Question

What is the angle between the vectors A=6i+8j+20k and B=6i+8j+10k

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

$A = 6i + 8j + 20k$

$B = 6i + 8j + 10k$

$A•B =(6i+8j+20k)•(6i+8j+10k)$

$\: \: \: = (6*6) + (8*8) + (20*10)$

$\: \: \: = (36 + 64 + 200)$

$\: \: \: = 300$

$|A| =\sqrt{6^2 + 8^2 + 20^2}$

$\: \: =\sqrt{10^2 + 20^2}$

$\: \: =\sqrt{10^2 + 4*10^2}$

$\: \: =\sqrt{5*10^2}$

$\: \: = 10\sqrt{5}$

$|B| =\sqrt{6^2 + 8^2 + 10^2}$

$\: \: =\sqrt{10^2 + 10^2}$

$\: \: =\sqrt{2*10^2}$

$\: \: = 10\sqrt{2}$

$Now$

$\small\boxed{CosO=\dfrac{A • B}{|A| |B|}}$

$CosO =\frac{300}{10\sqrt{5} • 10\sqrt{2}}$

$CosO = \frac{300}{100\sqrt{10}}$

$π^2 = 10$

$CosO =\frac{3}{\sqrt{10}}=\frac{3}{π}$

$O = Cos^-¹(\frac{3}{π})$