Question

Can you relate the expression

6 (x^{2} + y^{2} + z^{2})

to three-dimensional shapes?

Accepted Answer

The expression $6 (x^{2} + y^{2} + z^{2})$ is related to three-dimensional shapes, particularly to the concept of distance or magnitude in three-dimensional space. This expression represents six times the sum of the squares of the coordinates $x, y, and z .$ In the context of three-dimensional geometry, $x, y, and z$ are typically the coordinates of a point in 3D space.

This expression can be associated with the square of the distance (or magnitude) of a point $(x, y, z)$ from the origin $(0, 0, 0)$ in a 3D coordinate system. The distance d between two points $(x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2})$ in 3D space is given by the formula:

$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$

If we consider the origin $(0, 0, 0)$ as one of the points and another point $(x, y, z)$ , the distance formula becomes:

$d = \sqrt{x^{2} + y^{2} + z^{2}}$

Now, if we square this distance, we get $x^{2} + y^{2} + z^{2}$ .

Multiplying this by $6$ gives $6 (x^{2} + y^{2} + z^{2}),$ which can be interpreted as six times the squared distance of a point from the origin.

In the context of three-dimensional shapes, this expression might be relevant in various geometric calculations involving distances or magnitudes, such as finding the squared distance between points, calculating potentials in physics, or expressing the squared norm of vectors in linear algebra.

EXERCISE 6.4

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Questions from EXERCISE 6.4 with Hints & Solutions

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