Find the angle between the pair of lines frac | vedclass

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Question

(A) The direction ratios of the first line are $a_1 = 3, b_1 = 5, c_1 = 4$. The direction ratios of the second line are $a_2 = 1, b_2 = 1, c_2 = 2$. I

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{8 \sqrt{3}}{15}\right)$

Vedclass · Answer

(A) The direction ratios of the first line are $a_1 = 3, b_1 = 5, c_1 = 4$. The direction ratios of the second line are $a_2 = 1, b_2 = 1, c_2 = 2$. If $	heta$ is the angle between them,then the formula is: $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$ Substituting the values: $\cos 	heta = \left| \frac{(3)(1) + (5)(1) + (4)(2)}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 1^2 + 2^2}} ight|$ $\cos 	heta = \left| \frac{3 + 5 + 8}{\sqrt{9 + 25 + 16} \sqrt{1 + 1 + 4}} ight|$ $\cos 	heta = \frac{16}{\sqrt{50} \sqrt{6}} = \frac{16}{5 \sqrt{2} \sqrt{6}} = \frac{16}{5 \sqrt{12}} = \frac{16}{5 	imes 2 \sqrt{3}} = \frac{8}{5 \sqrt{3}}$ Rationalizing the denominator: $\cos 	heta = \frac{8 \sqrt{3}}{5 	imes 3} = \frac{8 \sqrt{3}}{15}$ Therefore,$	heta = \cos ^{-1}\left(\frac{8 \sqrt{3}}{15}ight)$.

Find the angle between the pair of lines $\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$.

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Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
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