If two lines L_1 and L_2 in space are defined | vedclass

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(D) For line $L_1$,we have $x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1)$ and $z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda}$.Rewriting these in symmetric

Vedclass · Accepted Answer

$\lambda = \mu$

Vedclass · Answer

(D) For line $L_1$,we have $x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1)$ and $z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda}$.Rewriting these in symmetric form,we get $\frac{x - (\sqrt{\lambda} - 1)}{\sqrt{\lambda}} = y = \frac{z - \sqrt{\lambda}}{\sqrt{\lambda} - 1}$.The direction vector of $L_1$ is $\vec{v_1} = (\sqrt{\lambda}, 1, \sqrt{\lambda} - 1)$.Similarly,for line $L_2$,we have $x = \sqrt{\mu} y + (1 - \sqrt{\mu})$ and $z = (1 - \sqrt{\mu})y + \sqrt{\mu}$.Rewriting these in symmetric form,we get $\frac{x - (1 - \sqrt{\mu})}{\sqrt{\mu}} = y = \frac{z - \sqrt{\mu}}{1 - \sqrt{\mu}}$.The direction vector of $L_2$ is $\vec{v_2} = (\sqrt{\mu}, 1, 1 - \sqrt{\mu})$.Since $L_1 \perp L_2$,the dot product of their direction vectors must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.$(\sqrt{\lambda})(\sqrt{\mu}) + (1)(1) + (\sqrt{\lambda} - 1)(1 - \sqrt{\mu}) = 0$.$\sqrt{\lambda\mu} + 1 + (\sqrt{\lambda} - \sqrt{\lambda\mu} - 1 + \sqrt{\mu}) = 0$.$\sqrt{\lambda} + \sqrt{\mu} = 0$.Since $\lambda, \mu \geq 0$,this implies $\sqrt{\lambda} = 0$ and $\sqrt{\mu} = 0$,which means $\lambda = 0$ and $\mu = 0$. Thus,$\lambda = \mu$.

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