Vector geometry question?

Question

Find the vector u = [a,b,c]^T such that:
(i) u is perpendicular to both the vectors [-6,-2,-6 ]^T and [22,-18,-16 ]^T;

(ii) length of vector u = sqrt{14};

(iii) a is greater or equal to 0

Anyone know how to do this question , i cant figure it out.

Answer 1

Find the vector u = [a,b,c]^T such that:

(i) u is perpendicular to both the vectors [-6,-2,-6 ]^T and [22,-18,-16 ]^T;

The cross product of the two vectors will yield a vector that is perpendicular to both of them.

u = <-6, -2, -6> X <22, -18, -16> = <-76, -228, 152>

(ii) length of vector u = √14;

As long as the vector has the same direction we can change the magnitude of it and it will still be perpendicular to both original vectors. Let's calculate the magnitude of the vector u.

|| u || = √[(-76)² + (-228)² + 152²) = √(5776 + 51984 + 23104)
|| u || = √80864 = 76√14

Divide u by 76 to give it magnitude √14.

u = <-76/76, -228/76, 152/76> = <-1, 3, 2>

(iii) a is greater or equal to 0

u = <-1, 3, 2>
We can reverse the direction by multiplying by a negative number and it will still be perpendicular to the two given vectors. Multiply by -1.

u = <1, -3, -2> = [1, -3, -2]^T

The vector u now satisfies all conditions.

Answer 2

Let * stand for dot product, and d = [-6,-2,-6] and f = [22,-18,-16]. Then u*d = 0 and u*f = 0, which is equivalent to perpendicularity.

So -6a - 2b - 6c = 0 and 22a - 18b - 16c = 0. Since length of u is sqrt(14), u*u = 14, or a^2 + b^2 + c^2 = 14. You now have three equations in three unknowns, plus the fact a >= 0, so you should be able to solve it. Let me know if you need more help.

Steve

Answer 3

a vector that is perpendicular to two other vectors is the cross product of those two vectors. that will give you a unit vector in the direction you require. multiply that unit vector by a sqrt(14) to get the required length. if a is negative in the cross product, multiply every term by -1 to reverse the direction. it will still satisfy all conditions.