Question

Let $\overset{ \rightarrow }{q}$ and $\overset{ \rightarrow }{r}$ be non-collinear vectors. If $\overset{ \rightarrow }{p}$ is a vector such that $\overset{ \rightarrow }{p}.\left(\overset{ \rightarrow }{q} + \overset{ \rightarrow }{r}\right)=4$ and $\overset{ \rightarrow }{p}\times \left(\overset{ \rightarrow }{q} \times \overset{ \rightarrow }{r}\right)=\left(x^{2} - 2 x + 6\right)\overset{ \rightarrow }{q}+\left(sin y\right)\overset{ \rightarrow }{r}$ , then $\left(x , y\right)$ lies on the line

• A. $x+y=0$
• B. $x-y=0$
• C. $x=1$
• D. $y=\pi $

Answer 1

Answer: (C)

$\left(\overset{ \rightarrow }{p} . \overset{ \rightarrow }{r}\right)\overset{ \rightarrow }{q}-\left(\overset{ \rightarrow }{p} . \overset{ \rightarrow }{q}\right)\overset{ \rightarrow }{r}=\left(x^{2} - 2 x + 6\right)\overset{ \rightarrow }{q}+sin y\left(\overset{ \rightarrow }{r}\right)$
on comparison, we get,
$\overset{ \rightarrow }{p}.\overset{ \rightarrow }{r}=\left(x^{2} - 2 x + 6\right)$ and
$-\overset{ \rightarrow }{p}.\overset{ \rightarrow }{q}=sin y$
$\because \overset{ \rightarrow }{p}.\overset{ \rightarrow }{q}+\overset{ \rightarrow }{p}.\overset{ \rightarrow }{r}=4$ (given)
$\Rightarrow \left(x^{2} - 2 x + 6\right)-sin y=4$
$\Rightarrow \left(x - 1\right)^{2}+1=sin y$
$\because \left(x - 1\right)^{2}+1\geq 1,sin y\leq 1$
$\therefore $ Both sides are equal for $x=1,sin y=1$