A plane cuts intercepts 2, 2, 1 on the coordinate axes. What are the direction cosines of the normal to the plane?

  1. A. \langle 2/3, 2/3, 1/3 \rangle
  2. B. \langle 1/3, 2/3, 2/3 \rangle
  3. C. \langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \rangle
  4. D. \langle \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \rangle

Correct Answer: C. \langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \rangle

Explanation

The intercept form of the plane is \frac{x}{2} + \frac{y}{2} + \frac{z}{1} = 1, which simplifies to x + y + 2z = 2. The normal vector has direction ratios (1, 1, 2). The direction cosines are obtained by dividing by the magnitude \sqrt{1^2+1^2+2^2} = \sqrt{6}, giving \langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \rangle.

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