Vector-Vector Product

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

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?

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

CORRECT

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

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

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x})$ 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?

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

CORRECT

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

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

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

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

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 $\bf{B}$ .

CORRECT

$\bf{C}$ 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 $\bf{B}$ .

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

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

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

CORRECT

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

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

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

The degree of a term of a polynomial is the sum of the exponents of its variables; the individual terms have degrees 0, 2, 4, 6, 8, 10, in that order. the degree of the polynomial is the highest of these, which is 10.

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

CORRECT

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

Explanation

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

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

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

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

$\bf{u}\times \bf{v} &= -2i + 24j +16k$

CORRECT

$\bf{u}\times \bf{v} &= -12i + 24j -16k$

$\bf{u}\times \bf{v} &= -2i - 24j +24k$

$\bf{u}\times \bf{v} &= -16i + 2j +8k$

The cross product does not exist.

Explanation

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

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

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

and are differentiable functions.

$A = \begin{bmatrix} f &g\end{bmatrix}$

$AB = \begin{bmatrix} \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{f}{g} \right )\end{bmatrix}$

Which value of makes this statement true?

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \ \ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

CORRECT

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \ \- \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \ \ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \ \ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \ \ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

Explanation

Recall the quotient rule of differentiation:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}}$

This can be rewritten as

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x}+ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g }{g^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}}+\frac{ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g }{g^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =f \left (- \frac{ \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}} \right )+g\left ( \frac{ \frac{\mathrm{d}f }{\mathrm{d} x} }{g^{2}} \right )$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =f \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x} +g\left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x}$

If $A = \begin{bmatrix} f &g\end{bmatrix}$ and $B = \begin{bmatrix} a \ b \end{bmatrix}$ ,

then multiply corresponding elements and add the products to get the sole element in :

$AB = \begin{bmatrix} f \cdot a + g \cdot b \end{bmatrix}$

Since we want

$AB = \begin{bmatrix} \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{f}{g} \right )\end{bmatrix}=\begin{bmatrix} f \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x} +g\left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x} \end{bmatrix}$ ,

It follows that, of the given choices, $a= \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x}$ and $b= \left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x}$ , and

$B = \begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \ \ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$ .

$\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ ; $\textbf{w}, \textbf{x} \in \mathbb{R}^{4}$

Which of the following applies to $(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x})$ , where " $\times$ " refers to the cross product of two vectors, and "" refers to either scalar or vector addition, as applicable?

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

CORRECT

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

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

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{2}$

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x}) \in M_{2 \times 3}$

Explanation

The cross product of two vectors is defined only if both vectors are in $\mathbb{R}^{3}$ . $\textbf{w}$ and $\textbf{x}$ are vectors in $\mathbb{R}^{4}$ , so $\textbf{w} \times \textbf{x}$ is undefined; consequently, so is $(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x})$ .

$\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ ; $\textbf{w}, \textbf{x} \in \mathbb{R}^{4}$

Which of the following applies to $(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x})$ , where " $\cdot$ " refers to the dot product of two vectors, and "" refers to either scalar or vector addition, as applicable?

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

CORRECT

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

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

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}^{4}$

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in M_{3 \times 4 }$

Explanation

The dot product of two vectors in the same vector space is a scalar quantity. $\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ , so $\textbf{u}$ and $\textbf{v}$ are in the same vector space; their dot product is defined, and $\textbf{u} \cdot \textbf{v} \in \mathbb{R}$ . For similar reasons, $\textbf{w} \cdot \textbf{x} \in \mathbb{R}$ . Therefore, their sum is defined, and $(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}$ .

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

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

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

CORRECT

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

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

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

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.

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