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Given a basis vector, basis set v1, v2, v3 of basis vectors, the gramsmith process gives us a basis w1, w to w3 such that, an orthogonal basis such that, w1 is simply v1, w2, minus the projection of v2 onto w1, which is given by v2 .w1 over w1 .w1.
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W1.
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2 .w .1.
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And w3 equals v3 minus the projection of v3 onto w1 minus the projection of v3 onto w2.
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So to find w1, w2 and w3, we first have to calculate all these terms.
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Now here, we were given v1 is 0 .0 .6.
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V2 is the vector 0 .1 .1 .1 and v3 is the vector 1 .1 .1.
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Now because w1, we simply take as v1, we have w1 equals v1 equals 0 .0 .6.
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Now, v2 .w1 is 0 .1 .1.
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Dot product, 0 .0 .6, which turns out to be 0 plus 0 plus 6, that is 6.
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And w1 .w1 turns out to be 0 .6 .0 .0 .6, which is 36.
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So we can now find w2.
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Therefore, w2 equals v2 minus the projection of v2 onto w1, which is given by this expression.
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That is w2 equals 0 .1 .1 which is v2 minus v2.
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V2 .w1 is 6.
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So minus 6 over w1 .w1 is 36 and w1 is times w1 which is 0 .0.
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0 .6.
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So this is w2 equals 0 .1 .1 minus 6 over 36 is 1 over 6.
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Multiplying that by 6.
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We get 0 .0 .1.
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That is w2 equals the vector 0 .1 .0.
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Now we have found out w1 and w2.
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Next we'll find w3.
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So finally, first we'll find out the projection of v3 onto w1 and the projection of v3 onto w2.
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So we have v3 .w1.
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Is 1 .1 .1 .1...