Statement-1: The shortest distance between th | vedclass

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(B) For Statement-$1$: The lines are $L_1: \vec{r} = (-3, 6, 0) + \lambda(-4, 3, 2)$ and $L_2: \vec{r} = (-3, 0, 7) + \mu(-4, 1, 1)$.The shortest dist

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Statement-$1$ is false,Statement-$2$ is true

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(B) For Statement-$1$: The lines are $L_1: \vec{r} = (-3, 6, 0) + \lambda(-4, 3, 2)$ and $L_2: \vec{r} = (-3, 0, 7) + \mu(-4, 1, 1)$.The shortest distance $d$ between two skew lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.Here,$\vec{a}_2 - \vec{a}_1 = (-3 - (-3), 0 - 6, 7 - 0) = (0, -6, 7)$.$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & 3 & 2 \ -4 & 1 & 1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(-4+8) + \hat{k}(-4+12) = (1, -4, 8)$.$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{1^2 + (-4)^2 + 8^2} = \sqrt{1 + 16 + 64} = \sqrt{81} = 9$.$|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)| = |(0, -6, 7) \cdot (1, -4, 8)| = |0 + 24 + 56| = 80$.So,$d = \frac{80}{9} 
eq 9$. Thus,Statement-$1$ is false.For Statement-$2$: By definition,skew lines are lines in three-dimensional space that are neither parallel nor intersecting,meaning they do not lie in the same plane. Thus,Statement-$2$ is true.

Statement-$1$: The shortest distance between the skew lines $\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{x+3}{-4} = \frac{y}{1} = \frac{z-7}{1}$ is $9$.
Statement-$2$: Two lines are skew lines if there exists no plane passing through them.

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Statement-$1$: The shortest distance between the skew lines $\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{x+3}{-4} = \frac{y}{1} = \frac{z-7}{1}$ is $9$.Statement-$2$: Two lines are skew lines if there exists no plane passing through them.