Direction angles of a vector, equation of a plane
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Direction angles of a vector, equation of a plane

[From: ] [author: ] [Date: 11-05-11] [Hit: ]
Letting * represent the dot product,y is the angle with respect to k,But then I followed that procedure to find a and b (angles with respect to i and j) and got 1/sqrt(3) for both of them as well, which doesnt make sense. What am I doing wrong?d) find the equation of the plane containing O,......
Consider the unit cube in the first octant with a vertex at the origin. P is the point (1, 1, 1) and Q is at the centre of the face of the cube on the plane x = 1.

A) write the vectors OP and OQ in terms of the unit vectors i, j, and k.
For this I got OP = i + j + k and OQ = i + 1/2 j + 1/2 k.

b) find the direction cosines of OP.
This is where I'm having trouble. Letting * represent the dot product, I was doing this:
y is the angle with respect to k,
cos y = (OP * k) / |OP||k|
I got cos y = 1/sqrt(3)

But then I followed that procedure to find a and b (angles with respect to i and j) and got 1/sqrt(3) for both of them as well, which doesn't make sense. What am I doing wrong?

c) find the cosine of the angle between OP and OQ

d) find the equation of the plane containing O, P, and Q


help would be appreciated

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(A) You are correct.
(B) Your answer is correct. The three will all be the same, by symmetry. That is, if you exchange the x and y axes or the x and z axes or the y and z axes, you should get the same angles each time.
(C) Since you could actually do (B) fine, I won't show my work. My answer is sqrt(2).
(D) The equation of a plane is ax + by + cz = d, where (a, b, c) is the vector normal to the plane and d is a constant. Since the plane passes through O, d=0. We know that P and Q are contained on the plane, so the normal vector is just their cross product. Using the usual formula, this is just (0, 1/2, -1/2), so the equation of the plane is 1/2 y - 1/2 z = 0, or, in a prettier form, y = z. You could also have derived this geometrically by "looking at" the situation.
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