Question

how do I find an equation for the plane that passes through the point ( 2,-1 3) and is perpendicular to to the line v= (1,-2,2,)=+t(3,-2,4)

Answer 1

Question 23292

First, let us rewrite the equation of the line $x(t) = 1 + 3t$ ; $y(t) = -2 - 2t$ ; $z(t) = 2 + 4t$ in symmetrical form: $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 2}{4}$ . So, vector $\vec{a}(3; -2; 4)$ goes along this line. The equation of plane, which goes through point $(x_0, y_0, z_0)$ , with normal vector to it $\vec{n}(A, B, C)$ is $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$ . This plane must be perpendicular to $\vec{a}$ , it means that $\vec{a} \| \vec{n}$ , so in order to find $\vec{n}$ , we need to make $\vec{a}$ unit vector. The length of $\vec{a}$ is $|\vec{a}| = \sqrt{9 + 4 + 16} = \sqrt{29}$ . Hence, $\vec{n}\left(\frac{3}{\sqrt{29}}; -\frac{2}{\sqrt{29}}; \frac{4}{\sqrt{29}}\right)$ , and equation of plane is $\frac{3}{\sqrt{29}}(x - 2) - \frac{2}{\sqrt{29}}(y + 1) + \frac{4}{\sqrt{29}}(z - 3) = 0$ , or $3(x - 2) - 2(y + 1) + 4(z - 3) = 0$ , or in canonical form $3x - 2y + 4z - 20 = 0$ .