Find a vector of magnitude 7 which is perpendicular to both 2i + j - k and i - 2j + 7k .

In three dimensional space there are two answers because if you have one answer than multiplying it by -1 gives you second.

Are you familar with the vector cross product?

I assume not. The vector cross product is an operation denoted ^ (sometimes x). It has the useful property that if a and b are vectors then a ^b is a vector perpendicular to a and b .

The cross product satisfies the following relations:

a ^(b +c ) = a ^b + a ^c
a ^b = -b ^c

Also if i , j , k are a set of right handed orthonormal basis vectors (i.e. the standard unit vectors):

i ^j = k
(derive similar relations for j ^i , j ^k etc)

Now you know about the properties of this product can you get an answer?

Anonymous has given enough information for you to work out how to calculate vector products.
However, for those who would like a short-cut to the method, have a look at the entry in thesaurus.maths.org .

Once you've found a vector perpendicular to the two given, find its magnitude, and then you should be able to work out what to multiply by to get the right magnitude.

It is possible to show that:

(ai+bj+ck).(di+ej+fk) = a d+b e+c f

So the solution to this problem is given by saying the vector you're after is r=xi+yj+zk, and saying r.a=0, r.b=0, and |r|=7, where a and b are the given vectors, then solve simultaneously for x, y, z.

Slightly differently, you can use the two dot products to find two of x, y, z in terms of the third. This tells you the ratio of the three components of your vector (in this case -1:3:1), and hence a vector in the right direction. Then find its magnitude as above, and multiply by something appropriate.

For those wishing to check that they've got the idea, the answer to this question is ±(7/sqrt(11)(-i +3j +k ).