Find vector whose magnitude = 3 and whose component in the i direction is = to the component in the j directio

Answer 1

Well, let v = ai + bj.
From the Pythagorean theorem, ||v||^2 = 3^2 = a^2 + b^2
a = b, so 3^2 = 2a^2 = 9
a = b = + or - sqrt(9/2).
So, there are four vectors that fit the description. Just put in sqrt(4.5) for a and b, and give every possible combination of signs between the i and j components.

Answer 2

Generalising the vector in 3 dimensions,we can have the vector as
A=xi+xj+zk(since components in x and y directions are same)

hence /A/=3;i.e,2x^2+z^2=9;clearly x=2 and z=1;

So the vector is A=2i+2j+k

Answer 3

Mag = sqrt(i^2 +j^2)
now if i = j
Mag = sqrt(i^2 + i^2) = sqrt(2*i^2)
So solve for i
i = Mag/sqrt(2) = 3/sqrt(2)
So the vector is
= <3/sqrt(2),3/sqrt(2)>