Question

Find the bisector of the acute angle between the planes : 3x 4y + 12z = 26 ; x + 2y - 2z = 9

Answer 1

Answer:

The bisector of acute angle given by 4 x + 14 y - 62 z  - 39 = 0  

Step-by-step explanation:

Given as :

The equation of two plane are

3 x + 4 y + 12 z = 26              .....1

x + 2 y - 2 z = 9                      .....2

Now, The bisector of acute angle between the planes

$\dfrac{3x + 4y +12z -26}{\sqrt{3^{2} +4^{2} +12^{2} } }$ = $\pm$ $\dfrac{x +2 y - 2z -9}{\sqrt{1^{2}+2^{2}+(-2)^{2} } }$

or, $\dfrac{3x + 4y + 12 z -26}{\sqrt{169} }$ = $\pm$ $\dfrac{x + 2y - 2z -9}{\sqrt{9} }$

Or, $\dfrac{3x + 4y + 12z -26}{13}$ = $\pm$ $\dfrac{x +2 y -2 z -9}{3}$

Taking + ve sign

Or, 3 ( 3 x + 4 y + 12 z -26) = 13 (x + 2 y - 2 z -9)

Or,  9 x + 12 y + 36 z - 78= 13 x + 26 y - 26 z - 117

or, 4 x + 14 y - 62 z  - 39 = 0  

Again

Taing -ve sign

3 ( 3 x + 4 y + 12 z -26) =  - 13 (x + 2 y - 2 z -9)

Or,  9 x + 12 y + 36 z - 78= - 13 x - 26 y + 26 z + 117

or, 22 x + 38 y + 10 z  - 195 = 0  

Now, Again

$a_1 a_2 + b_1 b_2 + c_1 c_2$ = 4 × 22 + 14 × 38 + ( - 62 ) × 10

I.e $a_1 a_2 + b_1 b_2 + c_1 c_2$ = 88 + 532 - 620

Or, $a_1 a_2 + b_1 b_2 + c_1 c_2$ = 620 - 620

So, $a_1 a_2 + b_1 b_2 + c_1 c_2$ = 0

Hence, The bisector of acute angle given by 4 x + 14 y - 62 z  - 39 = 0   Answer