Question

which vector is parallel to i^+j^+k^

Answer 1

Explanation:

The vector (i + j + k) is (1, 1, 1) which has a magnitude R of

R = sqrt(1^2 + 1^2 + 1^2) = sqrt(3) = 1.732

so the unit vector u parallel to (1, 1, 1) is found by dividing this magnitude by 1.732

u = (1, 1, 1)/1.732 = (1/1.732, 1/1.732, 1/1.732) = (0.577, 0.577, 0.577)