If the lines frac{x - 5}{5m + 2} = frac{2 - y | vedclass

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(A) Step $1$: Rewrite the equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x - 5}{5m + 2} =

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(A) Step $1$: Rewrite the equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x - 5}{5m + 2} = \frac{y - 2}{-5} = \frac{z - 1}{1}$. The direction vector is $\vec{v_1} = (5m + 2, -5, 1)$.Step $2$: For the second line: $\frac{x}{1} = \frac{y + 1/2}{2m} = \frac{z - 1}{3}$. The direction vector is $\vec{v_2} = (1, 2m, 3)$.Step $3$: Since the lines are perpendicular, their dot product must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.$(5m + 2)(1) + (-5)(2m) + (1)(3) = 0$.Step $4$: Solve for $m$: $5m + 2 - 10m + 3 = 0$.$-5m + 5 = 0 \implies 5m = 5 \implies m = 1$.

If the lines $\frac{x - 5}{5m + 2} = \frac{2 - y}{5} = \frac{1 - z}{-1}$ and $\frac{x}{1} = \frac{2y + 1}{4m} = \frac{1 - z}{-3}$ are perpendicular to each other, then the value of $m$ is ...

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