Three Dimensional Geometry

Find the vector equation of the plane which is at a distance of from the origin and its normal vector from the origin is Also find its cartesian form. - Wired Faculty

Question

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Find the vector equation of the plane which is at a distance of $fraction numerator 6 over denominator square root of 29 end fraction$ from the origin and its normal vector from the origin is $2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space space 4 space straight k with hat on top.$ Also find its cartesian form.

Solution

Here $straight p space equals space fraction numerator 6 over denominator square root of 29 end fraction$
and $straight n with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top$
$therefore space space space space open vertical bar straight n with rightwards arrow on top close vertical bar space equals space square root of left parenthesis 2 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root space equals space square root of 4 plus 9 plus 16 end root space equals space square root of 29 therefore space space space space space space straight n with hat on top space equals space fraction numerator straight n with rightwards arrow on top over denominator open vertical bar straight n with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 29 end fraction left parenthesis 2 straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis$
The required equation of plane is
$straight r with rightwards arrow on top. space straight n with hat on top space equals space straight p$
or $straight r with rightwards arrow on top. space fraction numerator 1 over denominator square root of 29 end fraction space left parenthesis 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis space equals space fraction numerator 6 over denominator square root of 29 end fraction$
or $stack straight r. with rightwards arrow on top left parenthesis 2 space straight i with hat on top space minus space space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis space equals space 6$
Taking $straight r with rightwards arrow on top space equals space straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top comma space we space get$
$left parenthesis straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top right parenthesis. space space space open parentheses 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses space equals space 6$
or 2x – 3 y + 4 z = 6, which is cartesian equation of plane.

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