The angle between the lines 2x = 3y = -z and | vedclass

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(B) First,express the lines in standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line $2x = 3y = -z$,divide

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) First,express the lines in standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line $2x = 3y = -z$,divide by $6$ to get $\frac{x}{3} = \frac{y}{2} = \frac{z}{-6}$. The direction ratios are $\vec{v_1} = (3, 2, -6)$.For the second line $6x = -y = -4z$,divide by $12$ to get $\frac{x}{2} = \frac{y}{-12} = \frac{z}{-3}$. The direction ratios are $\vec{v_2} = (2, -12, -3)$.The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$.Calculate the dot product: $a_1a_2 + b_1b_2 + c_1c_2 = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$.Since the dot product is $0$,the lines are perpendicular,so $	heta = 90^o$.

The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is ......... $^o$.

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