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Three Dimensional Geometry

Question
CBSEENMA12033404

Find the angle between the two planes
2 x + y – 2 z = 5 and 3x – 6 y – 2 z = 7 using vector method.

Solution

The equations of two planes are
2x + y – 2 z = 5 and 3x – 6y –2 z = 7
These equations can be written as
             open parentheses 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 close parentheses. space space open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses space equals space 5
and        open parentheses 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 close parentheses. space space open parentheses 3 space straight i with hat on top space minus space 6 space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses space equals space 7
or             straight r with rightwards arrow on top. space stack straight n subscript 1 with rightwards arrow on top space equals space 5 space space and space space straight r with rightwards arrow on top. space stack straight n subscript 2 with rightwards arrow on top space equals space 7
therefore space space space space space space space space space space stack straight n subscript 1 with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space straight j with hat on top space space minus space 2 space straight k with hat on top space space and space stack straight n subscript 2 with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 6 space straight j with hat on top space minus space 2 space straight k with hat on top
therefore space space space space space space open vertical bar stack straight n subscript 1 with rightwards arrow on top close vertical bar space equals space square root of left parenthesis 2 right parenthesis squared plus left parenthesis 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root space equals space square root of 4 plus 1 plus 4 end root space equals space square root of 9 space equals space 3
and space space space space open vertical bar stack straight n subscript 2 with rightwards arrow on top close vertical bar space equals space square root of left parenthesis 3 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root space equals space square root of 9 plus 36 plus 4 end root space equals square root of 49 space equals space 7
Also, stack straight n subscript 1 with rightwards arrow on top. space stack straight n subscript 2 with rightwards arrow on top space equals space left parenthesis 2 right parenthesis thin space left parenthesis 3 right parenthesis space plus space left parenthesis 1 right parenthesis space left parenthesis negative 6 right parenthesis space plus space left parenthesis negative 2 right parenthesis thin space left parenthesis negative 2 right parenthesis space equals space 6 minus 6 plus 4 space equals space 4
Let  θ be the angle between the planes
therefore space space space space space cos space straight theta space equals space fraction numerator stack straight n subscript 1 with rightwards arrow on top. space stack straight n subscript 2 with rightwards arrow on top over denominator open vertical bar stack straight n subscript 1 with rightwards arrow on top close vertical bar space open vertical bar stack straight n subscript 2 with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 4 over denominator left parenthesis 3 right parenthesis thin space left parenthesis 7 right parenthesis end fraction space equals space 4 over 21
therefore space space space space space space space straight theta space equals space cos to the power of negative 1 end exponent open parentheses 4 over 21 close parentheses.

Some More Questions From Three Dimensional Geometry Chapter

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