If the direction cosines \langle l, m, n \rangle of a line are connected by relation l+2m+n=0, 2l-2m+3n=0, then what is the value of l^{2}+m^{2}-n^{2}?

  1. A. \frac{1}{101}
  2. B. \frac{29}{101} ✓
  3. C. \frac{41}{101}
  4. D. \frac{92}{101}

Correct Answer: B. \frac{29}{101}

Explanation

From the first equation, l = -2m - n. Substitute this into the second equation: 2(-2m - n) - 2m + 3n = 0 \implies -4m - 2n - 2m + 3n = 0 \implies n = 6m. Thus, l = -2m - 6m = -8m. The direction ratios are proportional to \langle -8m, m, 6m \rangle, or simply \langle -8, 1, 6 \rangle. Since l^2+m^2+n^2=1, we normalize them by dividing by \sqrt{(-8)^2+1^2+6^2} = \sqrt{101}. So, l = -\frac{8}{\sqrt{101}}, m = \frac{1}{\sqrt{101}}, n = \frac{6}{\sqrt{101}}. Then l^2+m^2-n^2 = \frac{64+1-36}{101} = \frac{29}{101}.

Related questions on 3D Geometry

Practice more NDA Mathematics questions