Try assuming an arbitrary k vector: ai + bj + ck (or whatever). Then see what happens.Originally posted by jlmac2001
... But I don't know what k is. What's k?
I got 2k (where k is the constant vector) but I did it quick. (So my answer differs by a minus sign from yours.) My main point was that you didn't have to know the value of k (the a, b, & c) to do the problem.Originally posted by jlmac2001
The answer I got is: -2ai-2bj-2ck
By "2k" I mean: 2ai + 2bj + 2ck. (The same as you, but with a + instead of -.) [itex]2\vec{k}[/itex] is what I should have written. Make sense?Originally posted by jlmac2001
how did you just get 2k? i don't understand
A constant vector is a vector that has a fixed magnitude and direction. It does not change in value or direction, regardless of the coordinate system it is being referenced in.
A constant vector is represented by a boldface letter with an arrow on top, such as v or F. It can also be represented by its components, (a,b), where a represents the horizontal component and b represents the vertical component.
A cross product is used to find a vector that is perpendicular to two given vectors. It is also used to calculate the area of a parallelogram formed by two vectors.
A cross product is calculated by taking the determinant of a 3x3 matrix formed by the two given vectors and the unit vectors i, j, and k. The resulting vector is perpendicular to the given vectors and has a magnitude equal to the area of the parallelogram formed by the two vectors.
Yes, a constant vector and cross product can both have negative values. The direction of a vector can be negative if it points in the opposite direction of a given reference point. The magnitude of a cross product can also be negative if the two given vectors form a clockwise rotation, rather than a counterclockwise rotation, as seen from a reference point.