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Three Dimensional Geometry

Question
CBSEENMA12033390

Find the D.C.’s of the perpendicular from origin to the plane  straight r with rightwards arrow on top. space open parentheses negative 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top close parentheses plus 14 space equals space 0.  Find also the distance of the plane from the origin. 

Solution
The equation of plane is
                        straight r with rightwards arrow on top. space left parenthesis negative 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis space plus space 14 space equals space 0
or                 straight r with rightwards arrow on top. space left parenthesis 2 straight i with hat on top space plus space 3 straight j with hat on top space minus space 6 straight k with hat on top right parenthesis space equals space 14
Now,      open vertical bar 2 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space 6 space straight k with hat on top close vertical bar space equals space square root of 4 plus 9 plus 36 end root space equals space square root of 49 space equals space 7
therefore         2 over 7 straight i with hat on top space plus space 3 over 7 straight j with hat on top space minus space 6 over 7 straight k with hat on top space is space straight a space unit space vector
therefore space space space equation space left parenthesis 1 right parenthesis space of space plane space can space be space written space as space straight r with rightwards arrow on top. space open parentheses 2 over 7 straight i with hat on top space plus space 3 over 7 straight j with hat on top space minus space 6 over 7 straight k with hat on top close parentheses space equals space 1
which is of the form space straight r with rightwards arrow on top. space straight n with rightwards arrow on top space equals space straight p
therefore space space space straight n with rightwards arrow on top space equals space 2 over 7 straight i with hat on top space space plus 3 over 7 straight j with hat on top space minus space 6 over 7 straight k with hat on top comma which is perpendicular vector from the origin
∴ direction cosines of straight n with rightwards arrow on top are 2 over 7 comma space 3 over 7 comma space minus 6 over 7
Perpendicular distance = p = 1

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.