Question

The distance of the point $(1,2,-1)$ from the plane $x-2y+4z+10=0$ is

• A. $\dfrac{3}{\sqrt{7}}$ units
• B. $\dfrac{\sqrt{3}}{7}$ units
• C. $\sqrt{\dfrac{7}{3}}$ units
• D. $\sqrt{\dfrac{3}{7}}$ units

Hint:

Always simplify radicals after applying the distance formula.

Answer 1

The Correct Option is D

Step 1: Using the point-to-plane distance formula.
The distance of point $(x_1,y_1,z_1)$ from plane $Ax+By+Cz+D=0$ is \[ \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} \]
Step 2: Substituting given values.
Here, $A=1$, $B=-2$, $C=4$, $D=10$, and $(x_1,y_1,z_1)=(1,2,-1)$. \[ \text{Distance}=\frac{|1-4-4+10|}{\sqrt{1+4+16}} =\frac{3}{\sqrt{21}} \]
Step 3: Simplifying.
\[ \frac{3}{\sqrt{21}}=\sqrt{\frac{3}{7}} \]
Step 4: Conclusion.
The required distance is $\sqrt{\dfrac{3}{7}}$ units.