The minimum value of $(x_1 - x_2)^2 + (\sqrt{2 - x_1^2} - \frac{9}{x_2})^2$ where $x_1 \in (0, \sqrt{2})$ and $x_2 \in R^+$.

  • A
    $8$
  • B
    $6$
  • C
    $4$
  • D
    $2$

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Let $O$ be the origin and $\overline{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\overline{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\overline{OC} = \frac{1}{2}(\overline{OB} - \lambda\overline{OA})$ for some $\lambda > 0$. If $|\overline{OB} \times \overline{OC}| = \frac{9}{2}$,then which of the following statements is (are) $TRUE$?
$(A)$ Projection of $\overline{OC}$ on $\overline{OA}$ is $-\frac{3}{2}$
$(B)$ Area of the triangle $OAB$ is $\frac{9}{2}$
$(C)$ Area of the triangle $ABC$ is $\frac{9}{2}$
$(D)$ The acute angle between the diagonals of the parallelogram with adjacent sides $\overline{OA}$ and $\overline{OC}$ is $\frac{\pi}{3}$

Match the statements given in Column $I$ with the values given in Column $II$.
Column $I$ Column $II$
$(A)$ If $\vec{a}=\hat{j}+\sqrt{3} \hat{k}, \vec{b}=-\hat{j}+\sqrt{3} \hat{k}$ and $\vec{c}=2 \sqrt{3} \hat{k}$ form a triangle,then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is $(p)$ $\frac{\pi}{6}$
$(B)$ If $\int_a^b(f(x)-3 x) d x=a^2-b^2$,then the value of $f\left(\frac{\pi}{6}\right)$ is $(q)$ $\frac{2 \pi}{3}$
$(C)$ The value of $\frac{\pi^2}{\ln 3} \int_{1 / 6}^{5 / 6} \sec (\pi x) d x$ is $(r)$ $\frac{\pi}{3}$
$(D)$ The maximum value of $|\operatorname{Arg}(\frac{1}{1-z})|$ for $|z|=1, z \neq 1$ is given by $(s)$ $\pi$
$(t)$ $\frac{\pi}{2}$

The vector that is parallel to the vector $2 \hat{i} - 2 \hat{j} - 4 \hat{k}$ and coplanar with the vectors $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ is

Let $A$ and $B$ be two points. The position vector of $A$ is $6b - 2a$. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,what is the position vector of $B$?

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$, then $k$ is equal to

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