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
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.
(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.