Question

The shortest distance from the point $(-10, 10, -10)$ to the $z$-axis, is

• A. $2\sqrt{10}$
• B. $10\sqrt{3}$
• C. 10
• D. $3\sqrt{10}$
• E. $10\sqrt{2}$

Hint:

Distance to $x$-axis: $\sqrt{y^2 + z^2}$.
Distance to $y$-axis: $\sqrt{x^2 + z^2}$.
Distance to $z$-axis: $\sqrt{x^2 + y^2}$.
Just omit the coordinate of the axis you are measuring to!

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The shortest distance from a point in 3D space to the z-axis is the perpendicular distance.
Any point on the z-axis has coordinates \((0, 0, z)\). The projection of point \((x, y, z)\) onto the z-axis is exactly \((0, 0, z)\).
Step 2: Key Formula or Approach:
The perpendicular distance from a point \(P(x, y, z)\) to the z-axis is given by the distance formula in the xy-plane: \(d = \sqrt{x^2 + y^2}\).
Step 3: Detailed Explanation:
The given point is \((-10, 10, -10)\), meaning \(x = -10\) and \(y = 10\).
The coordinates of the foot of the perpendicular on the z-axis are \((0, 0, -10)\).
Use the distance formula between \((-10, 10, -10)\) and \((0, 0, -10)\):
\[ d = \sqrt{(-10 - 0)^2 + (10 - 0)^2 + (-10 - (-10))^2} \] \[ d = \sqrt{(-10)^2 + (10)^2 + 0^2} \] \[ d = \sqrt{100 + 100} = \sqrt{200} \] Simplify the radical:
\[ d = \sqrt{100 \times 2} = 10\sqrt{2} \] Step 4: Final Answer:
The shortest distance is \(10\sqrt{2}\).