The angle between the lines with direction ra | vedclass

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(A) Let the direction ratios of the two lines be $a_1, b_1, c_1$ and $a_2, b_2, c_2$.Here,$(a_1, b_1, c_1) = \left( \frac{\sqrt{3}}{4}, \frac{1}{4}, \

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Let the direction ratios of the two lines be $a_1, b_1, c_1$ and $a_2, b_2, c_2$.Here,$(a_1, b_1, c_1) = \left( \frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2} ight)$ and $(a_2, b_2, c_2) = \left( \frac{\sqrt{3}}{4}, \frac{1}{4}, -\frac{\sqrt{3}}{2} ight)$.First,we calculate the magnitudes of the vectors:$|v_1| = \sqrt{(\frac{\sqrt{3}}{4})^2 + (\frac{1}{4})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{3}{16} + \frac{1}{16} + \frac{3}{4}} = \sqrt{\frac{3+1+12}{16}} = \sqrt{\frac{16}{16}} = 1$.$|v_2| = \sqrt{(\frac{\sqrt{3}}{4})^2 + (\frac{1}{4})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{3}{16} + \frac{1}{16} + \frac{3}{4}} = 1$.Since the magnitudes are $1$,the given ratios are actually direction cosines.The cosine of the angle $	heta$ between the lines is given by $\cos 	heta = |a_1a_2 + b_1b_2 + c_1c_2|$.$\cos 	heta = |(\frac{\sqrt{3}}{4})(\frac{\sqrt{3}}{4}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2})|$$\cos 	heta = |\frac{3}{16} + \frac{1}{16} - \frac{3}{4}| = |\frac{4}{16} - \frac{12}{16}| = |-\frac{8}{16}| = |-\frac{1}{2}| = \frac{1}{2}$.Thus,$	heta = \cos^{-1}(\frac{1}{2}) = 60^o$. However,if we consider the angle between the vectors,$\cos 	heta = -\frac{1}{2}$,which gives $	heta = 120^o$.

The angle between the lines with direction ratios $\left( \frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2} \right)$ and $\left( \frac{\sqrt{3}}{4}, \frac{1}{4}, -\frac{\sqrt{3}}{2} \right)$ is ......... $^o$.

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