If the direction ratios (d.r.'s) of two lines | vedclass

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(A) Given the relations between the direction ratios $(a, b, c)$ of two lines: $a - b + c = 0$  $(i)$ $a^2 - b^2 + 2c^2 = 0$  $(ii)$ From $(i)$, we ha

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$\frac{2}{\sqrt{7}}$

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(A) Given the relations between the direction ratios $(a, b, c)$ of two lines: $a - b + c = 0$  $(i)$ $a^2 - b^2 + 2c^2 = 0$  $(ii)$ From $(i)$, we have $c = b - a$. Substituting this into $(ii)$: $a^2 - b^2 + 2(b - a)^2 = 0$ $a^2 - b^2 + 2(b^2 - 2ab + a^2) = 0$ $a^2 - b^2 + 2b^2 - 4ab + 2a^2 = 0$ $3a^2 - 4ab + b^2 = 0$ $(3a - b)(a - b) = 0$ This gives two cases: Case $1$: $b = 3a \implies a:b = 1:3$. Then $c = b - a = 3a - a = 2a$. So, the direction ratios are $(1, 3, 2)$. Case $2$: $b = a \implies a:b = 1:1$. Then $c = b - a = a - a = 0$. So, the direction ratios are $(1, 1, 0)$. The angle $	heta$ between the lines with direction ratios $(a_1, b_1, c_1) = (1, 3, 2)$ and $(a_2, b_2, c_2) = (1, 1, 0)$ is given by: $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$ $\cos 	heta = \frac{|(1)(1) + (3)(1) + (2)(0)|}{\sqrt{1^2 + 3^2 + 2^2} \sqrt{1^2 + 1^2 + 0^2}}$ $\cos 	heta = \frac{|1 + 3 + 0|}{\sqrt{1 + 9 + 4} \sqrt{1 + 1 + 0}} = \frac{4}{\sqrt{14} \sqrt{2}} = \frac{4}{\sqrt{28}} = \frac{4}{2\sqrt{7}} = \frac{2}{\sqrt{7}}$.

If the direction ratios $(d.r.'s)$ of two lines are connected by the relations $a-b+c=0$ and $a^2-b^2+2c^2=0$, and $\theta$ is the angle between these lines, then $\cos \theta = $

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