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  • In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. (a) 7 x + y + 6 z + 30 = 0 and 3 x – y – 10 z + 4 = 0

13 In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 

    (a)      7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Answers (1)

best_answer

Two planes

L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2},

are said to Parallel: 

If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

and Perpendicular:

If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

And the angle between L_{1}\ and\ L_{2} is given by the relation,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
So, given two planes  7x + 5y + 6z + 30 = 0\ and\ 3x -y - 10z + 4 = 0

Here,

a_{1} = 7,b_{1} = 5, c_{1} = 6   and   a_{2} = 3,b_{2} = -1, c_{2} = -10

So, applying each condition to check:

Parallel check:   \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\Rightarrow \frac{a_{1}}{a_{2}} =\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1},\frac{c_{1}}{c_{2}} = \frac{6}{-10}

Clearly, the given planes are NOT parallel.\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

\Rightarrow 7(3)+5(-1)+6(-10) = 21-5-60 = -44 \neq 0.

Clearly, the given planes are NOT perpendicular.

Then find the angle between them,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |

     = \cos^{-1}\left | \frac{-44}{\sqrt{7^2+5^2+6^2}.\sqrt{3^2+(-1)^2+(-10)^2}} \right |

     = \cos^{-1}\left | \frac{-44}{\sqrt{110}.\sqrt{110}} \right |

    = \cos^{-1}\left ( \frac{44}{110} \right )

    = \cos^{-1}\left ( \frac{2}{5} \right )

Posted by

Divya Prakash Singh

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