part a of the questions asks us to find all vectors V such that <1,2,1>xV=<3,1,-
ID: 3215343 • Letter: P
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part a of the questions asks us to find all vectors V such that <1,2,1>xV=<3,1,-5>. I set up the cross product making vector v= . then I got it down to 3=2c-b 1=-c+a -5=b-2a. from there I don't know how to solve it. part b of this problem asks us to prove the <1,2,1>xV=<3,1,5> is not possible. this is problem number 44, section 12.4 in james stewart multivariable calculus, 7th editionExplanation / Answer
Let V = ai + bj + ck (i + 2j + k) x (ai + bj + ck) = 3i + j - 5k ---> bk - cj - 2ak + 2ci + aj - bi = 3i + j - 5k ---> 2c - b = 3, a - c = 1, 2a - b = 5. Now, from eq(ii): c = a - 1 so replacing c by a - 1, we get: 2a - 2 - b = 3 so 2a - b = 5, which = eq(iii) we can derive eq(iii) from (i) and (ii) which effectively gives us only 2 equations and 3 unknowns so this set of equations have infinite solutions.
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