DavidLiew said:Finally I find it is equal to [tex]\sqrt{212}[/tex]. Am I right?
The formula for finding ||2w-v|| is ||w-v|| = ||w|| - 2||v||cos(theta), where theta is the angle between vector w and vector v.
To find the value of
No, the Pythagorean theorem only applies to right triangles, while ||2w-v|| is the magnitude of a vector.
The units for ||2w-v|| are the same as the units for ||v|| and ||w||, since ||2w-v|| is a scalar quantity. In this case, the units would be the same as the units for ||v|| and ||w||, such as meters or seconds.
No, ||2w-v|| is not equal to ||v-w||. The order of subtraction matters when finding the magnitude of a vector, so ||2w-v|| is not equivalent to ||v-w||.