Answer:
a.[tex]x=6,y=-6t,z=6t[/tex]
b.[tex]\theta=29.5^{\circ}[/tex]
Step-by-step explanation:
We are given that
[tex]x+y+z=6[/tex]
[tex]x+7y+7z=6[/tex]
a.Substitute z=0
[tex]x+y=6[/tex]...(1)
[tex]x+7y=6[/tex]..(2)
Subtract equation (1) from equation (2)
[tex]6y=0[/tex]
[tex]y=0[/tex]
Substitute y=0 in equation(1)
[tex]x=6[/tex]
The point (6,0,0) lie on a line.
[tex]r_0=(x_0,y_0,z_0)=(6,0,0)[/tex]
Let [tex]A=<1,1,1>[/tex]
[tex]B=<1,7,7>[/tex]
[tex]A\times B=\begin{vmatrix}i&j&k\\1&1&1\\1&7&7\end{vmatrix}[/tex]
[tex]A\times B=i(7-7)-j(7-1)+k(7-1)=-6j+6k[/tex]
Therefore, the vector [tex]a'=(a,b,c)=<0,-6,6>[/tex]
Line is parallel to vector a' and passing through the point (6,0,0).
The parametric equation is given by
[tex]x=x_0+at,y=y_0+bt,z=z_0+ct[/tex]
Using the formula
The parametric equation is given by
[tex]x=6,y=-6t,z=6t[/tex]
Angle between two plane
[tex]a_1x+b_1y+c_1z=d_1[/tex]
and [tex]a_2x+b_2y+c_2z=d_2[/tex]
[tex]cos\theta=\frac{(a_1,b_1,c_1)\cdot (a_2,b_2,c_2)}{\sqrt{a^2_1+b^2_1+c^2_1}\cdot \sqrt{a^2_2+b^2_2+c^2_2}}[/tex]
Using the formula
[tex]cos\theta=\frac{(1,1,1)\cdot(1,7,7)}{\sqrt{1+1+1}\times \sqrt{1+7^2+7^2}}[/tex]
[tex]cos\theta=\frac{1+7+7}{\sqrt 3\times 3\sqrt{11}}[/tex]
[tex]cos\theta=\frac{15}{3\sqrt{33}}}=\frac{5}{\sqrt{33}}[/tex]
[tex]\theta=cos^{-1}(0.87)=29.5^{\circ}[/tex]
Where [tex]\theta[/tex] in degree.
