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

Question

If $\overrightarrow{AB}=\hat{j}+\hat{k}$ and $\overrightarrow{AC}=3\hat{i}-\hat{j}+4\hat{k}$ represent the two vectors along the sides $AB$ and $AC$ of $\triangle ABC$, prove that the median $\overrightarrow{AD}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$ where $D$ is the midpoint of $BC$. Hence, find the length of the median $AD$.

Answer 1

The direction cosines of a line are denoted by $ l, m, n $ and adhere to the condition $ l^2 + m^2 + n^2 = 1 $. Given that $ l = \sqrt{3}k $, $ m = \sqrt{3}k $, and $ n = \sqrt{3}k $, we substitute these into the equation: $ (\sqrt{3}k)^2 + (\sqrt{3}k)^2 + (\sqrt{3}k)^2 = 1 $. Simplifying the terms yields $ 3k^2 + 3k^2 + 3k^2 = 1 $, which further simplifies to $ 9k^2 = 1 $. Solving for $ k^2 $ gives $ k^2 = \frac{1}{9} $. Taking the square root of both sides results in $ k = \pm \frac{1}{3} $. Therefore, the value of $ k $ is $ \pm \frac{1}{3} $.

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

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The Correct Option is D

Solution and Explanation

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