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,
, 0)
The line intersects the xz plane at (-13, 0, -11)
The line intersects the yz plane at (0,
, 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
<em>Substitute z = 0 into equation (iii)</em>
0 = 3 + 7t
t = ![\frac{-3}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B-3%7D%7B7%7D)
<em>Now substitute t = </em>
<em> into equation(i)</em>
x = 1 + 7(
)
x = -3
<em>Also, substitute t = </em>
<em> into equation (ii)</em>
y = 2 + 1(
)
y = ![\frac{11}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B7%7D)
Therefore, the line intersects the xy plane at (-3,
, 0)
(ii) The line will intersect the xz plance when y = 0
<em>Substitute y = 0 into equation (ii)</em>
0 = 2 + 1t
t = -2
<em>Now substitute t = -2 into equation(i)</em>
x = 1 + 7(-2)
x = 1 - 14
x = -13
<em>Also, substitute t = -2 into equation (iii)</em>
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
<em>Substitute x = 0 into equation (i)</em>
0 = 1 + 7t
t = ![\frac{-1}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B7%7D)
<em>Now substitute t = </em>
<em> into equation(ii)</em>
y = 2 + 1(
)
y = ![\frac{13}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B13%7D%7B7%7D)
<em>Also, substitute t = </em>
<em> into equation (iii)</em>
z = 3 + 7(
)
z = 3 - 1
z = 2
Therefore, the line intersects the yz plane at (0,
, 2)