Write the condition for the lines` vec r= vec a_1+lambda vec b_1 a n d\ vec r= vec a_2+mu vec b_2dot\ ` to be intersecting.

Write the formula for the shortest distance between the lines vec r= vec a_1+lambda vec b_1 a n d\ vec r= vec a_2+mu vec b_2dot\

Write the equation of the plane containing the lines vec r= vec a+lambda vec b\ a n d\ vec r= vec a+mu vec c

Statement 1: Lines vec r= hat i+ hat j- hat k+lambda(3 hat i- hat j)a n d vec r=4 hat i- hat k+mu(2 hat i++3 hat k) intersect. Statement 2: vec bxx vec d=0 , then lines vec r= vec a+lambda vec ba n d vec r= vec c+lambda vec d do not intersect.

Write the equation of the plane vec r= vec a+lambda vec b+mu vec c\ in scalar product form.

Write the position vector of the point where the line vec r= vec a+lambda vec b meets the plane vec r dot (vec n)=0.

Let A( vec a)a n dB( vec b) be points on two skew lines vec r= vec a+lambda vec pa n d vec r= vec b+u vec q and the shortest distance between the skew lines is 1, w h e r e vec pa n d vec q are unit vectors forming adjacent sides of a parallelogram enclosing an area of 1/2 units. If angle between A B and the line of shortest distance is 60^0, then A B= a. 1/2 b. 2 c. 1 d. lambda R={10}

Let A( vec a)a n dB( vec b) be points on two skew lines vec r= vec a+lambda vec pa n d vec r= vec b+u vec q and the shortest distance between the skew lines is 1, w h e r e vec pa n d vec q are unit vectors forming adjacent sides of a parallelogram enclosing an area of 1/2 units. If angle between A B and the line of shortest distance is 60^0, then A B= a. 1/2 b. 2 c. 1 d. lambda R={0}

Distance of the point P( vec p) from the line vec r= vec a+lambda vec b is a. |( vec a- vec p)+((( vec p- vec a)dot vec b) vec b)/(| vec b|^2)| b. |( vec b- vec p)+((( vec p- vec a)dot vec b) vec b)/(| vec b|^2)| c. |( vec a- vec p)+((( vec p- vec b)dot vec b) vec b)/(| vec b|^2)| d. none of these

Which of the following is/are correct? (a)If vec a= hat i+ hat j and vec b=2 hat i- hat k then the point of intersection of the lines vec r x vec a = vec b x vec a and vec a x vec b = vec a x vec b is 3 vec i + vec j - veck (b)( vec r dot hat i )( hat i x vec r )+( vec r dot hat j )( hat j x vec r )+( vec r dot hat k )( hat k x vec r )= vec0 (c)If | vec a|=3,| vec b|=5,| vec c|=7 and vec a+ vec b+ vec c= vec0 then the angle between vec aa n d vec b is pi/3 (d)If the vectors vec a , vec b& vec c from the sides B C ,C A&A B respectively of a triangle A B C , then vec a x vec b = vec b x vec c x vec a

For any vector vec a\ a n d\ vec b prove that | vec a+ vec b|lt=| vec a|+| vec b|dot