Given the following vector:
a = i + 2j -2k
Find a possible set of values for y and z such that the unit vector shown below is perpendicular to vector a.
b = (y)j + (z)k
I've tried using both of these rules [ (i) |a x b| = |a||b|sin90 (ii)a.b = 0 ]to form an equation but i can't seem to get an answer.
I suspect it has something to do with matrices also, as this is part of a matrices question.
ok. but i'm still not sure of how to go about solving x and y. what i've done is:
|a| = 3
|b| = 1 (since b is unit vector)
a x b = (2z + 2y)i - (z)j + (y)k
|a x b| = square root of (2z + 2y)^2 + z^2 + y^2
since y^2 + z^2 = 1,
|a x b| = square root of 1 + 4(1) + 6zy
Using |a x b| = |a||b|sin90,
square root of 1 + 4(1) + 6zy = 3
y = -z/3
How do i form the other eqn?