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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 (e) 4x + 8y + z – 8 = 0 and y + 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.

     (e)      4x + 8y + z – 8 = 0 and y + z – 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  4x + 8y + z -8 = 0\ and\ y + z - 4 = 0

Here,

a_{1} = 4,b_{1} = 8, c_{1} = 1   and   a_{2} = 0,b_{2} = 1, c_{2} = 1

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{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1},\frac{c_{1}}{c_{2}} = \frac{1}{1}

Clearly, the given planes are NOT parallel as  \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 4(0)+8(1)+1(1) =0+8+1 = 9 \neq 0.

Clearly, the given planes are NOT perpendicular.

Then finding 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{9}{\sqrt{4^2+8^2+1^2}.\sqrt{0^2+1^2+1^2}} \right |

     = \cos^{-1}\left | \frac{9}{\sqrt{81}.\sqrt{2}} \right |

    = \cos^{-1}\left ( \frac{9}{9\sqrt{2}} \right ) = \cos^{-1}\left ( \frac{1}{\sqrt{2}} \right )

   = 45^{\circ}

   

Posted by

Divya Prakash Singh

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