a. Let \(\mathbf{v} = (1,0,1),\) and let p be the projection of v onto the plane \(2x + 3y + z = 0.\).

Compute v-p

b. Compute p (continuation of a)

c. Let \(V = (1,1,2),\) and let P be the point in the plane \(4x + y - 3z = 1\) closest to V.

godmathguy May 20, 2023

#1**0 **

The equation of the plane can be rewritten as z = -\frac{2}{3}x - \frac{1}{3}y. The normal vector to the plane is (2/3, 1/3, -1). The projection of v onto the plane is given by

p = \frac{v \cdot \hat{n}}{\hat{n} \cdot \hat{n}} \hat{n}

where n^ is the unit normal vector to the plane. Plugging in the values, we get

p = \frac{(1,0,1) \cdot (2/3, 1/3, -1)}{(2/3)^2 + (1/3)^2 + (-1)^2} (2/3, 1/3, -1) = \frac{1}{2} (2/3, 1/3, -1) = \boxed{\left(\frac{1}{3},\frac{1}{6},-\frac{1}{2}\right)}

Guest May 20, 2023