Show that the points A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) are the vertices of a rhombus.

Given:

The points A (1, 3, 4), B (-1, 6, 10), C (-7, 4, 7) and D (-5, 1, 1)

We know that, all sides of both square and rhombus are equal.

But diagonals of a rhombus are not equal whereas they are equal for square.

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

The distance between the points A (1, 3, 4), B (-1, 6, 10) is AB,

\( \sqrt{1-(-1)^2+(3-6)^2+(4-10)^2}\)

\( \sqrt{4+9+36}=\sqrt{49}=7\)

The distance between the points B (-1, 6, 10) and C (-7, 4, 7) is BC,

 \( \sqrt{1-(-7)^2+(6-4)^2+(10-7)^2}\)

\( \sqrt{36+4+9}=\sqrt{49}=7\)

 The distance between the points C (-7, 4, 7) and D (-5, 1, 1) is CD,

 \( \sqrt{7-(-5)^2+(4-1)^2+(7-1)^2}\)

\( \sqrt{36+4+9}=\sqrt{49}=7\)

The distance between the points A (1, 3, 4) and D (-5, 1, 1) is AD

  \( \sqrt{7-(-5)^2+(4-1)^2+(7-1)^2}\)

\( \sqrt{36+4+9}=\sqrt{49}=7\)

It is clear that,

AB = BC = CD = AD

So, all sides are equal

Now, let us find the length of diagonals

It is clear that,

AC ≠ BD

The diagonals are not equal but all sides are equal.

So we can say that quadrilateral formed by ABCD is a rhombus but not square.

Hence Proved.