Find the vector perpendicular to two others...?

Question

Find a vector of modulus 6 and perpendicular to vectors u = i - 2k and v = 2i + j+ k

Answer 1

u X v is given by :-

i____j____k

1___0___-2

2___1___1

2i - 5 j + k

Unit vector = (1 / √30)(2 i - 5 j + k)

Requred vector = (6 / √30) (2 i - 5 j + k )

Answer 2

The cross product of u and v will be perpendicular to both. And you can make the norm 6 by dividing it by the appropriate constant.

Answer 3

call the vector x*i +y*j+z*k=w

The dot product u*w and v*w must be zero

so

x-2z=0

2x+y+z=0

Also sqrt(x^2+y^2+z^2) =6

s ox=2z

and y = -5z so 4z^2+25z^2+z^2=36  

z^2= 1.2

you can do the calculations

Answer 4

You need a cross product. It produces a vector that is perpendicular to both vectors and has a magnitude equal to the volume of the parallelogram formed by the two vectors. It's defined like this:

u = (x1,y1,z1)

v = (x2,y2,z2)

u x v = (y1 * z2 - z1 * y2, z1 * x2 - x1 * z2, x1 * y2 - y1 * x2)

which is a horrible definition. It doesn't even seem to follow much of a pattern. However, if you know about determinants of 3x3 matrices, the cross product can be defined like this:

u = (x1,y1,z1)

v = (x2,y2,z2)

.... ( .. i .. j .. k .. )

det( . x1 y1  z1 . )

.... ( . x2 y2 z2 . )

That will produce the same vector as the one defined above, but if you don't know about determinants yet, don't worry about. You'll just have to rote-learn the above definition. :(