MCQ
The vector equation of $X Y$-plane is
- A$\overrightarrow{ r } \cdot \hat{ k }=0$
- B$\vec{r} \cdot \hat{j}=0$
- C$\overrightarrow{ r } \cdot \hat{ i }=0$
- D$\overrightarrow{ r } \cdot \vec{n}=1$
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| 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 $\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right|$ for $|z|=1, \quad z \neq 1$ is given by | $(s)$ $\pi$ |
| $(t)$ $\frac{\pi}{2}$ |