b and c are non collinear vectors. a is a vector such that a .(b+c )=4 and a×(b×c)=(x2-2x+6)b+(sin y)c and 0<y≤π. Then the point (x, y) is

a.b+a.c=4a.cb-a.bc=x2-2x+6b+(sin y>c)a.c=x2-2x+6=sin y+4x2-2x+2=sin y⇒x-12+1=sin yx=1, y=π2

b and c are non collinear vectors. a is a vector such that a .(b+c )=4 and a×(b×c)=(x2-2x+6)b+(sin y)c and 0<y≤π. Then the point (x, y) is

a.b+a.c=4a.cb-a.bc=x2-2x+6b+(sin y>c)a.c=x2-2x+6=sin y+4x2-2x+2=sin y⇒x-12+1=sin yx=1, y=π2

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$b$ and $c$ are non collinear vectors. $a$ is a vector such that $a . (b + c) = 4$ and $a \times (b \times c) = (x^{2} - 2 x + 6) b + (\sin y) c$ and $0 < y \leq π .$ Then the point (x, y) is

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$(1, π)$

$(1, \frac{π}{2})$

$(\frac{π}{2}, 1)$

(1, 1)

answer is B.

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Detailed Solution

$\begin{array}{rcl} a . b + a . c & = & 4 \\ (a . c) b - (a . b) c & = & (x^{2} - 2 x + 6) b + (\sin y > c) \\ a . c & = & x^{2} - 2 x + 6 = \sin y + 4 \\ x^{2} - 2 x + 2 & = & \sin y \Rightarrow {(x - 1)}^{2} + 1 = \sin y \\ x & = & 1, y = \frac{π}{2} \end{array}$