Vector-Vector Product - Linear Algebra

Card 0 of 184

Compute $A\cdot B$ , where

$A=\begin{bmatrix} 2\ 3\ 4\ \end{bmatrix}$

$B=\begin{bmatrix} 1 & 3 & 5 \end{bmatrix}$

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Before we compute the product of , and , we need to check if it is possible to take the product. We will check the dimensions. is $3\times1$ , and is $1\times3$ , so the dimensions of the resulting matrix will be $3\times3$ . Now let's compute it.

$A\cdot B=\begin{bmatrix} 2\cdot1 & 2\cdot3 & 2\cdot 5\ 3 \cdot 1& 3\cdot3 & 3 \cdot 5 \ 4 \cdot 1 & 4\cdot3 & 4\cdot 5 \end{bmatrix} =\begin{bmatrix} 2 & 6& 10\ 3& 9& 15 \ 4& 12& 20 \end{bmatrix}$

Find the vector-vector product of the following vectors.

$A=[1 \ 2]$

$B=\begin{bmatrix} 3\ 5 \end{bmatrix}$

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$A\cdot B=1\cdot3+2\cdot5=3+10=13$

Calculate $\bf{A} \times \bf{B}$ , given

$\textbf{A}= 4\hat{i}-2\hat{j}$

$\textbf{B}= 1\hat{i}+3\hat{j}$

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By definition,

$\textbf{A} \times \textbf{B} = \begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \ 4& -2& 0\ 1& 3& 0 \end{vmatrix} =\hat{i}(0)-\hat{j}(0)+\hat{k}[4(3)-1(-2)] = 14 \hat{k}$ .

What is the physical significance of the resultant vector $\bf{C}$ , if $\bf{C}= \bf{A} \times \bf{B}$ ?

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By definition, the resultant cross product vector (in this case, $\bf{C}$ ) is orthogonal to the original vectors that were crossed (in this case, $\bf{A}$ and $\bf{B}$ ). In , this means that $\bf{C}$ is a vector that is normal to the plane containing $\bf{A}$ and $\bf{B}$ .

$\begin{align}&$\text{Find the product }$A\times B \&$\text{Where }$A=\begin{bmatrix}-18&12&-11&4&-19&13\end{bmatrix}$\text{ and }$B=\begin{bmatrix}-6\-2\-17\-20\-8\10\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since A has dimensions: }$1\times6\&$\text{and B has dimensions: }$6\times1\&$\text{The two vectors can be multiplied:}$\&\begin{bmatrix}-18&12&-11&4&-19&13\end{bmatrix}\times\begin{bmatrix}-6\-2\-17\-20\-8\10\end{bmatrix}\&(-18)(-6)+(12)(-2)+(-11)(-17)+(4)(-20)+(-19)(-8)+(13)(10)\&473\end{align}$

$\begin{align}&$\text{Find the vector product }$\&\begin{bmatrix}-7&-6\end{bmatrix}\times\begin{bmatrix}11\-9\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since our vectors have dimensions: }$1\times2$\text{ and }$2\times1\&$\text{The two can be multiplied to find a product:}$\&\begin{bmatrix}-7&-6\end{bmatrix}\times\begin{bmatrix}11\-9\end{bmatrix}\&(-7)(11)+(-6)(-9)\&-23\end{align}$

$\begin{align}&$\text{Find the product }$A\times B \&$\text{Where }$A=\begin{bmatrix}18&-15&10&-15&-14&11\end{bmatrix}$\text{ and }$B=\begin{bmatrix}6\-5\-13\9\10\-8\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since A has dimensions: }$1\times6\&$\text{and B has dimensions: }$6\times1\&$\text{The two vectors can be multiplied:}$\&\begin{bmatrix}18&-15&10&-15&-14&11\end{bmatrix}\times\begin{bmatrix}6\-5\-13\9\10\-8\end{bmatrix}\&(18)(6)+(-15)(-5)+(10)(-13)+(-15)(9)+(-14)(10)+(11)(-8)\&-310\end{align}$

$\begin{align}&$\text{Find the value of }$C$\text{ if }$C=A\times B \&$\text{Where }$A=\begin{bmatrix}16&-1&6&-20\end{bmatrix}$\text{ and }$B=\begin{bmatrix}10\19\-19\1\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since A has dimensions: }$1\times4\&$\text{and B has dimensions: }$4\times1\&$\text{It is possible to find a value for }$C:\&C=\begin{bmatrix}16&-1&6&-20\end{bmatrix}\times\begin{bmatrix}10\19\-19\1\end{bmatrix}\&C=(16)(10)+(-1)(19)+(6)(-19)+(-20)(1)\&C=7\end{align}$

$\begin{align}&$\text{Find the product }$A\times B \&$\text{Where }$A=\begin{bmatrix}13&8\end{bmatrix}$\text{ and }$B=\begin{bmatrix}-12\7\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since A has dimensions: }$1\times2\&$\text{and B has dimensions: }$2\times1\&$\text{The two vectors can be multiplied:}$\&\begin{bmatrix}13&8\end{bmatrix}\times\begin{bmatrix}-12\7\end{bmatrix}\&(13)(-12)+(8)(7)\&-100\end{align}$

$\begin{align}&$\text{Find the vector product }$\&\begin{bmatrix}-11&19&13&15&-18&6\end{bmatrix}\times\begin{bmatrix}3\-14\9\-15\-16\10\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since our vectors have dimensions: }$1\times6$\text{ and }$6\times1\&$\text{The two can be multiplied to find a product:}$\&\begin{bmatrix}-11&19&13&15&-18&6\end{bmatrix}\times\begin{bmatrix}3\-14\9\-15\-16\10\end{bmatrix}\&(-11)(3)+(19)(-14)+(13)(9)+(15)(-15)+(-18)(-16)+(6)(10)\&-59\end{align}$

$\begin{align}&$\text{Find the product }$A\times B \&$\text{Where }$A=\begin{bmatrix}13&-7&-14\end{bmatrix}$\text{ and }$B=\begin{bmatrix}-11\12\-4\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since A has dimensions: }$1\times3\&$\text{and B has dimensions: }$3\times1\&$\text{The two vectors can be multiplied:}$\&\begin{bmatrix}13&-7&-14\end{bmatrix}\times\begin{bmatrix}-11\12\-4\end{bmatrix}\&(13)(-11)+(-7)(12)+(-14)(-4)\&-171\end{align}$

$\begin{align}&$\text{Find the value of }$C$\text{ if }$C=A\times B \&$\text{Where }$A=\begin{bmatrix}-5&5&-16&-19&8\end{bmatrix}$\text{ and }$B=\begin{bmatrix}18\13\-11\-11\-12\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since A has dimensions: }$1\times5\&$\text{and B has dimensions: }$5\times1\&$\text{It is possible to find a value for }$C:\&C=\begin{bmatrix}-5&5&-16&-19&8\end{bmatrix}\times\begin{bmatrix}18\13\-11\-11\-12\end{bmatrix}\&C=(-5)(18)+(5)(13)+(-16)(-11)+(-19)(-11)+(8)(-12)\&C=264\end{align}$

$\begin{align}&$\text{Find the vector product }$\&\begin{bmatrix}-18&-4&-20\end{bmatrix}\times\begin{bmatrix}-9\-11\-11\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since our vectors have dimensions: }$1\times3$\text{ and }$3\times1\&$\text{The two can be multiplied to find a product:}$\&\begin{bmatrix}-18&-4&-20\end{bmatrix}\times\begin{bmatrix}-9\-11\-11\end{bmatrix}\&(-18)(-9)+(-4)(-11)+(-20)(-11)\&426\end{align}$

$\begin{align}&$\text{Find the vector product }$\&\begin{bmatrix}-8&-19&5&-9&2\end{bmatrix}\times\begin{bmatrix}-5\-14\0\10\12\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since our vectors have dimensions: }$1\times5$\text{ and }$5\times1\&$\text{The two can be multiplied to find a product:}$\&\begin{bmatrix}-8&-19&5&-9&2\end{bmatrix}\times\begin{bmatrix}-5\-14\0\10\12\end{bmatrix}\&(-8)(-5)+(-19)(-14)+(5)(0)+(-9)(10)+(2)(12)\&240\end{align}$

$\begin{align}&$\text{Find the vector product }$\&\begin{bmatrix}1&-7&8&9&-9\end{bmatrix}\times\begin{bmatrix}-20\-8\-7\-2\-20\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since our vectors have dimensions: }$1\times5$\text{ and }$5\times1\&$\text{The two can be multiplied to find a product:}$\&\begin{bmatrix}1&-7&8&9&-9\end{bmatrix}\times\begin{bmatrix}-20\-8\-7\-2\-20\end{bmatrix}\&(1)(-20)+(-7)(-8)+(8)(-7)+(9)(-2)+(-9)(-20)\&142\end{align}$

$\begin{align}&$\text{Find the vector product }$\&\begin{bmatrix}-3&-11&-9&-8&3\end{bmatrix}\times\begin{bmatrix}8\-3\4\-20\8\end{bmatrix}\end{align}$

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$\begin{align}&$\text{In order to multiply two vectors, }$A\times b$\text{, the respective dimensions of each}$\&$\text{must be of the form }$1\times n$\text{ and }$n\times 1\&$\text{Note that order does matter:}$\&(A\times b \neq b \times A)\&$\text{Since our vectors have dimensions: }$1\times5$\text{ and }$5\times1\&$\text{The two can be multiplied to find a product:}$\&\begin{bmatrix}-3&-11&-9&-8&3\end{bmatrix}\times\begin{bmatrix}8\-3\4\-20\8\end{bmatrix}\&(-3)(8)+(-11)(-3)+(-9)(4)+(-8)(-20)+(3)(8)\&157\end{align}$

Let $\bf{a}$ and $\bf{b}$ be vectors defined by

$\bf{a} = [ 9, 0, 9, 5], \bf{b} = [4, 3, 0, 9]$ .

Find the dot product $\bf{a}\cdot\bf{b}$ .

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Vectors $\bf{a}$ and $\bf{b}$ are both of length 4. The dimensions match and the dot product exists.

$\bf{a}\cdot\bf{b} = [ 9, 0, 9, 5]\cdot [4, 3, 0, 9] = (9)(4)+(0)(3)+(9)(0)+(5)(9)=81$

Let $\bf{a}$ and $\bf{b}$ be vectors defined by

$\bf{a} = [8, -1, 4], \bf{b} = [7, 4, 3]$ .

Find the cross product $\bf{a}\times\bf{b}$ .

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We can find the cross product by calculating the determinant of the following matrix

$\bf{a}\times\bf{b} = \det \begin{bmatrix} i& j & k\ 8 & -1 & 4\ 7&4&3 \end{bmatrix}$

$\begin{align} \bf{a}\times\bf{b} &= i ((-1)(3)-(4)(4))-j((8)(3)-(4)(7))+k((8)(4)-(-1)(7)) \ \bf{a}\times\bf{b} &= -19i+4j+39k \end{align}$

Let $\bf{u}$ and $\bf{v}$ be vectors defined by

$\bf{u} = [4,-3, 5], \bf{v}=[8, -2, 4]$ .

Find the cross product $\bf{u}\times\bf{v}$ .

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We find the cross product by finding the determinant of the following matrix

$\bf{u}\times \bf{v}=\det \begin{bmatrix} i & j & k\ 4 & -3 & 5\ 8 & -2 & 4 \end{bmatrix}$

$\begin{align} \bf{u}\times \bf{v} &= i ((-3)(4)-(5)(-2)) - j((4)(4)-(5)(8)) + k((4)(-2)-(-3)(8)) \ \bf{u}\times \bf{v} &= -2i + 24j +16k \end{align}$

$\textbf{u} = (1, x, $x^{2}$, x ^{3}, $x^{4}$, $x^{5}$)$

$\textbf{v} = $(y^{5}$, $y^{4}$, $y^{3}$, $y^{2}$, y, 1)$

The expression $\textbf{u} \cdot \textbf{v}$ yields a polynomial of what degree?

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The dot product $\textbf{u} \cdot \textbf{v}$ is the sum of the products of entries in corresponding positions, so

$\textbf{u} \cdot \textbf{v} = 1 $(y^{5}$) + x $(y^{4}$)+ $x^{2}$$(y^{3}$)+ $x^{3}$$(y^{2}$)+ $x^{4}$ (y)+ $x^{5 }$(1)$

$\textbf{u} \cdot \textbf{v} = $y^{5}$ + x $y^{4}$ + $x^{2}$$y^{3}$+ $x^{3}$$y^{2}$+ $x^{4}$ y+ $x^{5 }$$

The degree of a term of a polynomial is the sum of the exponents of its variables. Each term in this polynomial has exponent sum 5, so each term has degree 5. The degree of the polynomial is the greatest of the degrees, so the polynomial has degree 5.