Hello Subash!

Here is the solution for the question you are asking for, I solved it in procedural way but if you are among the one who prefer OOP style then you can still ask it for me cause I have solved it from both methods but here I am just going to leave procedural one....

//author:Manish Acharya

import java.util.Scanner;

import java.util.*;

public class idgenerator {

public static void main(String[] args) {

String small_name="", long_name="", new_small_name="", new_long_name="";

char lr='a',...

Program to input any number and display number of odd numbers in it;

#include<stdio.h>

#include<stdlib.h>

#include<conio.h>

#include<string.h>

int main() {

int i, number, num1, num2=0, num3, num4 =0, rem, rem1, rem2, rev = 0;

printf("Enter your number ==> ");

scanf("%d", &number);

num1 = number;

while(num1 != 0) {

rem = num1%10;

rem1 = rem%2;

if (rem1 != 0) {

num2 = num2*10 + rem;

}

num1 /= 10;

}

num3 = num2;

while (num2 != 0) {

...

No, if three vectors do not lie in a plane, they cannot give zero resultant.

Explanation:

Let A, B and C be three vectors. If they give zero resultant, then

     A+B+C=0

or, A= -(B+C)

Hence, they will produce zero resultant, if A is equal to negative of vector (B+C). The vector (B+C) lies in the plane of B and C. Hence, A will be equal to negative of (B+C) if A, B and C all lie in a plane.