If L is the line with direction ratios \lt 3,-2, 6\gt and passing through (1,-1,1), then what are the coordinates of the points on L whose distance from (1,-1,1) is 2 units?

  1. A. (-\frac{11}{7},\frac{13}{7},\frac{19}{7}) and (\frac{1}{7},\frac{3}{7},\frac{5}{7})
  2. B. (\frac{19}{7},-\frac{11}{7},\frac{13}{7}) and (-\frac{1}{7},\frac{3}{7},-\frac{5}{7})
  3. C. (\frac{13}{7},\frac{11}{7},\frac{19}{7}) and (-\frac{1}{7},-\frac{3}{7},\frac{5}{7})
  4. D. (\frac{13}{7},-\frac{11}{7},\frac{19}{7}) and (\frac{1}{7},-\frac{3}{7},-\frac{5}{7})

Correct Answer: D. (\frac{13}{7},-\frac{11}{7},\frac{19}{7}) and (\frac{1}{7},-\frac{3}{7},-\frac{5}{7})

Explanation

The direction cosines l, m, n are derived from the direction ratios by dividing by \sqrt{3^2+(-2)^2+6^2} = 7. Thus, l, m, n = \frac{3}{7}, -\frac{2}{7}, \frac{6}{7}. Coordinates of a point at distance r from (x_1, y_1, z_1) are (x_1 \pm lr, y_1 \pm mr, z_1 \pm nr). Using r=2, the points are (1 \pm \frac{6}{7}, -1 \mp \frac{4}{7}, 1 \pm \frac{12}{7}). This yields (\frac{13}{7}, -\frac{11}{7}, \frac{19}{7}) and (\frac{1}{7}, -\frac{3}{7}, -\frac{5}{7}).

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