diketahui vektor â=4î+2j,b=î+pj dan sudut antara â dan b adalah [tex] \frac{1}{4} \pi[/tex] tentukan nilai p

Sudut yang di bentuk oleh vektor [tex]\bar u[/tex] dan [tex]\bar v[/tex] didefinisikan sebagai:

[tex]\cos\theta=\dfrac{\bar{u}\cdot\bar{v}}{\left\Vert \bar{u}\right\Vert \cdot\left\Vert \bar{v}\right\Vert }[/tex]

Karena:

[tex]\bar{u}=4i+2j=\left(4,2\right) \\ \bar{v}=i+pj=\left(1,p\right) \\ \\ \bar{u}\cdot\bar{v}=\left(4,2\right)\cdot\left(1,p\right)=4+2p \\ \left\Vert \bar{u}\right\Vert =\sqrt{4^{2}+2^{2}}=\sqrt{20} \\ \left\Vert \bar{v}\right\Vert =\sqrt{1^{2}+p^{2}}[/tex]

Selanjutnya diperoleh:

[tex]\cos\left(\frac{\pi}{4}\right)=\dfrac{4+2p}{\sqrt{20}\cdot\sqrt{1+p^{2}}} \\ \dfrac{\sqrt{2}}{2}=\dfrac{4+2p}{\sqrt{20}\cdot\sqrt{1+p^{2}}} \\ \left(\dfrac{\sqrt{2}}{2}\right)^{2}=\left(\dfrac{4+2p}{\sqrt{20}\cdot\sqrt{1+p^{2}}}\right)^{2} \\ \frac{2}{4}=\frac{\left(4+2p\right)^{2}}{20\left(1+p^{2}\right)}[/tex]

[tex]40\left(1+p^{2}\right)=4\left(4+2p\right)^{2} \\ 40+40p^{2}=4\left(4p^{2}+16p+16\right) \\ 40+40p^{2}=16p^{2}+64p+64 \\ 40p^{2}-16p^{2}-64p+40-64=0 \\ 24p^{2}-64p-24=0 \\ 8\, \left( 3\,p+1 \right) \left( p-3 \right)=0[/tex]

Jadi solusinya [tex]p=3\quad atau p=- \frac{1}{3} [/tex]