The cartesian equation of a line AB is `(3-x)/1=(y+2)/-2=(z-5)/4`. Find the direction ratios of a line parallel to AB.

If the cartesian equation of a line AB is (x-1)/(2)=(2y-1)/(12)=(z+5)/(3) , then the direction cosines of a line parallel to AB are -

The Cartesian equation of a line is (x-3)/2=(y+1)/(-2)=(z-3)/5 . Find the vector equation of the line.

The cartesian equation of a line is (2x-5)/(3)=(6-3y)/(2)=(z+1)/(6) . Find the direction ratios of the given line.

The equation of a line are (4-x)/(2)=(y+3)/(2)=(z+2)/(1) . Find the direction cosines of a line parallel to the above line.

The cartesian equation of a line is (x+3)/(2)=(y-5)/(4)=(z+6)/(2) . the vector equation for the line.

The cartesian equation of a line is 3x+2=5y-4=3-z . Find a point on the line and its direction ratios, hence rewrite this equation in symmetric from and then reduce it to vector form.

The cartesian equation of a line is given by : 6x-1=3y+2=2x-2 . Find its direction ratios and reduce it to vector form.

The Cartesian equations of a line are 6x-2=3y+1=2z-2. Find its direction ratios and also find a vector equation of the line.

The cartesian equation of a straight line is (x-3)/(4)=(y+2)/(5)=(z-4)/(3) , its vector form will be -

(i) Find the vector equation of a line passing through a point with position vector 2hat(i)-hat(j)+hat(k) , and parallel to the line joining the points -hat(i)+4hat(j)+hat(k) and hat(i)+2hat(j)+2hat(k) . Also find the cartesian equivalent of this equation. (ii) The cartesian equations of a line are 6x-2=3y+1=2z-2 . Find its direction ratios and also find vector equation of the line.