Question:

If the direction cosines of a line are 3k,3k,3k \sqrt{3}k, \sqrt{3}k, \sqrt{3}k , then the value of k k is:

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Direction cosines of a line always satisfy the equation l2+m2+n2=1 l^2 + m^2 + n^2 = 1 . Use this to determine unknown values.
Updated On: Jan 18, 2025
  • ±1 \pm 1
  • ±3 \pm \sqrt{3}
  • ±3 \pm 3
  • ±13 \pm \frac{1}{3}
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The Correct Option is D

Solution and Explanation

The direction cosines of a line satisfy the relation: l2+m2+n2=1, l^2 + m^2 + n^2 = 1,  
where l,m,n l, m, n are the direction cosines. 

Here: l=3k,m=3k,n=3k. l = \sqrt{3}k, \, m = \sqrt{3}k, \, n = \sqrt{3}k.  

Substitute into the equation: (3k)2+(3k)2+(3k)2=1. (\sqrt{3}k)^2 + (\sqrt{3}k)^2 + (\sqrt{3}k)^2 = 1.  

Simplify: 3k2+3k2+3k2=1    9k2=1    k2=19. 3k^2 + 3k^2 + 3k^2 = 1 \implies 9k^2 = 1 \implies k^2 = \frac{1}{9}. Thus: k=±13. k = \pm \frac{1}{3}.  

The correct answer is (D) ±13 \pm \frac{1}{3} .

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