Let
$\begin{gathered}\lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right. \\ \left.+\cdots \cdots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right) \text { be } \frac{\pi}{k}\end{gathered}$
using only...

Let $\overrightarrow{\mathrm{a}}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\imath}+4 \hat{\jmath}-5 \hat{\mathrm{k}}$, and a vector $\vec{c}$ be such that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{\imath}+8 \hat{\jmath}+13 \hat{k}$ If $...

Let three vectors $\vec{a}=\alpha \hat{\imath}+4 \hat{\jmath}+2 \hat{k}, \vec{b}=5 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}, \vec{c}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$ from a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of...

The shortest distance between the line $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is :

Let $\vec{a}=9 \hat{\imath}-13 \hat{\jmath}+25 k, b=3 \hat{\imath}+7 \hat{\jmath}-13 k$ and $\overrightarrow{\mathrm{c}}=17 \hat{\imath}-2 \hat{\jmath}+\mathrm{k}$ be three given vectros. If $\overrightarrow{\mathrm{r}}$ is a vector such that $...

Let $\vec{a}=\hat{i}-3 \hat{j}+7 k \hat{k}=2 \hat{i}-\hat{j}+k$ and $\vec{c}$ be a vector such that $(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$. If $\vec{a} \cdot \vec{c}=130$, then $...

Let ABC be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{\mathrm{AB}}=\hat{\imath}+2 \hat{\jmath}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=a \hat{\imath}+b \hat{\jmath}+c \hat{k}$ and $...

Let $\overrightarrow{\mathrm{OA}}=2 \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{~b}}$ and $\overrightarrow{\mathrm{OC}}=3 \overrightarrow{\mathrm{~b}}$, where O is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and...

If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$ and $\int_0^{\mathrm{k}}\left[\mathrm{x}^2\right] \mathrm{dx}=\alpha- \sqrt{\alpha}$, where $[\mathrm{x}]$ denotes the greatest integer function,...

If the line (2-x)/3=(3y-2)/(4λ+1)=4-z makes a right angle with the line (x+3)/3μ=(1-2y)/6=(5-z)/7, then 4λ+9μ is equal to :