Question:medium
The distance between the \( x \)-axis and the point \( (3, 12, 5) \) is:
The distance between the \( x \)-axis and the point \( (3, 12, 5) \) is:
Show Hint
The formula for distance from the $x$-axis is simply $\sqrt{y^2 + z^2}$. Similarly, distance from the $y$-axis is $\sqrt{x^2 + z^2}$.
Updated On: May 6, 2026
- \( 3 \)
- \( 13 \)
- \( 14 \)
- \( 12 \)
- \( 5 \)
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on Distance of a Point from a Plane
- An equation of a plane parallel to the plane \(x-2y+2z-5=0\) and at a unit distance from the origin is?
- JEE Main - 2024
- Mathematics
- Distance of a Point from a Plane
- The square of the distance of the point \(\left( \frac{15}{7}, \frac{32}{7}, 7 \right)\) from the line \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) in the direction of the vector \(\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\) is:
- JEE Main - 2025
- Mathematics
- Distance of a Point from a Plane
- Let the plane x+3y–2z+6=0 meet the co-ordinate axes at the points A,B,C. If the orthocentre of the triangle ABC is\( (α,β,\frac{6 }{7} ), \) then 98(α + β)2 is equal to___.
- JEE Main - 2023
- Mathematics
- Distance of a Point from a Plane
- Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line \(\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}\). Then the distance of the point
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.- JEE Main - 2023
- Mathematics
- Distance of a Point from a Plane
- Want to practice more? Try solving extra ecology questions todayView All Questions
