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Question

The points $$A(1, 2, -1), B(2, 5, -2), C(4, 4, -3)$$ and $$D(3, 1, -2)$$ are


A
collinear
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B
vertices of a rectangle
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C
vertices of a square
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D
vertices of a rhombus
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Solution

The correct option is C vertices of a rectangle
Given, $$A(1,2,-1), B(2,5,-2), C(4,4,-3),$$ and $$D(3,1,-2)$$

$$AB$$ $${=}$$ $$\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$$ $${=}$$ $$\sqrt{11}$$

$$BC$$ $${=}$$ $$\sqrt{{(4-2)}^{2}+{(4-5)}^{2}+{(-3+2)}^{2}}$$ $${=}$$ $$\sqrt{6}$$

$$CD$$ $${=}$$ $$\sqrt{{(3-4)}^{2}+{(1-4)}^{2}+{(-2+3)}^{2}}$$ $${=}$$ $$\sqrt{11}$$

$$DA$$ $${=}$$ $$\sqrt{{(1-3)}^{2}+{(2-1)}^{2}+{(-1+2)}^{2}}$$ $${=}$$ $$\sqrt{6}$$

$$AB{=}CD$$ and $$BC{=}DA$$ 

i.e. opposite sides are equal.

Now Direction ratio of $$AB =(1,3,-1)$$ and Direction ratio of $$BC= (2,-1,-1)$$, 

$$\vec {AB}=\hat i+\hat j+\hat k$$     &     $$\vec {BC}=2\hat i-\hat j-\hat k $$

$$\vec {AB}\cdot \vec {BC}=2-1-1=0$$

As the dot product between $$AB$$ and $$BC$$ is $$0$$, that means angle between them is $$90^o$$. 

Similarly check for $$BC$$ and $$CD$$, then you find that all angles are equal to $$90^o$$ and opposite sides are equal.

Hence, it is a rectangle.

Mathematics

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