Explanation Given: The vectors are v = 〈 1 , 2 , 1 〉 and w = 〈 1 , 0 , − 1 〉 . Formula used: The dot product of the two vectors is a scalar. Formula for the Dot Product of the vectors: 〈 a , b , c 〉 ⋅ 〈 d , e , f 〉 = a d + b e + c f The two vectors are orthogonal if and only if their dot product is zero. Calculation: The vectors are v = 〈 1 , 2 , 1 〉 and w = 〈 1 , 0 , − 1 〉 . Let the vector which is to be find u = 〈 x , y , z 〉 . Now, find the dot product. u ⋅ v = 〈 x , y , z 〉 ⋅ 〈 1 , 2 , 1 〉 = x ( 1 ) + y ( 2 ) + z ( 1 ) = x + 2 y + z The dot product is zero because the vector is orthogonal. x + 2 y + z = 0 … (1) Now, find the dot product...

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Author: Rogawski

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Chapter 12.3, Problem 32E

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Find a vector that is orthogonal to $v = 〈 1, 2, 1 〉$ but not to $w = 〈 1, 0, - 1 〉$ .

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Chapter 12.3, Problem 31E

Chapter 12.3, Problem 33E

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4. Let F RNR be a mapping. (i) x ЄRN ? (ii) : What does it mean to say that F is differentiable at a point [1 Mark] In Theorem 5.4 in the Lecture Notes we proved that if F is differentiable at a point x E RN then F is continuous at x. Proof. Let (n) CRN be a sequence such that xn → x ЄERN as n → ∞. We want to show that F(xn) F(x), which means F is continuous at x. Denote hnxn - x, so that ||hn|| 0. Thus we find ||F(xn) − F(x)|| = ||F(x + hn) − F(x)|| * ||DF (x)hn + R(hn) || (**) ||DF(x)hn||+||R(hn)||| → 0, because the linear mapping DF(x) is continuous and for all large nЄ N, (***) ||R(hn) || ||R(hn) || ≤ → 0. ||hn|| (a) Explain in details why ||hn|| → 0. [3 Marks] (b) Explain the steps labelled (*), (**), (***). [6 Marks]

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Chapter 12 Solutions

Calculus: Early Transcendentals

Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E Ch. 12.1 - Prob. 4E

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Find a vector that is orthogonal to v = 〈 1 , 2 , 1 〉 but not to w = 〈 1 , 0 , − 1 〉 .

Find a vector that is orthogonal to v=〈1,2,1〉 but not to w=〈1,0,−1〉.

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Chapter 12 Solutions

Find a vector that is orthogonal to $v = 〈 1, 2, 1 〉$ but not to $w = 〈 1, 0, - 1 〉$ .