Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non-coplanar vectors. If $m$ and $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$,then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$

  • A
    $\vec{a}-\vec{d}$
  • B
    $\vec{a}+\vec{d}$
  • C
    $\vec{0}$
  • D
    $\vec{b}+\vec{c}+2 \vec{d}$

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The position vectors of the vertices of a quadrilateral $ABCD$ are $a, b, c$ and $d$ respectively. The area of the quadrilateral formed by joining the midpoints of its sides is

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Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$,$\vec{b}+6\vec{c}$ is collinear with $\vec{a}$,and $\vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0}$,then $\alpha+\beta$ is equal to

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors in the $xyz$-space such that $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \neq 0$. If $A, B, C$ are points with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively,then the number of possible positions of the centroid of $\triangle ABC$ is

Between the following two statements :
Statement $-I$ : Let $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r}=0$ is of magnitude $\sqrt{10}$.
Statement $-II$ : In a triangle $ABC$,$\cos 2A+\cos 2B+\cos 2C \geq -\frac{3}{2}$.

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 $(s)$ $\pi$
$(t)$ $\frac{\pi}{2}$

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