Let $$P(x,y,z)$$ be a point in the first octant, whose projection in the $$xy$$-plane is the point $$Q$$. Let $$OP=\gamma$$; the angle between $$OQ$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$OP$$ and the positive $$z$$-axis be $$\varphi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is

If the shortest distance between the lines $$\begin{array}{ll}{L}_1:\vec{r}=(2+\lambda )\hat{i}+(1-3\lambda )\hat{j}+(3+4\lambda )\hat{k},& \lambda \in \mathbb{R}\\ L_2:\vec{r}=2(1+\mu )\hat{i}+3(1+\mu )\hat{j}+(5+\mu )\hat{k},& \mu \in \mathbb{R}\end{array}$$ is $$\frac{m}{\sqrt{n}}$$, where $$\gcd (m,n)=1$$, then the value of $$m+n$$ equals

If the line $$\frac{2-x}{3}=\frac{3y-2}{4\lambda +1}=4-z$$ makes a right angle with the line $$\frac{x+3}{3\mu }=\frac{1-2y}{6}=\frac{5-z}{7}$$, then $$4\lambda +9\mu$$ is equal to :

Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $${\mathrm{d}}^2+6$$ is equal to :

Let $$(\alpha ,\beta ,\gamma )$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha +\beta +\gamma$$ is equal to :

Let $$\mathrm{P}(\alpha ,\beta ,\gamma )$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $${\lambda }^2=14\mathrm{K}$$, then $$\mathrm{K}$$ is equal to :

If $$A(3,1,-1),B\left(\frac{5}{3},\frac{7}{3},\frac{1}{3}\right),C(2,2,1)$$ and $$D\left(\frac{10}{3},\frac{2}{3},\frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$ABCD$$, then its area is

The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is

Consider 10 observations $x_1,x_2,\dots ,x_{10}$ such that $\sum _{i=1}^{10}\left(x_i-\alpha \right)=2$ and $\sum _{i=1}^{10}{\left(x_i-\beta \right)}^2=40$, where $\alpha ,\beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{\beta }{\alpha }$ is equal to :

Let the median and the mean deviation about the median of 7 observation $170,125,230,190,210$, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is :

Let ${\mathrm{a}}_1,{\mathrm{a}}_2,\dots {\mathrm{a}}_{10}$ be 10 observations such that $\sum _{\mathrm{k}=1}^{10}{\mathrm{a}}_{\mathrm{k}}=50$ and $\sum _{\forall \mathrm{k}<\mathrm{j}}{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{a}}_{\mathrm{j}}=1100$. Then the standard deviation of ${\mathrm{a}}_1,{\mathrm{a}}_2,\dots ,{\mathrm{a}}_{10}$ is equal to :

Let the mean and the variance of 6 observations $$a,b,68,44,48,60$$ be $$55$$ and $$194$$, respectively. If $$a>b$$, then $$a+3b$$ is

If the mean and variance of five observations are $$\frac{24}{5}$$ and $$\frac{194}{25}$$ respectively and the mean of the first four observations is $$\frac{7}{2}$$, then the variance of the first four observations in equal to

Let M denote the median of the following frequency distribution .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Frequency 3 9 10 8 6 Then 20M is equal to :

If the variance of the frequency distribution .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} $$x$$ $$c$$ $$2c$$ $$3c$$ $$4c$$ $$5c$$ $$6c$$ $$f$$ 2 1 1 1 1 1 is 160, then the value of $$c\in N$$ is

The frequency distribution of the age of students in a class of 40 students is given below. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-1wig{font-weight:bold;text-align:left;vertical-align:top} .tg .tg-baqh{text-align:center;vertical-align:top} Age 15 16 17 18 19 20 No of Students 5 8 5 12 $$x$$ $$y$$ If the mean deviation about the median is 1.25, then $$4x+5y$$ is equal to :

Let $$\alpha ,\beta \in \mathbf{R}$$. Let the mean and the variance of 6 observations $$-3,4,7,-6,\alpha ,\beta$$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then ${\mathrm{S}}_{15}-{\mathrm{S}}_5$ is equal to :

Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :