Question

Find an equation for the plane that passes through the point (-7,10,8), is parallel to the line x= 6−t, y= 9 +t, z=−5+ 2t,and intersects the plane 2x+y−z=e^pi^sqrt2at a 30◦angle.

Answer 1

The equation of the plane will be

$$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$$

Given $x_1=-7, y_1=10, z_1=8.$

$\vec n=\langle a, b, c \rangle$

The plane is parallel to the line $x= 6−t, y= 9 +t, z=−5+ 2t$

$$\vec n\cdot\langle -1, 1, 2\rangle=0$$

$$-a+b+2c=0$$

The plane intersects the plane $2x+y−z=e\cdot\pi\cdot\sqrt{2}$ at a $30\degree.$ angle.

$$\cos 30\degree=\dfrac{2a+b-c}{\sqrt{a^2+b^2+c^2}\sqrt{4+1+1}}$$

$$9a^2+9b^2+9c^2=2(2a+b-c)^2$$

Let $a=1.$ Then

$$b=1-2c$$

$$9+9(1-2c)^2+9c^2=2(2+1-2c-c)^2$$

$$9(1+1-4c+4c^2+c^2)=2(9)(1-2c+c^2)$$

$$5c^2-4c+2=2c^2-4c+2$$

$$3c^2=0=>c=0$$

$$b=1$$

Substitute

$$1(x-(-7))+1(y-10)+0(z-8)=0$$ $$x+7+y-10=0$$

The equation of the plane will be

$$x+y=3$$