If the line passing through the points (5, 1, | 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}

Vedclass · Accepted Answer

$a = 6, b = 4$

Vedclass · Answer

(C) The equation of the line passing through $(5, 1, a)$ and $(3, b, 1)$ is given by: $\frac{x - 5}{3 - 5} = \frac{y - 1}{b - 1} = \frac{z - a}{1 - a} = k$ $\frac{x - 5}{-2} = \frac{y - 1}{b - 1} = \frac{z - a}{1 - a} = k$ Any point on the line is $(5 - 2k, 1 + k(b - 1), a + k(1 - a))$. Since the line passes through $(0, \frac{17}{2}, -\frac{13}{2})$,we equate the coordinates: $5 - 2k = 0 \implies k = \frac{5}{2}$ For the $y$-coordinate: $1 + \frac{5}{2}(b - 1) = \frac{17}{2}$ $1 + \frac{5}{2}b - \frac{5}{2} = \frac{17}{2} \implies \frac{5}{2}b = \frac{17}{2} + \frac{3}{2} = 10 \implies b = 4$ For the $z$-coordinate: $a + \frac{5}{2}(1 - a) = -\frac{13}{2}$ $a + \frac{5}{2} - \frac{5}{2}a = -\frac{13}{2} \implies -\frac{3}{2}a = -\frac{13}{2} - \frac{5}{2} = -\frac{18}{2} = -9$ $a = 6$ Thus,$a = 6$ and $b = 4$.

If the line passing through the points $(5, 1, a)$ and $(3, b, 1)$ intersects the plane at the point $(0, \frac{17}{2}, -\frac{13}{2})$,then find the values of $a$ and $b$.

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