Statement -1: The point A(1, 0, 7) is the mir | vedclass

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(D) Let the line be $L: \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = k$. Any point on the line is $P(k, 2k + 1, 3k + 2)$.For $A(1, 0, 7)$ to be t

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Statement $-1$ is true,Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$.

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(D) Let the line be $L: \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = k$. Any point on the line is $P(k, 2k + 1, 3k + 2)$.For $A(1, 0, 7)$ to be the mirror image of $B(1, 6, 3)$ in line $L$,the line $AB$ must be perpendicular to $L$ and the midpoint of $AB$ must lie on $L$.$1$. Midpoint of $AB$ is $M = (\frac{1+1}{2}, \frac{0+6}{2}, \frac{7+3}{2}) = (1, 3, 5)$.Checking if $M$ lies on $L$: $\frac{1}{1} = \frac{3-1}{2} = \frac{5-2}{3} \implies 1 = 1 = 1$. Thus,$M$ lies on $L$.$2$. Direction ratios of $AB$ are $(1-1, 6-0, 3-7) = (0, 6, -4)$.Direction ratios of $L$ are $(1, 2, 3)$.Dot product: $(0)(1) + (6)(2) + (-4)(3) = 0 + 12 - 12 = 0$.Since the dot product is $0$,$AB$ is perpendicular to $L$.Both conditions are satisfied,so Statement $-1$ is true. Statement $-2$ is also true because the line $L$ passes through the midpoint of $AB$ and is perpendicular to $AB$,which is the definition of a mirror image in a line. Thus,Statement $-2$ is the correct explanation for Statement $-1$.

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

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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 $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.