urgent
let b and c be non collinear vectors. If a is a vector such that a.(b+c) = 4 and a*(b*c) = (x²-2x+6)b+siny.c , then (x,y) lies on the line
Ans: x=1

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{By} \mathrm{definition} \mathrm{of} \mathrm{vector} \mathrm{triple} \mathrm{product}, \mathrm{we} \mathrm{have}: \\ \overrightarrow{\mathrm{a}}\times \left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)=\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}\right)\overrightarrow{\mathrm{b}}-\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}\right)\overrightarrow{\mathrm{c}} \\ \mathrm{According} \mathrm{to} \mathrm{question}: \\ \left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}\right)\overrightarrow{\mathrm{b}}-\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}\right)\overrightarrow{\mathrm{c}}=\left({\mathrm{x}}^2-2\mathrm{x}+6\right)\overrightarrow{\mathrm{b}}+\left(\mathrm{siny}\right)\overrightarrow{\mathrm{c}} \\ \mathrm{On} \mathrm{comparing} \mathrm{we} \mathrm{get}: \\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}={\mathrm{x}}^2-2\mathrm{x}+6 \\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=-\mathrm{siny} \\ \overrightarrow{\mathrm{a}}.\left(\overrightarrow{\mathrm{b}}+.\overrightarrow{\mathrm{c}}\right)=4 \\ \Rightarrow \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}=4 \\ \Rightarrow -\mathrm{siny}+{\mathrm{x}}^2-2\mathrm{x}+6=4 \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+6-4=\mathrm{siny} \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+2=\mathrm{siny} \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1+1=\mathrm{siny} \\ \Rightarrow {\left(\mathrm{x}-1\right)}^2+1=\mathrm{siny} \\ \mathrm{Minium} \mathrm{value} \mathrm{of} \mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{occurs} \mathrm{when} \mathrm{square} \mathrm{term} \mathrm{is} 0 \mathrm{i}.\mathrm{e}. \mathrm{x}-1=0 \\ \mathrm{i}.\mathrm{e}. \mathrm{x}=1 \mathrm{and} \mathrm{this} \mathrm{minimum} \mathrm{value} \mathrm{will} \mathrm{be} {\left(1-1\right)}^2+1=1 \\ \mathrm{Also} \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{siny} \mathrm{is} 1. \\ \mathrm{Hence} \\ {\left(\mathrm{x}-1\right)}^2+1=\mathrm{siny} \mathrm{is} \mathrm{possible} \\ \mathrm{when} \mathrm{x}=1 \mathrm{and} \mathrm{y}=\frac{\mathrm{\pi }}{2} \\ \left(\mathrm{x},\mathrm{y}\right)=\left(1,\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{There} \mathrm{will} \mathrm{be} \mathrm{infinite} \mathrm{lines} \mathrm{on} \mathrm{which} \left(\mathrm{x},\mathrm{y}\right)=\left(1,\frac{\mathrm{\pi }}{2}\right) \mathrm{will} \mathrm{lie} \mathrm{and} \mathrm{x}=1 \mathrm{is} \mathrm{one} \\ \mathrm{of} \mathrm{them}. \end{gathered}$$

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