Question:
If the sum of the squares of the distance of the point $\text{P}(x, y, z)$ from the co-ordinate axes is 242 , then the distance of the point P from the origin is units.
If the sum of the squares of the distance of the point $\text{P}(x, y, z)$ from the co-ordinate axes is 242 , then the distance of the point P from the origin is units.
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Sum of squares of distances from axes = $2(x^2+y^2+z^2)$.
Updated On: Apr 26, 2026
- 121
- 11
- 22
- $\frac{121}{2}$
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The Correct Option is B
Solution and Explanation
Concept:
Distance of point $P(x,y,z)$ from:
Step 2: Use given condition. \[ 2(x^2 + y^2 + z^2) = 242 \] \[ x^2 + y^2 + z^2 = 121 \]
Step 3: Distance from origin. \[ OP = \sqrt{x^2 + y^2 + z^2} = \sqrt{121} = 11 \]
Step 4: Conclusion. \[ {11} \]
Distance of point $P(x,y,z)$ from:
- X-axis $= \sqrt{y^2 + z^2}$
- Y-axis $= \sqrt{x^2 + z^2}$
- Z-axis $= \sqrt{x^2 + y^2}$
Step 2: Use given condition. \[ 2(x^2 + y^2 + z^2) = 242 \] \[ x^2 + y^2 + z^2 = 121 \]
Step 3: Distance from origin. \[ OP = \sqrt{x^2 + y^2 + z^2} = \sqrt{121} = 11 \]
Step 4: Conclusion. \[ {11} \]
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