Find the distance of the point (1, -5, 9) fro | vedclass

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Question

(B) The equation of the line passing through $P(1, -5, 9)$ and parallel to the line $x = y = z$ is given by $\frac{x-1}{1} = \frac{y+5}{1} = \frac{z-9

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$10\sqrt{3}$

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(B) The equation of the line passing through $P(1, -5, 9)$ and parallel to the line $x = y = z$ is given by $\frac{x-1}{1} = \frac{y+5}{1} = \frac{z-9}{1} = r$. Any point on this line is of the form $(r+1, r-5, r+9)$. Since this point lies on the plane $x - y + z = 5$,we substitute these coordinates into the plane equation: $(r+1) - (r-5) + (r+9) = 5$. $r + 1 - r + 5 + r + 9 = 5$. $r + 15 = 5 \Rightarrow r = -10$. The point of intersection $A$ is $(-10+1, -10-5, -10+9) = (-9, -15, -1)$. The distance $AP$ is the distance between $P(1, -5, 9)$ and $A(-9, -15, -1)$: $AP = \sqrt{(-9-1)^2 + (-15 - (-5))^2 + (-1-9)^2}$. $AP = \sqrt{(-10)^2 + (-10)^2 + (-10)^2} = \sqrt{100 + 100 + 100} = \sqrt{300} = 10\sqrt{3}$ units.

Find the distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$.

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