Need help on vector analysis :( Guys, I need ur help please... Can u help me to answer these problems? I'm very confused... 1. Show that the vector Ai + Bj + Ck is normal to the plane which equation is Ax + By + Cz = D, where A, B, C, D are constants 2. n = 0.5i + 0.5j + 0.7071k is the unit-normal for plane A. b = 4i + 5j + 2k, c = 2i + 3j + k. Calculate the area of parallelogram project from b x c to plane A. Calculate components of vectors b and c that are parallel to plane A. 3. New right hand coordinate axes are chosen at the same origin with e1' = (2e1 + 2e2 + e3)/3 and e2' = (e1 - e2) x 1.4. Express e3' in term of e1. If t = 10e1 + 10e2 - 20e3, express t in terms of the new basis ek'. Express the old coordinate xi in term of xk' , xi = f(xk') Please help me guys... :( Thanks a lot...
What are the relevant equations? How do you define "normal"? What is a projection? How do you define "parallel"?
Let [itex]\boldsymbol x_0 = (x_0, y_0, z_0)[/itex] be some specific point on the plane. Let [itex]\boldsymbol x = (x, y, z)[/itex] be any point on the plane. Can you write equations that describe the points [itex]\boldsymbol x_0[/itex] and [itex]\boldsymbol x[/itex]? What is the inner product of the vector from [itex]\boldsymbol x_0[/itex] to [itex]\boldsymbol x[/itex] with the vector [itex]\boldsymbol n = (A,B,C)[/itex]?
No. Where did you get the division? Yes, but this is what you need to prove. Expanding on my previous post: Because both [itex]\boldsymbol x_0[/itex] and [itex]\boldsymbol x[/itex] are on the plane, [itex]Ax_0 +By_0+Cz_0 = D[/itex] and [itex]Ax +By+Cz = D[/itex]. (1) What is the difference between these equations? (2) What is the vector from [itex]\boldsymbol x_0[/itex] to [itex]\boldsymbol x[/itex]? You are given that [itex]\boldsymbol n = (A,B,C) [/itex]. (3) What is the inner product between [itex]\boldsymbol n[/itex] and the vector from [itex]\boldsymbol x_0[/itex] to [itex]\boldsymbol x[/itex]? Don't guess. Use the answer to question 2. Finally, relate the answers to questions 1 and 3.
The vector from x0 to x is (x-x0) right? So should I do the inner product between (x-x0) . (n)? Then I won't get any number, just some equation in x n x0?
Yes. Do that. The reason for doing this is that the vector [itex]\boldsymbol x - \boldsymbol x_0[/itex] represents any arbitrary vector on the plane. If a vector is normal to every vector on some plane the vector is normal to the plane itself. Please use English, not TXT-speak. Is it really that much harder to type x and x0?
Hi, can anyone help me with the other question please...? I am so depressed... :( Need some hints... Thanks...
I calculated the cross product of b x c, buat I don't know what is the meaning of "projection to plane A" But for number 3, I can't figure out the meaning. Can you please give a hint? Thanks...
It tells you what to do in the next sentence: "Calculate components of vectors b and c that are parallel to plane A". Find the projection of b x c on the normal vector you found in 1, subtract it from b x c to find the part parallel to the plane. Are these new coordinate axes still orthogonal to one another? Check the inner product of e1' and e2' that you are given to make sure. Assuming they are orthogonal, then e3' must be orthogonal to both. How do you find a vector orthogonal to two given vectors? Of course, you have to be careful of the sign- this has to be a right hand coordinate system.
for number 3, i've found the e'3 vector which is 1/3 V2 e1 + 1/3 V2 e2 - 4/3 V2 e3, here V means the square root. But I still don't know how to make it to ei part...?
I'm sorry, the question should be like this: New right hand coordinate axes are chosen at the same origin with e1' = (2e1 + 2e2 + e3)/3 and e2' = (e1 - e2) x 1.4. Express e3' in term of ei (not e1, sorry...). If t = 10e1 + 10e2 - 20e3, express t in terms of the new basis ek'. Express the old coordinate xi in term of xk' , xi = f(xk')