Question

Component of 3i + 4j perpendicular to i + j and in the same plane as that of 3i + 4j is

Answer 1

First of all we have to find ,
Angle between given vectors A = (3i + 4j) and B = (i + j)
|A| = √{3² + 4²} = 5 and |B| = √{1² + 1²} = √2
cosФ = A.B/|A||B|
= (3i + 4j).(i + j)/5.√2
= 7/5√2
= 7/5√2

We know, component of A perpendicular to B = |A|sinФ.n
Where n is unit vector of perpendicular to B
So, sinФ = √{1 - cos²Ф} = √{1 - {7/5√2}²}
= √{1 - 49/50} = 1/5√2

∵ unit vector along B = B/|B| = 1/√2(i + j)
Now, unit vector perpendicular upon B = ± 1/√2(i - j) [you can easily assume perpendicular vector of two dimension when x and y components are equal in magnitude . Like a vector A = xi + xj then perpendicular vector of it = ±(xi - xj)]

Now, component of A perpendicular to B = 5 × 1/5√2 × ±1/√2(i - j)
= ±1/2(i - j)