Find the point of intersection of the curves `a r g(z-3i)=(3pi)/4 and arg(2z+1-2i)=pi//4.`

Find the coordinates of the points of intersection of the curves y=cosx , y=sin3x if-pi/2lt=xlt=pi/2

The point of intersection of the lines (x + 1)/(3) = (y + 3)/(5) = (z + 5)/(7) and (x - 2)/(1)= (y - 4)/(3) = (z - 6)/(5) is

The distance of the point (2,3,-4) from the point of intersection of the line (x-1)/-3 = (y-2)/2 = (z-2)/1 and the plane 2x-3y-z-20 =0 is

The distance of the point (1, 0, 2) from the point of intersection of the line (x-2)/3=(y+1)/4 =(z-2)/12 and the plane x-y+z=16, is :

The distance of the point (1,0,2) from the point of intersection of the line (x-2)/(3)=(y+1)/(4)=(z-2)/(12) and the plane x-y+z=16 is

The line of intersection of the planes vec r . (3i-j+k) =1 and vec r . (i+4j-2k) = 2 is parallel to the vector

Find the multiplicative inverse of z= 2-3i

Find the multiplicative inverse of z= 6-3i

Find the multiplicative inverse of z= 4 -3i

The point of intersection of the line (x-5)/3 = (y-7)/(-1) = (z+2)/1 , (x+3)/(-36) = (y-3)/2 = (z-6)/4 is