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

Answer:

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

b.$\theta=29.5^{\circ}$

Step-by-step 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 $A=$

$B=$

$A\times B=\begin{vmatrix}i&j&k\\1&1&1\\1&7&7\end{vmatrix}$

$A\times B=i(7-7)-j(7-1)+k(7-1)=-6j+6k$

Therefore, the vector $a'=(a,b,c)=$

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 $a_2x+b_2y+c_2z=d_2$

$cos\theta=\frac{(a_1,b_1,c_1)\cdot (a_2,b_2,c_2)}{\sqrt{a^2_1+b^2_1+c^2_1}\cdot \sqrt{a^2_2+b^2_2+c^2_2}}$

Using the formula

$cos\theta=\frac{(1,1,1)\cdot(1,7,7)}{\sqrt{1+1+1}\times \sqrt{1+7^2+7^2}}$

$cos\theta=\frac{1+7+7}{\sqrt 3\times 3\sqrt{11}}$

$cos\theta=\frac{15}{3\sqrt{33}}}=\frac{5}{\sqrt{33}}$

$\theta=cos^{-1}(0.87)=29.5^{\circ}$

Where $\theta$ in degree.