Find the Cartesian equation of a line passing through (1, –1, 2) and parallel to the line whose equations are (x - 3)/1 = (y - 1)/2 = (z + 1)/-2

Question

Find the Cartesian equation of a line passing through (1, –1, 2) and parallel to the line whose equations are \(\frac{x-3}{1}\) = \(\frac{y-1}{2}\) =\(\frac{z+1}{-2}.\) Also, reduce the equation obtained in vector form.

Answer 1

cartesian equation of the given line is  \(\frac{x-3}{1}\) = \(\frac{y-1}{2}\) =\(\frac{z+1}{-2}\)

Hence its direction ratios are‹1, 2, –2›

The cartesian equation of the line is given by

  \(\frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3}\) the point is (1, –1, 2) and the direction ratios are ‹1, 2, –2›

The cartesian equation of the line is

Let this be equal to \(\lambda\)

Hence comparing 1 & 2

The vector equation of the line is given as