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- Recall, 2 intersecting vectors $u$ and $v$ together with the resultant vetcor $w$; such that, $$ w = u - v $$
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- From the "Law of Cosines":
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- $$ Since, ||w||^2 = ||u||^2 + ||v||^2 - 2 * ||u|| * ||v|| * cos(\theta) $$
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- $$ Then, cos(\theta) = (||w||^2 - ||u||^2 - ||v||^2) / -2 * ||u|| * ||v||\ Lemma\.01 $$
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+ $$ Since, ||w||^2 = ||u||^2 + ||v||^2 - 2 * ||u||^2 * ||v||^2 * cos(\theta) $$
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+ $$ Then, cos(\theta) = (||w||^2 - ||u||^2 - ||v||^2) / -2 * ||u||^2 * ||v||^2 \ Lemma\.01 $$
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$$ ------- $$
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@@ -18,14 +18,23 @@ $$||u|| = \sqrt{{u_x}^2 + {u_y}^2 + {u_z}^2}$$
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$$ ||w|| = ||u-v|| = \sqrt{{(u_x - v_x)}^2 + {(u_y - v_y)}^2 + {(u_z - v_z)}^2} $$
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- $$ Then, 1)\ ||v||^2 = {v_x}^2 + {v_y}^2 + {v_z}^2 $$
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+ $$ Then,\ Lemma.02: $$
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- $$ 2 )\ ||u ||^2 = {u_x }^2 + {u_y }^2 + {u_z }^2$$
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+ $$ 1 )\ ||v ||^2 = {v_x }^2 + {v_y }^2 + {v_z }^2$$
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- $$ 3)\ ||w||^2 = {u_x - v_x}^2 + {u_y - v_y}^2 + {u_z - v_z}^2 $$
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+ $$ 2)\ ||u||^2 = {u_x}^2 + {u_y}^2 + {u_z}^2 $$
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+
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+ $$ 3)\ ||w||^2 = {u_x - v_x}^2 + {u_y - v_y}^2 + {u_z - v_z}^2 $$
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$$ ------- $$
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+ - By back-substitution in $Lemma.01$:
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+ $$ cos(\theta) = (||w||^2 - ||u||^2 - ||v||^2) / -2 * ||u||^2 * ||v||^2 = {(R.H.S)}_1 / {(R.H.S)}_2
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+
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+ $${(R.H.S)}_1 = (||w||^2 - ||u||^2 - ||v||^2) $$
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+ $$ {(R.H.S)}_2 = -2 * ||u||^2 * ||v||^2 = -2 * ({u_x}^2 + {u_y}^2 + {u_z}^2) * ({v_x}^2 + {v_y}^2 + {v_z}^2)
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+
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$$Hence, cos(\theta) = u.v / ||u|| * ||v|| $$
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$$ Thence, u.v = ||u|| * ||v|| * cos(\theta) $$
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