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

Answer 1

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, $\frac{11}{7}$, 0)

The line intersects the xz plane at (-13, 0, -11)

The line intersects the yz plane at (0, $\frac{13}{7}$, 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 = $\frac{-3}{7}$

Now substitute t = $\frac{-3}{7}$ into equation(i)

x = 1 + 7($\frac{-3}{7}$)

x = -3

Also, substitute t = $\frac{-3}{7}$ into equation (ii)

y = 2 + 1($\frac{-3}{7}$)

y = $\frac{11}{7}$

Therefore, the line intersects the xy plane at (-3, $\frac{11}{7}$, 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 = $\frac{-1}{7}$

Now substitute t = $\frac{-1}{7}$ into equation(ii)

y = 2 + 1($\frac{-1}{7}$)

y = $\frac{13}{7}$

Also, substitute t = $\frac{-1}{7}$ into equation (iii)

z = 3 + 7($\frac{-1}{7}$)

z = 3 - 1

z = 2

Therefore, the line intersects the yz plane at (0, $\frac{13}{7}$, 2)