Question

The distance of the point (1,-5,9) from the plane x-y+z=5 measured along a straight line

Answer 1

Answer:

$\text{The distance of the point }(1,-5,9)\text{ from the plane x-y+z-5=0 is}$

$\bf|\frac{10}{\sqrt{3}}|$.

Step-by-step explanation:

$\text{Concept:}$

$\text{The distance of the point }(x_1,y_1,z_1)\text{ from the plane ax+by+cz+d=0 is}$

$\bf|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|$

$\text{Now }$

$\text{The distance of the point }(1,-5,9)\text{ from the plane x-y+z-5=0 is}$

$|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|$

$=|\frac{1(1)-1(-5)+(9)-5}{\sqrt{1^2+(-1)^2+(1)^2}}|$

$=|\frac{1+5+9-5}{\sqrt{1+1+1}}|$

$=\bf|\frac{10}{\sqrt{3}}|$ units