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Halle the value of the constant k so u and w are perpendicular

* Function of several variables * given the vector u = (2, −2, 0) and v = (0.2, −2) and w = v. ku− (a) the value of the constant to halle to

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Ojete chulete Points 0
u= (2, -2,0) and v= (0,2, -2) and w= ku− v. To be perpendicular, their scalar product must be equal to 0: (2; -2; 0) \* (2k-0; -2k-2; 0k+2) = 0; 4k +4k+4=0; 8k = -4; \ #### k= -1/2 Corroboro: (2; -2; 0) \* (-1-0; 1-2; 2) = -2+2+0=0 For the angle between v and w, with k=-2: v=<0; 2; -2>; w=<-4-0; 4-2; 2> or: w = <-4; 2; 2> The area of the parallelogram formed by these two vectors: A = v\*w = |v|\*|w| \* cos T; (0+4-4) = √(0+4+4) \* √ (16+4+4) \* Cos T; 0 / \[√(0+4+4) \* √ (16+4+4)\] = Cos T; T = Cos^(-1) 0; T=90°or 270°

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