The line passing through the points (5, 1, a) | vedclass

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(C) The equation of the line passing through $(5, 1, a)$ and $(3, b, 1)$ is given by $\frac{x-5}{3-5} = \frac{y-1}{b-1} = \frac{z-a}{1-a} = \mu$.This

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$a=6, b=4$

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(C) The equation of the line passing through $(5, 1, a)$ and $(3, b, 1)$ is given by $\frac{x-5}{3-5} = \frac{y-1}{b-1} = \frac{z-a}{1-a} = \mu$.This simplifies to $\frac{x-5}{-2} = \frac{y-1}{b-1} = \frac{z-a}{1-a} = \mu$.The line crosses the $yz-$ plane where $x=0$. Substituting $x=0$ into the equation $\frac{x-5}{-2} = \mu$,we get $\frac{0-5}{-2} = \mu$,so $\mu = \frac{5}{2}$.Now,for the $y-$coordinate: $y = \mu(b-1) + 1 = \frac{17}{2}$.Substituting $\mu = \frac{5}{2}$: $\frac{5}{2}(b-1) + 1 = \frac{17}{2} \Rightarrow \frac{5}{2}(b-1) = \frac{15}{2} \Rightarrow b-1 = 3 \Rightarrow b = 4$.For the $z-$coordinate: $z = \mu(1-a) + a = -\frac{13}{2}$.Substituting $\mu = \frac{5}{2}$: $\frac{5}{2}(1-a) + a = -\frac{13}{2} \Rightarrow \frac{5}{2} - \frac{5}{2}a + a = -\frac{13}{2} \Rightarrow -\frac{3}{2}a = -\frac{13}{2} - \frac{5}{2} = -\frac{18}{2} = -9$.Thus,$a = 6$. Therefore,$a=6$ and $b=4$.

The line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $yz-$ plane at the point $(0, \frac{17}{2}, -\frac{13}{2})$. Then:

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