Question

5.3.1. Find the closest point in the plane spanned by $(1, 2, -1)^T$, $(0, -1, 3)^T$ to the point $(1, 1, 1)^T$. What is the distance between the point and the plane? 5.3.2. Redo Exercise 5.3.1 using (a) the weighted inner product $(v, w) = 2v_1w_1 + 4v_2w_2 + 3v_3w_3$; (b) the inner product $(v, w) = v^TCw$ based on the positive definite matrix $C = \begin{pmatrix} 2 & -1 & 0 \ -1 & 2 & -1 \ 0 & -1 & 2 \end{pmatrix}$.

          5.3.1. Find the closest point in the plane spanned by $(1, 2, -1)^T$, $(0, -1, 3)^T$ to the point $(1, 1, 1)^T$. What is the distance between the point and the plane?
5.3.2. Redo Exercise 5.3.1 using
(a) the weighted inner product $(v, w) = 2v_1w_1 + 4v_2w_2 + 3v_3w_3$; (b) the inner product $(v, w) = v^TCw$ based on the positive definite matrix $C = \begin{pmatrix} 2 & -1 & 0 \ -1 & 2 & -1 \ 0 & -1 & 2 \end{pmatrix}$.

5.3.1. Find the closest point in the plane spanned by (1, 2, -1)^T, (0, -1, 3)^T to the point (1, 1, 1)^T. What is the distance between the point and the plane?
5.3.2. Redo Exercise 5.3.1 using
(a) the weighted inner product (v, w) = 2v1w1 + 4v2w2 + 3v3w3; (b) the inner product (v, w) = v^TCw based on the positive definite matrix C =
< p m a t r i x >.

Added by Derrick V.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

07/20/2023

Step 1

To do this, we can take the cross product of the two vectors to find the normal vector to the plane. Show more…

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Please do 5.3.2 part b. This was the answer for 5.3.1: closest point = ( 6/7 , 38/35 , 36/35 ) distance between v & w = 1/sqrt(35) 5.3.1. Find the closest point in the plane spanned by 1,2,-1,0,-1,3 to the point 1,1,1. What is the distance between the point and the plane? 5.3.2. Redo Exercise 5.3.1 using the weighted inner product vw=2w+4w+3Wb the inner product 2 -1 0 vw=vCw based on the positive definite matrix C 2 -1 0 -1 2