Question

The distance from the origin to the plane
x+3y-2z+1=0 is
​

Answer 1

Step-by-step explanation:

Distance of a plane ax+by+cz+d=0 from origin is

= |d|/√a²+b²+c²

=1/√1+9+4

=1/√14

Answer 2

The distance from the origin to the plane x + 3y - 2z + 1 = 0 is 1/√14 unit

Given :

The plane x + 3y - 2z + 1 = 0

To find :

The distance from the origin to the plane x + 3y - 2z + 1 = 0

Formula Used :

The distance from the origin to the plane ax + by + cz + d = 0 is

$\displaystyle \sf{ = \bigg| \frac{d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } } \bigg| }$

Solution :

Step 1 of 2 :

Write down the given equation of the plane

Here the given equation of the plane is x + 3y - 2z + 1 = 0

Step 2 of 2 :

Calculate distance from the origin to the plane

Since distance from the origin to the plane ax + by + cz + d = 0 is

$\displaystyle \sf{ = \bigg| \frac{d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } } \bigg| }$

Hence the distance from the origin to the plane x + 3y - 2z + 1 = 0 is

$\displaystyle \sf{ = \bigg| \frac{1}{ \sqrt{ {1}^{2} + {3}^{2} + {( - 2)}^{2} } } \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \frac{1}{ \sqrt{1 + 9 + 4 } } \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \frac{1}{ \sqrt{14} } \bigg| } \: \: unit$

$\displaystyle \sf = \frac{1}{ \sqrt{14} } \: \: unit$

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