Linear Algebra - Vector-Vector Product

$\textbf{u},\textbf{v} , \textbf{w}, \textbf{x} \in \mathbb{R}^{3}$

Which of the following applies to $(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x})$ , where " $\cdot$ " and " $\times$ " refer to the dot product and the cross product of two vectors?

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{3}$

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}$

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in M_{2 \times 2}$

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x})$ is an undefined expression.

Explanation

The cross product of two vectors in $\mathbb{R}^{3}$ is also a vector in $\mathbb{R}^{3}$ . It follows that $\textbf{u} \times \textbf{v} \in \mathbb{R}^{3}$ and $\textbf{w} \times \textbf{x} \in \mathbb{R}^{3}$ ; it further follows that $(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{3}$ .

$\textbf{u} = (1, x, x^{2}, x ^{3}, x^{4}, x^{5}, x^{6})$

$(x+ 1)^{6} = \textbf{u} \cdot \textbf{v}$ , where $\textbf{v}$ is which vector?

$\textbf{v} = (1, 6, 15, 20, 15, 6, 1)$

$\textbf{v} = (1, 2, 3, 4, 5,6 )$

$\textbf{v} = (6, 5, 4, 3, 2, 1 )$

$\textbf{v} = (1, 6, 36, 216, 1296, 7776 , 46656 )$

$\textbf{v} = ( 46656, 7776, 1296, 216, 36, 6, 1 )$

Explanation

Let $\textbf{v}= (v_{0}, v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6})$

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

$\textbf{u} \cdot \textbf{v} = v_{0} + v_{1}x+ v_{2}x^{2} + v_{3}x^{3}+ v_{4} x^{4}+ v_{5}x^{5}+ v_{6}x^{6}$

Therefore, $\textbf{v}$ is the vector of coefficients of the powers of of $(x+ 1)^{6}$ , in ascending order of exponent.

By the Binomial Theorem,

$(x+ 1)^{6} = \sum_{k = 0}^{6} C(6, k)x^{k}$ .

Therefore, $\textbf{v}$ has as its entries the binomial coefficients for 6, which are:

It follows that $\textbf{v} = (1, 6, 15, 20, 15, 6, 1)$ .

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

$\bf{C}$ is orthogonal to both $\bf{A}$ and $\bf{B}$ .

$\bf{C}$ is a scalar.

$\bf{C}$ lies in the same plane that contains both $\bf{A}$ and $\bf{B}$ .

$\bf{C}$ is the projection of $\bf{A}$ onto $\bf{B}$ .

Explanation

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 $\bf{B}$ .

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

$\text{The multiplication cannot be performed.}$

Explanation

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

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

$\text{The multiplication cannot be performed.}$

Explanation

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

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

$\text{The multiplication cannot be performed.}$

Explanation

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

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

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

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

$\textbf{A} \times \textbf{B} = 14 \hat{k}$

$\textbf{A} \times \textbf{B} = 14 \hat{i}$

$\textbf{A} \times \textbf{B} = 14 \hat{j}$

$\textbf{A} \times \textbf{B} = 14$

Explanation

By definition,

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

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

$\text{The multiplication cannot be performed.}$

Explanation

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

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

$\text{The multiplication cannot be performed.}$

Explanation

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

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

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

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

None of the other choices gives a correct response.

Explanation

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

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

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

The degree of a term of a polynomial is the sum of the exponents of its variables; the individual terms have degrees 5, 4, 3, 7, 4, and 7, in that order. the degree of the polynomial is the highest of these, which is 7.

Vector-Vector Product

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