Answer:
a.![x=6,y=-6t,z=6t](https://tex.z-dn.net/?f=x%3D6%2Cy%3D-6t%2Cz%3D6t)
b.![\theta=29.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D29.5%5E%7B%5Ccirc%7D)
Step-by-step explanation:
We are given that
![x+y+z=6](https://tex.z-dn.net/?f=x%2By%2Bz%3D6)
![x+7y+7z=6](https://tex.z-dn.net/?f=x%2B7y%2B7z%3D6)
a.Substitute z=0
...(1)
..(2)
Subtract equation (1) from equation (2)
![6y=0](https://tex.z-dn.net/?f=6y%3D0)
![y=0](https://tex.z-dn.net/?f=y%3D0)
Substitute y=0 in equation(1)
![x=6](https://tex.z-dn.net/?f=x%3D6)
The point (6,0,0) lie on a line.
![r_0=(x_0,y_0,z_0)=(6,0,0)](https://tex.z-dn.net/?f=r_0%3D%28x_0%2Cy_0%2Cz_0%29%3D%286%2C0%2C0%29)
Let ![A=](https://tex.z-dn.net/?f=A%3D%3C1%2C1%2C1%3E)
![B=](https://tex.z-dn.net/?f=B%3D%3C1%2C7%2C7%3E)
![A\times B=\begin{vmatrix}i&j&k\\1&1&1\\1&7&7\end{vmatrix}](https://tex.z-dn.net/?f=A%5Ctimes%20B%3D%5Cbegin%7Bvmatrix%7Di%26j%26k%5C%5C1%261%261%5C%5C1%267%267%5Cend%7Bvmatrix%7D)
![A\times B=i(7-7)-j(7-1)+k(7-1)=-6j+6k](https://tex.z-dn.net/?f=A%5Ctimes%20B%3Di%287-7%29-j%287-1%29%2Bk%287-1%29%3D-6j%2B6k)
Therefore, the vector ![a'=(a,b,c)=](https://tex.z-dn.net/?f=a%27%3D%28a%2Cb%2Cc%29%3D%3C0%2C-6%2C6%3E)
Line is parallel to vector a' and passing through the point (6,0,0).
The parametric equation is given by
![x=x_0+at,y=y_0+bt,z=z_0+ct](https://tex.z-dn.net/?f=x%3Dx_0%2Bat%2Cy%3Dy_0%2Bbt%2Cz%3Dz_0%2Bct)
Using the formula
The parametric equation is given by
![x=6,y=-6t,z=6t](https://tex.z-dn.net/?f=x%3D6%2Cy%3D-6t%2Cz%3D6t)
Angle between two plane
![a_1x+b_1y+c_1z=d_1](https://tex.z-dn.net/?f=a_1x%2Bb_1y%2Bc_1z%3Dd_1)
and ![a_2x+b_2y+c_2z=d_2](https://tex.z-dn.net/?f=a_2x%2Bb_2y%2Bc_2z%3Dd_2)
![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}}](https://tex.z-dn.net/?f=cos%5Ctheta%3D%5Cfrac%7B%28a_1%2Cb_1%2Cc_1%29%5Ccdot%20%28a_2%2Cb_2%2Cc_2%29%7D%7B%5Csqrt%7Ba%5E2_1%2Bb%5E2_1%2Bc%5E2_1%7D%5Ccdot%20%5Csqrt%7Ba%5E2_2%2Bb%5E2_2%2Bc%5E2_2%7D%7D)
Using the formula
![cos\theta=\frac{(1,1,1)\cdot(1,7,7)}{\sqrt{1+1+1}\times \sqrt{1+7^2+7^2}}](https://tex.z-dn.net/?f=cos%5Ctheta%3D%5Cfrac%7B%281%2C1%2C1%29%5Ccdot%281%2C7%2C7%29%7D%7B%5Csqrt%7B1%2B1%2B1%7D%5Ctimes%20%5Csqrt%7B1%2B7%5E2%2B7%5E2%7D%7D)
![cos\theta=\frac{1+7+7}{\sqrt 3\times 3\sqrt{11}}](https://tex.z-dn.net/?f=cos%5Ctheta%3D%5Cfrac%7B1%2B7%2B7%7D%7B%5Csqrt%203%5Ctimes%203%5Csqrt%7B11%7D%7D)
![\theta=cos^{-1}(0.87)=29.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%280.87%29%3D29.5%5E%7B%5Ccirc%7D)
Where
in degree.
Hello,
(u+v)^4= u^4+4^3v+6u²v²+4uv^3+v^4
with u=2x² and v=y²
(2x²+y²)^4=
16x^8 + 32x^6*y² +24x^4y^4 +8x²y^6 + y^8
All you have to do is subtract the two degrees to get your answer
Answer:
60°.
Step-by-step explanation:
It will be 66 because if u divide 22 from 3 you will get 66