y + 5 = 0 is () a (3,4) b (4,-3) c (3/5,4/5) d (4/5,-3/5)

Direction vector of straight line A unit vector collinear with the direction vector of the line 3x + 4Y + 5 = 0 is () a (3,4) b (4,-3) c (3/5,4/5) d (4/5,-3/5)

Halo, the simple way is to substitute a, B, C, B into 3x + 4Y + 5 = 0, because it is a direction vector, so one of these numbers must be on this straight line. The result is
He made a mistake to deceive people

What is the relationship between the direction vector and the normal vector of the same line?

The direction vector and normal vector of the same line are perpendicular to each other

What do you think of the direction vector and normal vector of 2x - y + Z = 0
Can we give an example of the equation perpendicular to it and the equation parallel to it?

First of all, the normal vector generally refers to the normal vector of the surface, and the standard equation of the surface is ax + by + CZ + D = 0. The normal vector is (a, B, c). The direction vector generally refers to the direction vector of the line. The line can be composed of parametric equations or represented by two surfaces. The standard parametric equation of the line is x = LT + A, y = MT + B, z = NT + C. the direction vector is (L, m, n). A point multiplied by B = 0, two vectors are vertical. A cross multiplied by B = 0, two vectors are parallel, All the relations between face and line can be solved by vector
So the normal vector of 2x-y + Z = 0 is (2, - 1,1), and the equation perpendicular to it needs only to find a vector point multiplied by 0, such as (0,1,1), equation y + Z + D = O (D is any constant). The vector parallel to it (4, - 2,2), equation 4x-2y + 2Z + D = 0