Answer on Question #52356 – Mathematics – Vector Calculus
Question
For the following vectors: a = ( 1 , − 5 , 9 ) a = (1, -5, 9) a = ( 1 , − 5 , 9 ) , b = ( − 4 , 12 , − 6 ) b = (-4, 12, -6) b = ( − 4 , 12 , − 6 ) , c = ( − 3 , 5 , − 7 ) c = (-3, 5, -7) c = ( − 3 , 5 , − 7 ) . Calculate the following dot products:
a) a ⋅ b a \cdot b a ⋅ b ;
b) a ⋅ c a \cdot c a ⋅ c ;
c) b ⋅ c b \cdot c b ⋅ c .
Solution
Let us write the formula for the dot (or scalar) product in terms of vector components. For the two three-dimensional vectors
a ⃗ = ( a 1 , a 2 , a 3 ) = a 1 i ⃗ + a 2 j ⃗ + a 3 k ⃗ \vec{a} = (a_1, a_2, a_3) = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k} a = ( a 1 , a 2 , a 3 ) = a 1 i + a 2 j + a 3 k
and
b ⃗ = ( b 1 , b 2 , b 3 ) = b 1 i ⃗ + b 2 j ⃗ + b 3 k ⃗ , \vec{b} = (b_1, b_2, b_3) = b_1 \vec{i} + b_2 \vec{j} + b_3 \vec{k}, b = ( b 1 , b 2 , b 3 ) = b 1 i + b 2 j + b 3 k ,
where i ⃗ , j ⃗ , k ⃗ \vec{i}, \vec{j}, \vec{k} i , j , k – the standard unit vectors of orthogonal coordinate system, the dot product is defined as
( a ⃗ ⋅ b ⃗ ) = a 1 b 1 + a 2 b 2 + a 3 b 3 . (\vec{a} \cdot \vec{b}) = a_1 b_1 + a_2 b_2 + a_3 b_3. ( a ⋅ b ) = a 1 b 1 + a 2 b 2 + a 3 b 3 .
Hence, using (1)-(3) for the given vectors we get:
a)
( a ⃗ ⋅ b ⃗ ) = 1 ⋅ ( − 4 ) + ( − 5 ) ⋅ 12 + 9 ⋅ ( − 6 ) = − 4 − 60 − 54 = − 118 , (\vec{a} \cdot \vec{b}) = 1 \cdot (-4) + (-5) \cdot 12 + 9 \cdot (-6) = -4 - 60 - 54 = -118, ( a ⋅ b ) = 1 ⋅ ( − 4 ) + ( − 5 ) ⋅ 12 + 9 ⋅ ( − 6 ) = − 4 − 60 − 54 = − 118 ,
b)
( a ⃗ ⋅ c ⃗ ) = 1 ⋅ ( − 3 ) + ( − 5 ) ⋅ 5 + 9 ⋅ ( − 7 ) = − 3 − 25 − 63 = − 91 , (\vec{a} \cdot \vec{c}) = 1 \cdot (-3) + (-5) \cdot 5 + 9 \cdot (-7) = -3 - 25 - 63 = -91, ( a ⋅ c ) = 1 ⋅ ( − 3 ) + ( − 5 ) ⋅ 5 + 9 ⋅ ( − 7 ) = − 3 − 25 − 63 = − 91 ,
c)
( b ⃗ ⋅ c ⃗ ) = ( − 4 ) ⋅ ( − 3 ) + 12 ⋅ 5 + ( − 6 ) ⋅ ( − 7 ) = 12 + 60 + 42 = 114. (\vec{b} \cdot \vec{c}) = (-4) \cdot (-3) + 12 \cdot 5 + (-6) \cdot (-7) = 12 + 60 + 42 = 114. ( b ⋅ c ) = ( − 4 ) ⋅ ( − 3 ) + 12 ⋅ 5 + ( − 6 ) ⋅ ( − 7 ) = 12 + 60 + 42 = 114.
Answer: a) ( a ⃗ ⋅ b ⃗ ) = − 118 (\vec{a} \cdot \vec{b}) = -118 ( a ⋅ b ) = − 118 ; b) ( a ⃗ ⋅ c ⃗ ) = − 91 (\vec{a} \cdot \vec{c}) = -91 ( a ⋅ c ) = − 91 ; c) ( b ⃗ ⋅ c ⃗ ) = 114 (\vec{b} \cdot \vec{c}) = 114 ( b ⋅ c ) = 114 .
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