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Three Dimensional Geometry

Question
CBSEENMA12033387

Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector 3 straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top.

Solution

Here   p = 7
and    straight n with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top
therefore     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 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top over denominator square root of 9 plus 25 plus 36 end root end fraction space equals space fraction numerator 1 over denominator square root of 70 end fraction left parenthesis 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top right parenthesis
The vector 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 open parentheses fraction numerator 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top over denominator space square root of 70 end fraction close parentheses space equals space 7.