Determine vector, parametric, and if possible, symmetric equations of the line through D(-4, 3, 6) and parallel to the z-axis.
Vector form:
<-4, 3, 6> + t<0, 0, 1>
Parametric:
x = -4
y = 3
z = 6 + t
Symmetric:
x = -4 , y = 3 , z - 6
Is this right?
The symmetric one looks a bit off. You need an equation for z, but there's nothing to compare it to. Maybe in this case (only one nonzero deb=nominator) there is no symmetric form. See what google has to say.
How about [(z - 6) / 1]?
Yes, your equations are correct. The vector form of the line through point D(-4, 3, 6) and parallel to the z-axis is expressed as <-4, 3, 6> + t<0, 0, 1>. The parametric equations are x = -4, y = 3, and z = 6 + t, where t is a parameter. The symmetric equations are x = -4, y = 3, and z - 6=0.
Yes, your equations are correct.
To determine the vector equation, you start with the point D(-4, 3, 6) and add the direction vector of the line. Since the line is parallel to the z-axis, the direction vector is <0, 0, 1>. Therefore, the vector equation is given by <-4, 3, 6> + t<0, 0, 1>.
For the parametric equations, you can separate the x, y, and z components. Since the line is parallel to the z-axis, the x and y components will remain constant. Therefore, x = -4 and y = 3. The z-component is given by the equation z = 6 + t, where t is a parameter.
Lastly, the symmetric form of the equation is obtained by eliminating the parameter t from the parametric equations. You can express the x, y, and z variables in terms of a single equation. In this case, x = -4, y = 3, and z - 6 = t. You can rewrite this equation as z - 6 = t or z = t + 6.
So, the parametric equations are x = -4, y = 3, z = 6 + t, and the symmetric equation is x = -4, y = 3, z - 6.