The perpendicular distance in units of the plane x-2y-3z-3-14-0 from origin is equal to

Given plane : nbsp;x 2y 3z 3√(14)=0 Normal vector of the plane is : î 2ĵ 3k̂ Unit vector normal to plane is : î 2ĵ 3k̂√(14) Now, converting the given equation ...

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Standard XII

Mathematics

Visualizing Conics from Right Circular Cone

The perpendic...

Question

The perpendicular distance (in units) of the plane $x - 2 y - 3 z - 3 \sqrt{14} = 0$ from origin is equal to

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Solution

Given plane : $x - 2 y - 3 z - 3 \sqrt{14} = 0$
Normal vector of the plane is : $^i - 2^j - 3^k$
Unit vector normal to plane is : $\frac{^i - 2^j - 3^k}{\sqrt{14}}$
Now,
converting the given equation of plane in normal form :
$\Rightarrow \frac{x}{\sqrt{14}} - \frac{2 y}{\sqrt{14}} - \frac{3 z}{\sqrt{14}} = 3$
So, perpendicular distance from origin is $3 units .$

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The perpendicular distance (in units) of the plane x−2y−3z−3√14=0 from origin is equal to

Given plane : x−2y−3z−3√14=0 Normal vector of the plane is : ^i−2^j−3^k Unit vector normal to plane is : ^i−2^j−3^k√14 Now, converting the given equation of plane in normal form : ⇒x√14−2y√14−3z√14=3 So, perpendicular distance from origin is 3 units.

The perpendicular distance (in units) of the plane $x - 2 y - 3 z - 3 \sqrt{14} = 0$ from origin is equal to