System of linear and non-linear equations

AI Thread Summary
To find a vector with a magnitude of 5 that is perpendicular to the vectors 3i-2j+4k and 4i-3j-k, three equations are established: one for the magnitude and two from the dot product of the given vectors. The first equation states that x^2 + y^2 + z^2 = 25, while the other two equations arise from the conditions of perpendicularity. A suggested approach to solve these equations is to express z in terms of x and y using the third equation, then substitute it into the second equation to find y in terms of x. Utilizing the cross product method is also recommended as a simpler alternative to solve for the desired vector.
abdo799
Messages
168
Reaction score
4

Homework Statement



find the vector with magnitude 5 and perpendicular to 3i-2j+4k and 4i-3j-k

The Attempt at a Solution


lets name the vector component x for the i y for the j and z for the k
i got three equation
1- x^2+y^2+z^2=25 ( magnitude)
2-3x-2y+4z=0
3-4x-3y-z=0 (2, 3 from dot product by the 2 given vectors)
i am supposed to solve those simultaneously, but i can't do it
 
Physics news on Phys.org
abdo799 said:

Homework Statement



find the vector with magnitude 5 and perpendicular to 3i-2j+4k and 4i-3j-k

The Attempt at a Solution


lets name the vector component x for the i y for the j and z for the k
i got three equation
1- x^2+y^2+z^2=25 ( magnitude)
2-3x-2y+4z=0
3-4x-3y-z=0 (2, 3 from dot product by the 2 given vectors)
i am supposed to solve those simultaneously, but i can't do it

You could if you tried. Use equation 3 to solve for z in terms of x and y. Substitute that into equation 2 and solve for y in terms of x. Put it all into equation 1 and solve for x. But it would be a lot easier to use the cross product, if you know what that is.
 
  • Like
Likes abdo799
I tried it and it worked, thanks
 

Similar threads

Back
Top