Let {e⃗ 1, e⃗ 2, e⃗ 3, e⃗ 4, e⃗ 5, e⃗ 6}{e→1, e→2, e→3, e→4, e→5, e→6} be the standard basis in r6r6. find the length of the vector x⃗ =−2e⃗ 1+5e⃗ 2−5e⃗ 3+4e⃗ 4−3e⃗ 5+5e⃗ 6x→=−2e→1+5e→2−5e→3+4e→4−3e→5+5e→6.
The basis of the vector [tex]\vec{r}[/tex] is the set of unit vectors [tex]\{\hat{e_{1}}, \, \hat{e_{2}}, \, ..., \hat{e_{6}} \}[/tex]
The given vector is [tex]\vec{x} = -2\hat{e_{1}} +5\hat{e_{2}} - 5\hat{e_{3}} +4\hat{e_{4}} - 3\hat{e_{5}} +5\hat{e_{6}}[/tex]
By definition, the length, d, of the vector x is given by [tex]d= \sqrt{(-2)^{2} + (5)^{2} + (-5)^{2} +(4)^{2} + (-3)^{2} + (5)^{2} }}= \sqrt{104} [/tex]
