please kk bantu saya beserta caranya

u = B + C
[tex]u = \left[\begin{array}{c}6&2&-3\end{array}\right] +\left[\begin{array}{c}3&0&-1\end{array}\right] \\ \\ \\ u=\left[\begin{array}{c}(6+3)&(2+0)&(-3-1)\end{array}\right] \\ \\ \\ u=\left[\begin{array}{c}9&2&-4\end{array}\right] [/tex]

Proyeksi u pada A adalah :
[tex]\frac{u\ .\ A}{IAI}= \frac{\left[\begin{array}{c}9&2&-4\end{array}\right].\left[\begin{array}{c}3&1&-2\end{array}\right]}{ \sqrt{(3^{2}) +(1^{2})+(-2^{2}) } } \\ \\ \\ = \frac{((2.(-2))-((-4).1) + ((-4).3)-(9.-2))+((9.1)-(2.3))}{\sqrt{9+1+4} } \\ \\ \\ = \frac{((-4)+4)+((-12)+18)+(9-6)}{\sqrt{9+1+4} } \\ \\ = \frac{0+6+3}{ \sqrt{14} } \\ \\ \\ = \frac{9}{ \sqrt{14} }. \frac{ \sqrt{14}}{ \sqrt{14}} = \frac{9}{14} \sqrt{14}[/tex]

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|A| = 4; |B|=√3 ; |A + B| = √2

|A - B| = ?

[tex]|A + B| = \sqrt{|A|^{2}+2AB+|B|^{2}} \\ \\ \sqrt{2}= \sqrt{4^{2}+2AB+( \sqrt{3}) ^{2}} \\ \\ ( \sqrt{2})^{2}=16+2AB+ 3 \\ \\ 2=19+2AB \\ \\ 2AB=-17 \\ \\ \\ |A - B| = \sqrt{|A|^{2}-2AB+|B|^{2}} \\ \\ |A - B| = \sqrt{4^{2}-(-17) +( \sqrt{3}) ^{2}} \\ \\ |A - B| = \sqrt{16+17 +3} \\ \\ |A - B| = \sqrt{36} \\ \\ |A - B| =6[/tex]