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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.

 May 20, 2023
 #1
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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)}

 May 20, 2023
 #2
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b. p = (3/8, -2/8, 1/4).

 May 21, 2023
 #3
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c. P = (2/3, 1, 2/3).

 May 21, 2023

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