Linear Algebra and planes in Cartesian space

Question

I was asked this question from the course Linear Algebra and I need to show all working.

The question is in 5 parts:

Consider the xyz-space R³ with the origin O. Let l be the line given by the Cartesian equation $$x = \frac{z - 1}2, y = 1 $$ Let p be the plane given by the Cartesian equation $$2 x + y - z = 1$$

a) Find two unit vectors parallel to the line l.

b) Find the point Q which is the intersection of the plane p and z-axis.

c) Take n = 2 i + j - k as a normal vector of the plane p. Decompose the vector QO into the sum of two vectors: one of them is parallel to n and the other one is orthogonal to n.

d) The plane p divides R³ into two parts. Find the unit vector perpendicular to p and pointing into the part containing the origin O.

e) Let P(x, y, z) be a point on the line l. Letting x = t for some constant t, find the y and z coordinates of P. Calculate the distance from P to the plane p.

I would like to thank everyone who takes time in helping me with this problem and I really appreciate the help.

Thanks again.

Answer 1

for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n? and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction? like i said this could be total rubbish, but just a thought! :)

Answer 2

Here is my solutions. FYI I can do only half d). Will maybe continue working on it someday later. Hope it helps.

a) use parametric form of 3D-line, and the coefficients of the parameter 't' will be your direction vector: i + 2k. Then simply work out it's +/-ve two unit vectors by dividing its modulus: sqrt(1^2 + 2^2), and eventually I got: +-(i + 2K)/sqrt5

b) same.

c) same.

d) However you don't know what the two parts are, you should be able to find the unit vector perpendicular/normal/orthogonal to a plane. I don't know how to show which vector is pointing into where the origin is, but I can show you from which two you can choose, i.e. the unit vectors pair:

The normal vector to a plane is simply of the coefficients of the plane equation. In this case, our plane is given: 2x + y - z = 1 (Don't worry about the 1 at last, it can be any constant), → normal vectors pair: +-(2, 1, -1), or say: n = +-(2i + j - k).

Therefore the unit vector of n: n_hat = n/|n| = +-(2i + j - k)/sqrt(2^2 + 1^2 + (-1)^2) I.e. the unit vectors pair n_hat = +-(2i + j - k)/sqrt6

Then you may wish to find someone else to teach you which one is pointing in where the origin part is, either the +ve n_hat or the -ve.

e) I use your P and Q, which I agree with you: QP.n = (t, 0, 2t+1).(2, 1, -1) = -1 where |n| = sqrt6 therefore distance d = |QP.n|/|n| = 1/sqrt6