A line has Cartesian equations given by
x-1/3=y+2/4=z-4/5
a) Give the coordinates of the point on the line
b) Give the vector parallel to the line
c) Write down the equation of the line in parametric form
d) Determine the Cartesian equation of the plane that is perpendicular to the line and passes through the point (2,3,-1)
I am sure you meant:
(x-1)/3 = (y+2)/4 = (z-4)/5
a) Give the coordinates of the point on the line
this is a poorly worded question, since we have an infinite number of points on this line
one of them is (1,-2,4)
b) you MUST know how to find the answer to this!!!
c) find the cross product with <3, 4, 5> as your first step.
(check my taking the dot product of your answer with <3, 4, 5> , it must be zero.
After that, just write down the equation, you have a point and the direction numbers.
a line has cartesian equation x-1/3
To solve this problem, we'll go step by step.
a) To find the coordinates of a point on the line, we need to choose a value for one of the variables and solve for the other two. Let's choose a value for z, for example, z = 0. Plugging z = 0 into the equation, we have:
x - 1/3 = y + 2/4 = 0 - 4/5
Simplifying these equations, we get:
x - 1/3 = y + 1/2 = -4/5
To find the values of x and y, we can use any two of the equations above. Let's choose the first two equations:
x - 1/3 = y + 1/2
Rearranging this equation, we have:
x = y + 1/2 + 1/3
x = y + 5/6
So, the coordinates of the point on the line are (y + 5/6, y, 0).
b) To find the vector parallel to the line, we can use the coefficients of x, y, and z in the given equation. The coefficients of x, y, and z are 1, -1, and 1/5, respectively. Therefore, the vector parallel to the line is (1, -1, 1/5).
c) To write down the equation of the line in parametric form, we can express x, y, and z in terms of a parameter t. Using the coordinates of the point on the line (y + 5/6, y, 0) from part (a), we have:
x = y + 5/6
y = y
z = 0
So, the parametric equations of the line are:
x = t + 5/6
y = t
z = 0
d) To determine the Cartesian equation of the plane perpendicular to the line and passing through the point (2, 3, -1), we need the normal vector of the plane. The normal vector is the coefficients of x, y, and z in the line's equation, which are 1, -1, and 1/5, respectively.
Using the point-normal form of a plane equation, the equation of the plane is:
1(x - 2) + (-1)(y - 3) + (1/5)(z + 1) = 0
Expanding and simplifying this equation, we get:
x - 2 - y + 3 + (1/5)z + 1/5 = 0
x - y + (1/5)z + 10/5 = 0
5x - 5y + z + 10 = 0
Therefore, the Cartesian equation of the plane is 5x - 5y + z + 10 = 0.