What I did was normalized v, which gives $v=\sqrt{0^{2}+1^{2}+-1^{2}}$ then I divided that by the norm of the vector with the same direction so that $u=\sqrt{2}/3$ and multiply that by vector v's coordinates. Is there something I'm missing? I got only partial credit on this one.
2026-04-06 12:38:33.1775479113
Find the vector v that has norm equal to 3 and has the same direction as the vector <0,1,-1>
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actually,you have to multiply 3/√2 with the vector coordinates.we know that the vector is parallel to <0,1,-1>.so,let it be <0,k,-k>.now,we also now that the modulus of the vector is 3.so,√(k^2+k^2)=3.this implies that k=3/√2.therefore,the vector is <0,3/√2,-3/√2>.