For non-zero vectors →p,→q and →r, let (→p×→q)×→r+(→q⋅→r)→q=(x2+y2)→q+(14−4x−6y)→p and (→r⋅→r)→p=→r where →p and →q are non-collinear, and x and y are scalars. Then the value of (x+y) is

For non-zero vectors →p,→q and →r, let (→p×→q)×→r+(→q�

1.	For non-zero vectors →p,→q and →r, let (→p×→q)×→r+(→q⋅→r)→q=(x2+y2)→q+(14−4x−6y)→p and (→r⋅→r)→p=→r where →p and →q are non-collinear, and x and y are scalars. Then the value of (x+y) is
Answer» For non-zero vectors $\to p, \to q$ and $\to r,$ let $(\to p \times \to q) \times \to r + (\to q \cdot \to r) \to q = (x^{2} + y^{2}) \to q + (14 - 4 x - 6 y) \to p$ and $(\to r \cdot \to r) \to p = \to r$ where $\to p$ and $\to q$ are non-collinear, and $x$ and $y$ are scalars. Then the value of $(x + y)$ is