# Find the angle between​​ ​​vector A = i + j - 2k and vector B = i + j - k

8 views
in Physics
closed

Find the angle between $\vec A=\hat i+\hat j-2\hat k \,and\, \vec B=\hat i+\hat j-\hat k$

+1 vote
by (9.6k points)
selected by

We have

$\vec A.\vec B$= | A | | B | cos θ

Here  | A | = $\sqrt{1^2+1^2+(-2)^2}=\sqrt6$

| B | =$\sqrt{1^2+1^2+(-1)^2}=\sqrt3$

$\vec A.\vec B=(\hat i+\hat j-\hat k).(\hat i+\hat j-\hat k)$

= 1 × 1 + 1 × 1 + (-2)(-1)

= 1 + 1 + 2 = 4

∴   cos θ = $\frac{\vec A.\vec B}{|A||B|}=\frac{4}{\sqrt6\sqrt3}$

cos θ = $\frac{4}{\sqrt{18}}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}