In the following cases, find the distance of each of the given points from the corresponding given plane.
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0, 0)
(b) (3, −2, 1)
(c) (2, 3, −5)
(d) (−6, 0, 0)
It is known that the distance between a point, p(x1, y1, z1), and a plane, Ax + By + Cz = D, is given by,
$d=\left|\frac{A x_{1}+B y_{1}+C z_{1}-D}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|$ ...(1)
(a) The given point is (0, 0, 0) and the plane is
$\therefore d=\left|\frac{3 \times 0-4 \times 0+12 \times 0-3}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}}\right|=\frac{3}{\sqrt{169}}=\frac{3}{13}$
(b) The given point is (3, − 2, 1) and the plane is
$\therefore d=\left|\frac{2 \times 3-(-2)+2 \times 1+3}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}\right|=\left|\frac{13}{3}\right|=\frac{13}{3}$
(c) The given point is (2, 3, −5) and the plane is
$\therefore d=\left|\frac{2+2 \times 3-2(-5)-9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}\right|=\frac{9}{3}=3$
(d) The given point is (−6, 0, 0) and the plane is
$d=\left|\frac{2(-6)-3 \times 0+6 \times 0-2}{\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}}\right|=\left|\frac{-14}{\sqrt{49}}\right|=\frac{14}{7}=2$
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