# Show that the points $\left( {1,1,1} \right)$ and $\left( { - 3,0,1} \right)$ are equidistant from the plane $\overrightarrow r .\left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$

Answer

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Hint: Write the equation of the plane in Cartesian form and the use distance formula of a point from a plane.

The given plane equation is $\overrightarrow r .\left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$.

The Cartesian form the plane equation is:

$ \Rightarrow 3x + 4y - 12z + 13 = 0$

We have to compare the distance of points $\left( {1,1,1} \right)$ and $\left( { - 3,0,1} \right)$ from this plane.

We know that the distance of a point from a plane is given by the formula:

$ \Rightarrow D = \left| {\dfrac{{ax + by + cz + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$

Using this formula, the distance of point $\left( {1,1,1} \right)$from the plane $3x + 4y - 12z + 13 = 0$is:

$

\Rightarrow D = \left| {\dfrac{{3\left( 1 \right) + 4\left( 1 \right) - 12\left( 1 \right) + 13}}{{\sqrt {{3^2} + {4^2} + {{\left( {12} \right)}^2}} }}} \right|, \\

\Rightarrow D = \dfrac{{\left| {3 + 4 - 12 + 13} \right|}}{{\sqrt {25 + 144} }}, \\

\Rightarrow D = \dfrac{8}{{13}} \\

$

Similarly, the distance of point $\left( { - 3,0,1} \right)$from the plane $3x + 4y - 12z + 13 = 0$is:

$

\Rightarrow D = \left| {\dfrac{{3\left( { - 3} \right) + 4\left( 0 \right) - 12\left( 1 \right) + 13}}{{\sqrt {{3^2} + {4^2} + {{\left( {12} \right)}^2}} }}} \right|, \\

\Rightarrow D = \dfrac{{\left| { - 9 + 0 - 12 + 13} \right|}}{{\sqrt {25 + 144} }}, \\

\Rightarrow D = \dfrac{{\left| { - 8} \right|}}{{\sqrt {169} }}, \\

$

$ \Rightarrow D = \dfrac{8}{{13}}$

Therefore, the distance of points $\left( {1,1,1} \right)$ and $\left( { - 3,0,1} \right)$ from plane $\overrightarrow r .\left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$ are equal.

Note: In the distance formula used above, we used modulus sign just to ensure that the distance never comes out as negative. If we are getting its value negative, modulus will turn it positive.

The given plane equation is $\overrightarrow r .\left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$.

The Cartesian form the plane equation is:

$ \Rightarrow 3x + 4y - 12z + 13 = 0$

We have to compare the distance of points $\left( {1,1,1} \right)$ and $\left( { - 3,0,1} \right)$ from this plane.

We know that the distance of a point from a plane is given by the formula:

$ \Rightarrow D = \left| {\dfrac{{ax + by + cz + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$

Using this formula, the distance of point $\left( {1,1,1} \right)$from the plane $3x + 4y - 12z + 13 = 0$is:

$

\Rightarrow D = \left| {\dfrac{{3\left( 1 \right) + 4\left( 1 \right) - 12\left( 1 \right) + 13}}{{\sqrt {{3^2} + {4^2} + {{\left( {12} \right)}^2}} }}} \right|, \\

\Rightarrow D = \dfrac{{\left| {3 + 4 - 12 + 13} \right|}}{{\sqrt {25 + 144} }}, \\

\Rightarrow D = \dfrac{8}{{13}} \\

$

Similarly, the distance of point $\left( { - 3,0,1} \right)$from the plane $3x + 4y - 12z + 13 = 0$is:

$

\Rightarrow D = \left| {\dfrac{{3\left( { - 3} \right) + 4\left( 0 \right) - 12\left( 1 \right) + 13}}{{\sqrt {{3^2} + {4^2} + {{\left( {12} \right)}^2}} }}} \right|, \\

\Rightarrow D = \dfrac{{\left| { - 9 + 0 - 12 + 13} \right|}}{{\sqrt {25 + 144} }}, \\

\Rightarrow D = \dfrac{{\left| { - 8} \right|}}{{\sqrt {169} }}, \\

$

$ \Rightarrow D = \dfrac{8}{{13}}$

Therefore, the distance of points $\left( {1,1,1} \right)$ and $\left( { - 3,0,1} \right)$ from plane $\overrightarrow r .\left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$ are equal.

Note: In the distance formula used above, we used modulus sign just to ensure that the distance never comes out as negative. If we are getting its value negative, modulus will turn it positive.

Last updated date: 20th Sep 2023

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