Calculating Vectors and Distances in Cartesian Coordinate System
The question is in German and it is asking to perform a series of calculations with given points in a Cartesian coordinate system. Let's proceed step by step.
Given points are:
\( A(-1/-1/-1); B(1/2/1); C(6/2/3); D(-4/2/-1/2) \).
a) Calculate the following vectors: \( \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}, \overrightarrow{BC}, \overrightarrow{BD}, \overrightarrow{CD} \).
To find the vector from one point to another, subtract the coordinates of the starting point from the coordinates of the ending point.
\( \overrightarrow{AB} = B - A = (1 - (-1), 2 - (-1), 1 - (-1)) = (2, 3, 2) \)
\( \overrightarrow{AC} = C - A = (6 - (-1), 2 - (-1), 3 - (-1)) = (7, 3, 4) \)
\( \overrightarrow{AD} = D - A = (-4 - (-1), 2 - (-1), -0.5 - (-1)) = (-3, 3, 0.5) \)
\( \overrightarrow{BC} = C - B = (6 - 1, 2 - 2, 3 - 1) = (5, 0, 2) \)
\( \overrightarrow{BD} = D - B = (-4 - 1, 2 - 2, -0.5 - 1) = (-5, 0, -1.5) \)
\( \overrightarrow{CD} = D - C = (-4 - 6, 2 - 2, -0.5 - 3) = (-10, 0, -3.5) \)
b) Calculate the distances of point A from B, C, and D.
The distance between two points is the magnitude of the vector from one to the other.
Distance between A and B (using \( \overrightarrow{AB} \)):
\( \| \overrightarrow{AB} \| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \)
Distance between A and C (using \( \overrightarrow{AC} \)):
\( \| \overrightarrow{AC} \| = \sqrt{7^2 + 3^2 + 4^2} = \sqrt{49 + 9 + 16} = \sqrt{74} \)
Distance between A and D (using \( \overrightarrow{AD} \)):
\( \| \overrightarrow{AD} \| = \sqrt{(-3)^2 + 3^2 + (0.5)^2} = \sqrt{9 + 9 + 0.25} = \sqrt{18.25} \)
c) Calculate the magnitude of all the vectors from part a.
\( \| \overrightarrow{AB} \| = \sqrt{17} \) (as calculated above)
\( \| \overrightarrow{AC} \| = \sqrt{74} \) (as calculated above)
\( \| \overrightarrow{AD} \| = \sqrt{18.25} \) (as calculated above)
\( \| \overrightarrow{BC} \| = \sqrt{5^2 + 0^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \)
\( \| \overrightarrow{BD} \| = \sqrt{(-5)^2 + 0^2 + (-1.5)^2} = \sqrt{25 + 2.25} = \sqrt{27.25} \)
\( \| \overrightarrow{CD} \| = \sqrt{(-10)^2 + 0^2 + (-3.5)^2} = \sqrt{100 + 12.25} = \sqrt{112.25} \)
d) The distance from point P (1/x/1) to A is now set to 10 units. Calculate x and specify the point/points.
We use the formula for the distance between two points:
\(10 = \| \overrightarrow{AP} \| = \sqrt{(1 - (-1))^2 + (x - (-1))^2 + (1 - (-1))^2}\)
Now solve for x:
\(10 = \sqrt{4 + (x + 1)^2 + 4}\)
\(10 = \sqrt{(x + 1)^2 + 8}\)
\(100 = (x + 1)^2 + 8\)
\(92 = (x + 1)^2\)
\(x + 1 = \pm \sqrt{92}\)
\(x = -1 \pm \sqrt{92}\)
So there are two possible points for P:
- When subtracting:
\(P_1 = (1, -1 - \sqrt{92}, 1)\)
- When adding:
\(P_2 = (1, -1 + \sqrt{92}, 1)\)
