Let O be the origin, and M and be the points on the lines and respectively such that is the shortest distance between the given lines. Then is equal to _________. ...

Now Equation (1) and (2) I and II I and III Solve (3) and (4) we get ...

Let O be the origin and M and N be the points on the lines f: (Mathematics)

Let O be the origin, and M and $\mathrm{N}$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $\mathrm{MN}$ is the shortest distance between the given lines. Then $\overrightarrow{O M} \cdot \overrightarrow{O N}$ is equal to _________.

The integral int_0^1 \frac{1}{7^{\left[\frac{1}{x}\right]}} {dx}, where [.] denotes the greatest integer function is equal to

For integers n and r let n r nCr ifn ge r ge 0 0 otherwise

Let A = { 1,3,4,6,9} and B = { 2,4,5,8,10}. Let R be a relation defined on A X B such that R = {( ( a_{1},b_{1} ), ( a_{2},b_{2} )):a_{1} ? b_{2} and b_{1} ? a_{2}}. Then the number of elements in the set R is

The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :

The minimum number of elements that must be added to the relation R = {(a,b),(b,c),(b,d)} on the set { a,b,c,d} so that it is an equivalence relation is

Let Z be the set of integers. If A = { x in Z:2^{(x + 2) ( x^{2} - 5x + 6 )} = 1 } and = { x in Z : - 3 < 2x - 1 < 9}, then the . number of subsets of the set A X B is

The area of the region enclosed by the parabolas y=4x-x^2 and 3y=(x-4)^2 is equal to

Area of the region (x,y):x^2+(y-2)^2≤4,x^2≥2y is

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have atleast one letter repeated, is

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

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