Answer:
Step-by-step explanation:
Consider the following planes. x + y + z = 5, x + 3y + 3z = 5
the parametric equations for the line of intersection of the planes are determined as follows:
From the first plane, the normal of the first plane is [tex]n_1 = (1,1,1)[/tex]
from the second plane, the normal of the second plane is [tex]n_2 = (1,3,3)[/tex]
[tex]n_1 \times n_2 = \begin {vmatrix} \left \begin{array}{ccc}i&j&k\\1&1&1\\1&3&3 \end{array}\right \end {vmatrix}[/tex]
= i(3-3) -j(3-1)+k(3-1)
= i(0) - j(2) + k(2)
= -2j +2k
Suppose z = 0
x+y + 0 = 5 ---(1)
x+3y + 3(0) = 5 (2)
subtracting by elimination
-2y = 0
y = 0/-2
y = 0
The intersection on point of the plane is (5,0,0)
The equation of plane is r (t) =(5, 0, 0) + t(0, -2, 2)
∴
(x(t), y (t), z(t) ) = (5, -2t, 2t)
B) Find the angle between the planes
The angle between the planes can be represented by the equation:
[tex]cos \theta = \dfrac{a_1a_2+b_1b_2 + c_1c_2}{\sqrt{a^2_1+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}[/tex]
[tex]cos \theta = \dfrac{1 \times 1+1\times 3 +1 \times 3}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+3^2+3^2}}[/tex]
[tex]cos \theta = \dfrac{1+ 3 +3}{\sqrt{3}\sqrt{19}}[/tex]
[tex]cos \theta = \dfrac{7}{\sqrt{3}\sqrt{19}}[/tex]
[tex]\mathbf{\theta = cos ^{-1} (\dfrac{7}{\sqrt{57}})}[/tex]
