Question

The perpendicular distance between point (2,5,p) and plane 2x+4x+z+6=0 is equal to $\sqrt{21}$ .
Then p=?

• A. 9
• B. 51
• C. -9
• D. -51

Answer 1

Answer: (C), (D)

-51

The perpendicular distance between point (x1, y1, z1) and the plane ax + by + cz + d = 0 is given by the formula
$\mathrm{Distance}=\left(\frac{{\mathrm{ax}}_1+{\mathrm{by}}_1+{\mathrm{cz}}_1+d}{\sqrt{a^2+b^2+c^2}}\right|\sqrt{21}=\left(\frac{2\times 2+4\times 5+1\times p+6}{\sqrt{2^2+4^2+1^2}}\right|\sqrt{21}=\left(\frac{30+p}{\sqrt{21}}\right|\pm 21=30+pp=-9,-51$