Question

A plane makes intercepts \(-2, \frac{4}{3}, \frac{-4}{5}\) on the X, Y, Z axes respectively. Find the direction cosines of a normal to the plane.

• A. \( \left( \frac{-1}{3}, \frac{2}{3}, \frac{-2}{3} \right) \)
• B. \( \left( \frac{2}{3\sqrt{5}}, \frac{-4}{3\sqrt{5}}, \frac{-5}{3\sqrt{5}} \right) \)
• C. \( \left( \frac{-4}{\sqrt{57}}, \frac{-4}{\sqrt{57}}, \frac{-5}{\sqrt{57}} \right) \)
• D. \( \left( \frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}} \right) \)

Hint:

To find direction cosines of a plane’s normal, use reciprocals of intercepts and normalize the resulting vector.

Answer 1

The Correct Option is D

Equation of plane with intercepts \( a, b, c \) on axes is: \[ \begin{align} \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \Rightarrow \text{Normal vector } \vec{n} = \left( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \right) \] Given: \( a = -2,\ b = \frac{4}{3},\ c = \frac{-4}{5} \Rightarrow \vec{n} = \left( -\frac{1}{2}, \frac{3}{4}, -\frac{5}{4} \right) \) Find direction cosines: \[ \begin{align} \text{Magnitude} = \sqrt{ \left( -\frac{1}{2} \right)^2 + \left( \frac{3}{4} \right)^2 + \left( -\frac{5}{4} \right)^2 } = \sqrt{ \frac{1}{4} + \frac{9}{16} + \frac{25}{16} } = \sqrt{ \frac{4 + 9 + 25}{16} } = \sqrt{ \frac{38}{16} } = \frac{\sqrt{38}}{4} \] So direction cosines: \[ \begin{align} \left( \frac{-1/2}{\sqrt{38}/4}, \frac{3/4}{\sqrt{38}/4}, \frac{-5/4}{\sqrt{38}/4} \right) = \left( \frac{-2}{\sqrt{38}}, \frac{3}{\sqrt{38}}, \frac{-5}{\sqrt{38}} \right) \Rightarrow \left( \frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}} \right) \]