If $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar vectors and $\overline{p}=\frac{\overline{b} \times \overline{c}}{[\overline{a} \overline{b} \overline{c}]}, \overline{q}=\frac{\overline{c} \times \overline{a}}{[\overline{a} \overline{b} \overline{c}]}, \overline{r}=\frac{\overline{a} \times \overline{b}}{[\overline{a} \overline{b} \overline{c}]}, \quad$ then $2 \overline{a} \cdot \overline{p}+\overline{b} \cdot \overline{q}+\overline{c} \cdot \overline{r}=$

  • A
    $0$
  • B
    $3$
  • C
    $4$
  • D
    $1$

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$A, B, C, D$ are four points in a plane with position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ respectively such that $(\overline{a}-\overline{d}) \cdot(\overline{b}-\overline{c})=(\overline{b}-\overline{d}) \cdot(\overline{c}-\overline{a})=0$. Then the point $D$ is the $\dots$ of $\triangle ABC$.

If $\vec{a}=\frac{3}{2} \hat{k}$ and $\vec{b}=\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{2}$,then the angle between $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is: (in $^{\circ}$)

Assertion $(A)$: $a, b, c, d$ are position vectors of $4$ points such that $2a - 3b + 7c - 6d = 0 \Rightarrow a, b, c, d$ are coplanar.
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Which of the following is true?

The vectors $3 \vec{a}-5 \vec{b}$ and $2 \vec{a}+\vec{b}$ are mutually perpendicular and the vectors $\vec{a}+4 \vec{b}$ and $-\vec{a}+\vec{b}$ are also mutually perpendicular. Then the angle between the vectors $\vec{a}$ and $\vec{b}$ is:

If $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$,$|\vec{b}| = 5$,and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then the area of the triangle formed by these two vectors as two sides is

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