Find the equation of the line through the point (4,5,5) that meets the line ...?

Question

Find the equation of the line through the point (4,5,5) that meets the line (x -11/ 3) = (y + 8/ -1) = (z - 4/ 1) at right angle.


Please show all logical work and reasoning!

Answer 1

We have an equation of a line. I assume you mean:

t = (x -11)/3 = (y + 8)/-1 = (z - 4)/1

Let's put it into parametric form. And then into vector form.

x = 11 + 3t
y = 8 - t
z = 4 + t

L = <11, 8, 4> + t<3, -1, 1>
where t is a scalar ranging over the real numbers

To find the equation of the line thru the point R(4, 5, 5) that meets the line at a right angle we first need to find the point of intersection Q, on the line with the perpendicular line.

Let t = 0. Then we can find a point P(11, 8, 4) on the line. We also have the directional vector v of the line,
v = <3, -1, 1>.

Q = P + sv
where s is a constant

The line QR is perpendicular to the given line. So the dot product of v with the vector QR is zero.

v • QR = 0
v • = 0
v • = 0
v • - sv • v = 0
v • - s|| v ||² = 0

s|| v ||² = v •
s = v • / || v ||²

Plug in the numbers and solve for s.

s = <3, -1, 1> • <4-11, 5-8, 5-4> / || 3² + (-1)² + 1² ||
s = <3, -1, 1> • <-7, -3, 1> / || 9 + 1 + 1 ||
s = (-21 + 3 + 1) / 11 = -17/11

Now solve for Q.

Q = P + sv
Q = (11, 8, 4) - (17/11)<3, -1, 1>
Q = (70/11, 105/11, 27/11)

Now we can write the directional vector u, of the perpendicular line.

u = QR = = <4 - 70/11, 5 - 105/11, 5 - 27/11>
u = <-26/11, -50/11, 28/11>

Any non-zero multiple of u will also be a directional vector of the line. Multiply by -11/2.

u = <13, 25, -14>

The equation of the perpendicular line L2 is below.

L2 = R + tu
L2 = <4, 5, 5> + t<13, 25, -14>
where t is a scalar ranging over the real numbers

Answer 2

To get the equation of a line you need a point and a vector telling which direction the line goes from that point. They give you the point so you just need to find the vector.
Let (x,y,z) be the point where the 2 lines intersect. Then the vector through this point and (4,5,5) is (x-4,y-5,z-5). A vector that points in the direction of the given line is (3,-1,1).
Since you want the two lines to meet at a right angle take the dot product of the two vectors and set it equal to zero. Use the equation you get from this along with the equation for the given line to find what x,y and z are. Then you'll have 2 points and it should be a simple matter from there to find the equation of the line.

Answer 3

The line given by (x -11/ 3) = (y + 8/ -1) = (z - 4/ 1) can be be written as

r(t) = [11/3, 8/3, 4] + t [1, 1, 1].

Then the direction vector of a line perpendicular to this, [a, b, c], has zero scalar (dot/inner) product with [1, 1, 1], ie.

[a, b, c]•[1, 1, 1] = a + b + c = 0,

so we can write such a line going through [4, 5, 5] as

p(s) = [4, 5, 5] + s [a, b, -a - b].

Then put

[4, 5, 5] + s [a, b, -a - b] = [11/3, 8/3, 4] + t [1, 1, 1]

and solve accordingly.