If the direction cosines of a line are k, k and k,then

(a) k>0 (b)0<k<1

(c) k=1 (d) $k = \frac{1}{\sqrt{3}} o r - \frac{1}{\sqrt{3}}$

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Solution

(d) Since,direction cosines of a line are k,k and k.

$∴ l = k, m = k a n d n = k W e k n o w t h a t, l_{2} + m_{2} + n_{2} = 1 \Rightarrow (k_{2})^{2} + (k_{2})^{2} + (k_{2})^{2} = 1 \Rightarrow (k_{2})^{2} = \frac{1}{3} B u t k_{2} = k ∴ k = \pm \frac{1}{\sqrt{3}}$

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Similar questions

The direction cosines of a line are k, k, k, then
(a) k > 0
(b) 0 < k < 1
(c) k = 1
(d) $k = \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}$

\frac{1}{2}, \frac{1}{2}, k

k > 0

k = ?

- \frac{1}{2}, \frac{1}{3}

\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}

\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}

\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k} a n d \frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}

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