The angle between the straight lines frac{x + | vedclass

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

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) The direction ratios of the first line are $a_1 = 2, b_1 = 5, c_1 = 4$.The direction ratios of the second line are $a_2 = 1, b_2 = 2, c_2 = -3$.Let $	heta$ be the angle between the lines. The formula for the angle is $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the values: $\cos 	heta = \frac{|(2)(1) + (5)(2) + (4)(-3)|}{\sqrt{2^2 + 5^2 + 4^2} \sqrt{1^2 + 2^2 + (-3)^2}}$.$\cos 	heta = \frac{|2 + 10 - 12|}{\sqrt{4 + 25 + 16} \sqrt{1 + 4 + 9}} = \frac{|0|}{\sqrt{45} \sqrt{14}} = 0$.Since $\cos 	heta = 0$,we have $	heta = 90^o$.

The angle between the straight lines $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4}$ and $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{-3}$ is ......... $^o$

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