Download App

STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
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  ${1 + \vec a \cdot \vec b = 4 - 2x - \sin y}$

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

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 10. vector algebra→STD 12 - 10. vector algebra — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Area of the region enclosed between the curves $x = y^2 -1$ and $x = |y|$ $\sqrt {1 - {y^2}} $ is
→
If $ b$  and  $c$  are any two non-collinear unit vectors and $a$  is any vector, then $(a\,.\,b)\,b + (a\,.\,c)\,c + \frac{{a\,.\,(b \times c)}}{{|b \times c|}}\,(b \times c) = $
→
If $\vec p$ and $\vec q$ are unequal unit vectors such that $\left( {\vec p - \vec q} \right) . \left( {\left( {2\vec q + \vec p} \right) \times \left( {3\vec p - \vec q} \right)} \right) = \left| {\vec p + \vec q} \right|$ , then angle between $\vec p$ and $\vec q$ will be
→
Choose the correct answer from the given four options. The value of the expression $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$ is: Hint: $\bigg[\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\bigg]$
→
If $a\,.i\,=a\,.\,(i+j)=a\,.\,(i+j+k)$ , then $a = $
→
In a triangle $ ABC$ , if $2\overrightarrow {AC} = 3\overrightarrow {CB} ,$ then $2\overrightarrow {OA} + 3\overrightarrow {OB} $ equals
→
Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is $K > 0,$ is
→
Let $S = \{1, 2, 3, 4, 5\}$ and let $A = S \times S.$ Define the relation $R$ on $A$ as follows$:(a, b) R (c, d)$ if $ad = cb.$ Then$, R$ is;
→
Which of the following is not a decreasing function on the interval $\left( {0,{\pi \over 2}} \right)$
→
Thirty two persons $X_1, X_2, \ldots, X_{32}$ are randomly seated around a circular table at equal intervals. Two persons $X_i$ and $X_j$ are said to be within earshot of each other if there are at most three persons between them on the minor arc joining $X_i$ and $X_j$. The

probabiliky that $X_1$ and $X_3$ are within earshot of each other is, Here, $\left.{ }^n C_r=\frac{n !}{(n-r) ! r !}\right)$

→