Question

Consider the following planes. x + y + z = 6, x + 7y + 7z = 6 (a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.) (x(t), y(t), z(t)) = (b) Find the angle between the planes. (Round your answer to one decimal place.) °

Answer 1

a.
`x=6,y=-6t,z=6t`

b.
`\theta=29.5^(\circ)`

Explanation:

We are given that

`x+y+z=6`

`x+7y+7z=6`

a.Substitute z=0

`x+y=6`...(1)

`x+7y=6`..(2)

Subtract equation (1) from equation (2)

`6y=0`

`y=0`

Substitute y=0 in equation(1)

`x=6`

The point (6,0,0) lie on a line.

`r_0=(x_0,y_0,z_0)=(6,0,0)`

Let
`A=<1,1,1>`

`B=<1,7,7>`

`A* B=\begin{vmatrix}i&j&k\\1&1&1\\1&7&7\end{vmatrix}`

`A* B=i(7-7)-j(7-1)+k(7-1)=-6j+6k`

Therefore, the vector
`a'=(a,b,c)=<0,-6,6>`

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`

Using the formula

The parametric equation is given by

`x=6,y=-6t,z=6t`

Angle between two plane

`a_1x+b_1y+c_1z=d_1`

and
`a_2x+b_2y+c_2z=d_2`

`cos\theta=((a_1,b_1,c_1)\cdot (a_2,b_2,c_2))/(√(a^2_1+b^2_1+c^2_1)\cdot √(a^2_2+b^2_2+c^2_2))`

Using the formula

`cos\theta=((1,1,1)\cdot(1,7,7))/(√(1+1+1)* √(1+7^2+7^2))`

`cos\theta=(1+7+7)/(\sqrt 3* 3√(11))`

`cos\theta=(15)/(3√(33))}=(5)/(√(33))`

`\theta=cos^(-1)(0.87)=29.5^(\circ)`

Where
`\theta` in degree.