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

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(D) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.Substituting the given points $(5, 1, a)$ and $(3, b, 1)$,we get:$\frac{x - 5}{3 - 5} = \frac{y - 1}{b - 1} = \frac{z - a}{1 - a}$$\frac{x - 5}{-2} = \frac{y - 1}{b - 1} = \frac{z - a}{1 - a} = k$ (say).Since the line passes through $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$,we substitute these coordinates into the equation:For $x = 0$: $\frac{0 - 5}{-2} = k \implies k = \frac{5}{2}$.For $y = \frac{17}{2}$: $\frac{\frac{17}{2} - 1}{b - 1} = \frac{5}{2} \implies \frac{15/2}{b - 1} = \frac{5}{2} \implies \frac{15}{b - 1} = 5 \implies b - 1 = 3 \implies b = 4$.For $z = \frac{-13}{2}$: $\frac{\frac{-13}{2} - a}{1 - a} = \frac{5}{2} \implies \frac{-13 - 2a}{2(1 - a)} = \frac{5}{2} \implies -13 - 2a = 5 - 5a \implies 3a = 18 \implies a = 6$.Now,calculate $2a + 3b = 2(6) + 3(4) = 12 + 12 = 24$.

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

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