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Step 1: Identify the coordinates of the given points
From the position vectors, we note the coordinates of the points as follows:
• $A = (7,\, -4,\, 7)$
• $B = (1,\, -6,\, 10)$
• $C = (-1,\, -3,\, 4)$
• $D = (5,\, -1,\, 5)$
Step 2: Calculate the distance AB
The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is given by
$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $.
For $AB:$
$x_2 - x_1 = 1 - 7 = -6$,
$y_2 - y_1 = -6 - (-4) = -2$,
$z_2 - z_1 = 10 - 7 = 3$.
So $AB = \sqrt{(-6)^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Step 3: Calculate the distance BC
For $BC:$
$x_2 - x_1 = -1 - 1 = -2$,
$y_2 - y_1 = -3 - (-6) = 3$,
$z_2 - z_1 = 4 - 10 = -6$.
So $BC = \sqrt{(-2)^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Step 4: Calculate the distance CD
For $CD:$
$x_2 - x_1 = 5 - (-1) = 6$,
$y_2 - y_1 = -1 - (-3) = 2$,
$z_2 - z_1 = 5 - 4 = 1$.
So $CD = \sqrt{6^2 + 2^2 + 1^2} = \sqrt{36 + 4 + 1} = \sqrt{41}$.
Step 5: Calculate the distance DA
For $DA:$
$x_2 - x_1 = 7 - 5 = 2$,
$y_2 - y_1 = -4 - (-1) = -3$,
$z_2 - z_1 = 7 - 5 = 2$.
So $DA = \sqrt{2^2 + (-3)^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$.
Step 6: Analyze the shape of the quadrilateral ABCD
We observe that $AB = BC = 7$, while $CD = \sqrt{41}$ and $DA = \sqrt{17}$.
For a shape to be a parallelogram, opposite sides must be equal in length. That condition is not satisfied because $AB \neq CD$ and $BC \neq DA$.
For a rhombus or a square, all four sides must be equal, which is also not satisfied (we have two sides of length 7, one of length $\sqrt{41}$, and one of length $\sqrt{17}$).
Hence, none of the given options (parallelogram but not rhombus, square, or rhombus) correctly describe this quadrilateral.
Step 7: Conclusion
Therefore, the correct choice is "None".
