Let b and c be non-collinear vector satisfying a ( b c ) + ( a. b… - Vidyadip

MCQ

Let $\vec b$ and $\vec c$ be non-collinear vector satisfying $\vec a \times \left( {\vec b \times \vec c} \right) + \left( {\vec a.\vec b} \right)\vec b = \left( {4 - 2x - \sin y} \right)\vec b + \left( {{x^2} - 1} \right)\vec c$ and $\left( {\vec c.\vec c} \right)\vec a = \vec c$ , then $x$ is equal to

✓
$1$
B
$2$
C
$4$
D
$6$

✓

Answer

Correct option: A.

$1$

a
$(\overrightarrow{\mathrm{a} .}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}+(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{b}}$

$\quad = (4 - 2{\rm{x}} - \sin {\rm{y}})\overrightarrow {\rm{b}} + \left( {{{\rm{x}}^2} - 1} \right)\vec c$

$ \Rightarrow \overrightarrow a \cdot \overrightarrow {\rm{c}} + \overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = 4 - 2{\rm{x}} - \sin {\rm{y}},\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = 1 - {{\rm{x}}^2}$

Also, $\quad(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}) \overline{\mathrm{a}}=\overrightarrow{\mathrm{c}}$

$\Rightarrow \quad (\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}} $

$|\overrightarrow{\mathrm{c}}|^{2} \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|^{2} $

$ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=1$

Now ${1 + \vec a \cdot \vec b = 4 - 2x - \sin y}$

${ \Rightarrow \quad {x^2} - 2x + 1 = \sin y - 1 \le 0}$

${ \Rightarrow \quad x = 1,y = \pi /2}$

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