Mohon sekali dibantu yaa kak lagi belajar mandiri T^T

Mohon sekali dibantu yaa kak lagi belajar mandiri T^T

Jawab:

Penjelasan dengan langkah-langkah:

$4 = left| vec{a}right| cdot cos(theta) = dfrac{text{Large{$bold{overrightarrow{a} boldsymbol{cdot} overrightarrow{b}}$}}}{|vec{b}|}\\cos(theta) = dfrac{4}{sqrt{p^2 + 4^2 + 2^2}} = dfrac{text{Large{$bold{2p+2p+4cdot 1}$}}}{sqrt{2^2+p^2+1^2}cdotsqrt{p^2+4^2+2^2}}\$

$1 = dfrac{text{Large{$bold{p+1}$}}}{sqrt{2^2+p^2+1^2}}\\sqrt{5+p^2} = p+1\\p^2+5 = p^2+2p+1\\2p = 4\\boxed{text{Huge{$bold{p = 2}$}}}$

$tan(theta) = tan(a-b)\\tan(a) = dfrac{4}{8} = dfrac{1}{2} to a = tan^{-1}Big(dfrac{1}{2}Big)\\tan(b) = dfrac{-3}{4} to b = pi-tan^{-1}Big(dfrac{3}{4}Big)\\tan(theta) = tanBig( tan^{-1}Big(dfrac{1}{2}Big) - pi+tan^{-1}Big(dfrac{3}{4}Big)Big)\\tan(theta) = -tanBig(pi - tan^{-1}Big(dfrac{1}{2}Big) -tan^{-1}Big(dfrac{3}{4}Big)Big)\\tan(theta) = tanBig( tan^{-1}Big(dfrac{1}{2}Big) +tan^{-1}Big(dfrac{3}{4}Big)Big)\$

$tan(theta) = dfrac{cfrac{1}{2} + dfrac{3}{4}}{1-cfrac{1}{2}cdot cfrac{3}{4}}\\tan(theta) = dfrac{ dfrac{5}{4}}{cfrac{5}{8}} = 2\\theta = tan^{-1}(2) to cos(theta) = cos(tan^{-1}(2)) = dfrac{1}{sqrt{1^2+2^2}}$

panjang proyeksi :

$|vec{b_a} | = |a|cdot cos(theta)\\|vec{b_a}| = sqrt{8^2+4^2} cdot dfrac{1}{sqrt{5}}\\boxed{text{Huge{$boldsymbol{|vec{b_a}| = 4}$}}}$

$cos(theta) =dfrac{text{Large{$bold{overrightarrow{a} boldsymbol{cdot} overrightarrow{b}}$}}}{|vec{b}|cdot |vec{a}|}\\cos(theta) = dfrac{3cdot 1 + 4cdot 2 + 1cdot 1}{sqrt{3^2+4^2+1^2}cdotsqrt{1^2+2^2+1^2}}\\cos(theta) = dfrac{12}{sqrt{26}cdotsqrt{6}}\$

$cos(theta) = dfrac{6}{sqrt{39}}$

lalu tinggal aplikasikan panjang proyeksi = |a| * cos(theta)