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Linear Algebra : Vector-Vector Product

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Example Question #21 : Vector Vector Product

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

C(6,0) = c(6,6) = 1

C(6,1) = c(6,5) = 6

C(6,2) = c(6,4) = 15

C(6,3) = 20

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

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Example Question #22 : Vector Vector Product

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

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

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

Possible Answers:

None of the other choices gives a correct response.

Correct answer:

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.

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Example Question #23 : Vector Vector Product

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

$\textbf{u} \cdot \textbf{v}$ is equal to the fifth-degree Maclaurin series for $e^{2x}$ for:

Possible Answers:

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

$\textbf{v}=\left ( 1, 2, 1, \frac{1}{3}, \frac{1}{12}, \frac{1}{60} \right )$

$\textbf{v}=\left ( \frac{4}{15} , \frac{2}{3} , \frac{4}{3} , 2, 2, 1 \right )$

$\textbf{v}=\left ( \frac{1}{60} , \frac{1}{12} , \frac{1}{3}, 1, 2, 1\right )$

None of the other choices gives the correct response.

Correct answer:

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

Explanation:

The th-degree Maclaurin series for a function f(x) is the polynomial

$\frac{f(0)}{0!} + \frac{f'(0)}{1!}x+ \frac{f''(0)}{2!}x^{2}+ ... + \frac{f^{(n)}(0)}{n!}x^{n}$

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

then

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

Therefore, we want $\textbf{v}$ to be the vector of Maclaurin coefficients by ascending order of degree.

The fifth-degree Maclaurin series for $e^{x}$ is

$e^{x}=1+x+ \frac{1}{2}x^{2}+ \frac{1}{6}x^{3}+ \frac{1}{24}x^{4}+ \frac{1}{120}x^{5}$

The Maclaurin series for $e^{2x}$ can be derived from this by replacing with :

$e^{2x}=1+2x+ \frac{1}{2}(2x)^{2}+ \frac{1}{6}(2x)^{3}+ \frac{1}{24}(2x)^{4}+ \frac{1}{120}(2x)^{5}$

$e^{2x}=1+2x+ \frac{1}{2}(4x^{2})+ \frac{1}{6}(8x^{3})+ \frac{1}{24}(16x^{4})+ \frac{1}{120}(32x^{5})$

$e^{2x}=1+2x+ 2 x^{2}+ \frac{4}{3}x^{3}+ \frac{2}{3}x^{4}+ \frac{4}{15} x^{5}$

Therefore,

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

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Example Question #24 : Vector Vector Product

$\textbf{u} = (-3, 2, 4 ,1 )$

$\textbf{v} = (-2, 0, 1, 4)$

Which of the following is undefined, $\textbf{u} \cdot \textbf{v}$ or $\textbf{u} \times \textbf{v}$ ?

Possible Answers:

$\textbf{u} \cdot \textbf{v}$

Neither

$\textbf{u} \times \textbf{v}$

Both

Correct answer:

$\textbf{u} \times \textbf{v}$

Explanation:

$\textbf{u} \cdot \textbf{v}$ , the dot product of the vectors, is a defined quantity if and only if both vectors are elements of the same vector space. Each has four entries, so both are in $\mathbb{R}^{4}$ . Consequently, $\textbf{u} \cdot \textbf{v}$ is defined.

$\textbf{u} \times \textbf{v}$ , the cross product of the vectors, is a defined vector if and only if both vectors are elements in $\mathbb{R}^{3}$ . As previously mentioned, they are in $\mathbb{R}^{4}$ , so $\textbf{u} \times \textbf{v}$ is undefined.

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Example Question #25 : Vector Vector Product

$\textbf{u} = (3, 7, -2)$

$\textbf{v} = (1, -5 )$

Which of the following is undefined, $\textbf{u} \cdot \textbf{v}$ or $\textbf{u} \times \textbf{v}$ ?

Possible Answers:

$\textbf{u} \times \textbf{v}$

Neither

$\textbf{u} \cdot \textbf{v}$

Both

Correct answer:

Both

Explanation:

$\textbf{u} \cdot \textbf{v}$ , the dot product of the vectors, is a defined quantity if and only if both vectors are elements of the same vector space. $\textbf{u}$ has three entries, so $\textbf{u} \in \mathbb{R}^{3}$ ; $\textbf{v}$ has two entries, so $\textbf{v} \in \mathbb{R}^{2}$ . The two are in different vector spaces, so $\textbf{u} \cdot \textbf{v}$ is undefined.

$\textbf{u} \times \textbf{v}$ , the cross product of the vectors, is a defined vector if and only if both vectors are elements in $\mathbb{R}^{3}$ . As previously mentioned, $\textbf{v} \in \mathbb{R}^{2}$ , so $\textbf{u} \times \textbf{v}$ is undefined.

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Example Question #21 : Vector Vector Product

$\textbf{u} = (3, 7, -2)$

$\textbf{v} = (1, -5, 1 )$

Evaluate $\textbf{u} \times \textbf{v}$

Possible Answers:

(-3, 5, -22)

(3, 5,22)

( 17, 5, 8)

(-3, -5, -22)

(-17, 5, -8)

Correct answer:

(-3, -5, -22)

Explanation:

One way to determine the cross-product of two vectors is to set up a matrix with the first row $\textbf{i}, \textbf{j}, \textbf{k}$ , where these are the unit vectors (1,0,0), (0,1,0),(0,0,1) , respectively, and with the entries of the vectors as the other two rows:

$\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 3 & 7 & -2 \\ 1 & -5 & 1 \end{vmatrix}$

We can evaluate this as we would evaluate a determinant of a matrix with real entries. Take the products of the upper-left-to-lower-right diagonals, and subtract the products of the lower-left-to-upper-right diagonals:

Cross product

$\textbf{u} \times \textbf{v} = 7\textbf{i}+ ( - 2\textbf{j})+ (-15\textbf{k}) - 7\textbf{k}- 10\textbf{i} - 3 \textbf{j}$

$\textbf{u} \times \textbf{v} = -3\textbf{i}- 5\textbf{j}-22\textbf{k}$

$\textbf{u} \times \textbf{v} = -3(1,0,0 )- 5(0,1,0)-22 (0,0,1)$

$\textbf{u} \times \textbf{v} =(-3, -5, -22)$

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Example Question #27 : Vector Vector Product

$\textbf{u} = (\ln 2 , 1, -1 )$

$\textbf{v} = (2, 1, - 2)$

If $\textbf{u} \cdot \textbf{v} = \ln a$ , then evaluate .

Possible Answers:

a = 64

a = 4e

$a = \frac{4}{e}$

$a = 4e^{3}$

a = 12

Correct answer:

$a = 4e^{3}$

Explanation:

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

$\textbf{u} \cdot \textbf{v} = (\ln 2) \cdot 2 + 1 \cdot 1 + (-1 )(-2)$

$\textbf{u} \cdot \textbf{v} = 2\ln 2+ 1 + 2$

$\textbf{u} \cdot \textbf{v} = 3+ 2\ln 2$

Applying the properties of logarithms:

$\textbf{u} \cdot \textbf{v} = 3+ \ln 2^{2}$

$\textbf{u} \cdot \textbf{v} = 3+ \ln 4$

$\textbf{u} \cdot \textbf{v} = \ln e^{3}+ \ln 4$

$\textbf{u} \cdot \textbf{v} = \ln 4 e^{3}$

Therefore, $a = 4e^{3}$ .

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Example Question #28 : Vector Vector Product

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

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

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

Possible Answers:

None of the other choices gives a correct response.

Correct answer:

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.

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Example Question #29 : Vector Vector Product

A triangle has two sides of length and ; their included angle has measure $\theta$ . The measure of the third side can be obtained from the expression

$\sqrt{\textbf{u} \cdot \textbf{v}}$ ,

where $\textbf{u}= (a , b, 2ab)$ and:

Possible Answers:

$\textbf{v} = (a, b, \cos \theta)$

$\textbf{v} = (b, a, -\cos \theta)$

$\textbf{v} = (a, b, -\sin \theta)$

$\textbf{v} = (b, a, -\sin \theta)$

$\textbf{v} = (a, b, -\cos \theta)$

Correct answer:

$\textbf{v} = (a, b, -\cos \theta)$

Explanation:

Given the lengths and of two sides of a triangle, and the measure of their included angle, $\theta$ , the length of the third side of a triangle can be calculated using the Law of Cosines, which states that

$c^{2} = a^{2}+ b^{2} - 2ab\cos \theta$ .

The dot product $\textbf{u} \cdot \textbf{v}$ is equal to the sum of the products of their corresponding entries, and since $c= \sqrt{\textbf{u} \cdot \textbf{v}}$ , we can substitute $\textbf{u} \cdot \textbf{v}$ for $c^{2}$ :

$\textbf{u} \cdot \textbf{v} = a^{2}+ b^{2} - 2ab\cos \theta$

$\textbf{u} \cdot \textbf{v} = a \cdot a + b \cdot b + 2ab \cdot ( - \cos \theta)$

$\textbf{u} = (a , b , 2ab )$ ; it follows that $\textbf{v} = (a, b, -\cos \theta)$ .

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Example Question #30 : Vector Vector Product

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?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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Linear Algebra : Vector-Vector Product

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : Vector Vector Product

Example Question #22 : Vector Vector Product

Example Question #23 : Vector Vector Product

Example Question #24 : Vector Vector Product

Example Question #25 : Vector Vector Product

Example Question #21 : Vector Vector Product

Example Question #27 : Vector Vector Product

Example Question #28 : Vector Vector Product

Example Question #29 : Vector Vector Product

Example Question #30 : Vector Vector Product

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