The correct option is D π2
Given, 3p+2q=i+j+k ..... (i)
and 3p−2q=i−j−k .... (ii)
On adding Eqs. (i) and (ii), we get
6p=2i
Therefore, p=i3
On subtracting Eq. (ii) from Eq. (i), we get
4q−2(j+k)⇒q=12(j+k)
Let θ be the angle between p and q, then
p⋅q=|p||q|cosθ
⇒cosθ=p⋅q|p||q|
⇒cosθ=16i⋅(j+k)|p||q|=0=cos90∘
Thus θ=90∘=π2