Question:
Consider the line which passes through the point P(-5, 1, 4), and which is parallel to the line x=1+7t,y=2+1t,z=3+7t. Find the point of intersection of this new line with each of the coordinate planes: xy-plane, xz-plane and yz-plane.
Answer:
The line intersects the xy plane at (-3, [tex]\frac{11}{7}[/tex], 0)
The line intersects the xz plane at (-13, 0, -11)
The line intersects the yz plane at (0, [tex]\frac{13}{7}[/tex], 2)
Step-by-step explanation:
Vector form of a line can be written as;
r(t) = r₀ + vt
Where;
r₀ = position vector of the point where the line passes through
v = direction or slope vector of the line.
From the question, we have a line passing through the point P(-5, 1, 4). This means that;
r₀ = (-5, 1, 4)
Also, the line is parallel to another line with parametric equations:
x=1+7t ------------------(i)
y=2+1t ----------------(ii)
z=3+7t ---------------(iii)
Since the first line is parallel to the second line, then their gradient or slope vector must be the same. This means that;
v = (7, 1, 7) [which are the coefficients of t in the parametric equations]
Therefore, the first line can be written as;
r(t) = (-5, 1, 4) + (7, 1, 7)t
Now, to get the point of intersection of this line with each of the coordinates;
(i) The line will intersect the the xy plane when z = 0
Substitute z = 0 into equation (iii)
0 = 3 + 7t
t = [tex]\frac{-3}{7}[/tex]
Now substitute t = [tex]\frac{-3}{7}[/tex] into equation(i)
x = 1 + 7([tex]\frac{-3}{7}[/tex])
x = -3
Also, substitute t = [tex]\frac{-3}{7}[/tex] into equation (ii)
y = 2 + 1([tex]\frac{-3}{7}[/tex])
y = [tex]\frac{11}{7}[/tex]
Therefore, the line intersects the xy plane at (-3, [tex]\frac{11}{7}[/tex], 0)
(ii) The line will intersect the xz plance when y = 0
Substitute y = 0 into equation (ii)
0 = 2 + 1t
t = -2
Now substitute t = -2 into equation(i)
x = 1 + 7(-2)
x = 1 - 14
x = -13
Also, substitute t = -2 into equation (iii)
z = 3 + 7(-2)
z = 3 - 14
z = -11
Therefore, the line intersects the xz plane at (-13, 0, -11)
(iii) The line will intersect the yz plane when x = 0
Substitute x = 0 into equation (i)
0 = 1 + 7t
t = [tex]\frac{-1}{7}[/tex]
Now substitute t = [tex]\frac{-1}{7}[/tex] into equation(ii)
y = 2 + 1([tex]\frac{-1}{7}[/tex])
y = [tex]\frac{13}{7}[/tex]
Also, substitute t = [tex]\frac{-1}{7}[/tex] into equation (iii)
z = 3 + 7([tex]\frac{-1}{7}[/tex])
z = 3 - 1
z = 2
Therefore, the line intersects the yz plane at (0, [tex]\frac{13}{7}[/tex], 2)
