Calculate Vector with Magnitude & Direction Given

  • Thread starter Thread starter jimbo_durham
  • Start date Start date
  • Tags Tags
    Magnitude Vector
AI Thread Summary
To calculate a vector with a known magnitude and direction in the xy-plane, first determine the unit vector in the desired direction by dividing the vector components by its magnitude. For example, starting from the point (10,10,0) and aiming towards the origin (0,0,0) with a magnitude of 4, the direction vector is (-10,-10,0). After finding the unit vector, multiply it by the desired magnitude to obtain the final vector components. This method applies similarly to other points, such as (10,9,0), ensuring the direction is correctly accounted for. The approach effectively provides the necessary vector components for any specified direction and magnitude.
jimbo_durham
Messages
13
Reaction score
0
Hi, i have a known magnitude to give my vector in an xy plane, and i have a desired direction. I need the (vx, vy, vz=0) to describe my vector. I am sure this can be done easily.

an example is,

i have a point at (10,10,0) in cartesian (x,y,z) and will use this as the starting point of my vector. This vector must have a magnitude of 4, and must be in a direction along the line connecting the point (10,10,0) with the origin (0,0,0).

ie my vector has the magnitude 5 and direciton that of a vector (-10,-10,0).

in this example the solution is simply to write

<br /> z^{2}=x^{2}+y^{2}, and as x=y,

<br /> \sqrt{ \frac{z^{2}}{2} }=x=3.5<br /> ish.

giving me vector with components (3.5,3.5,0)

this however is in the wrong direction (need (-3.5,-3.5,0))

however if x=/=y, how is this solved? and how is the direction accounted for (+ve or -ve)?
 
Last edited:
Mathematics news on Phys.org
for example a point at (10,9,0) with magnitude 5 along the vector from the point to the origin as before, which is now (-10,-9,0).

how do i find its components (vx, vy, vz=0)?
 
Find the unit vector first

First, find the unit vector in your desired direction; then multiply by your desired magnitude.

The vector (-10, -10, 0) lies on your direction; Now find the unit vector in that direction. i.e. divide by the magnitude. Then multiply by 4.
 
excelent, works like a dream. thanks sennyk
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Back
Top