Let ` vec pa n d vec q` be any two orthogonal vectors of equal magnitude 4 each. Let ` vec a , vec b ,a n d vec c` be any three vectors of lengths `7sqrt(15)a n d2sqrt(33),` mutually perpendicular to each other. Then find the distance of the vector `( vec adot vec p) vec p+( vec adot vec q) vec q+( vec adot( vec pxx vec q))( vec pxx vec q)+( vec bdot vec p) vec p( vec bdot vec q) vec q+( vec bdot( vec pxx vec q))( vec pxx vec q)+( vec cdot vec p) vec p+( vec cdot vec q) vec q+( vec cdot( vec pxx vec q))( vec pxx vec q)` from the origin.

Options

A. `veca+vecb+vecc`
B. `veca/|veca|+vecb/|vecb|+vecc/|vecc|`
C. `veca/|veca|^(2)+vecb/|vecb|^(2)+vecc/|vecc|^(2)`
D. `|veca|veca-|vecb|vecb+|vecc|vecc`

Answer

b

Explanation

Let `vecalpha= veca/|veca|+vecb/|vecb|+vecc/|vecc|`
since, `veca , vecb and vecc` ar mutually perpendicular vectors, if `vecalpha` makes angles `theta, phi,psi" with " veca,vecb and vecc,` respectively, then
`vecalpha.veca= (veca.veca)/|veca|`
`or |vecalpha|.|veca|costheta = |veca|`
`or cos theta= 1/|vecalpha|`
similarly, `cos theta = 1/|vecalpha| . cos psi = 1/|vecalpha|`
`theta= phi = psi`