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Linear Algebra : Linear Algebra

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Example Questions

Example Question #32 : Vector Vector Product

$\textbf{u} \cdot \textbf{v} = 20$

True or false: It follows that $\textbf{v} \cdot \textbf{u} = 20$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

One property of vector dot products is commutativity - that is,

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

Therefore, if $\textbf{u} \cdot \textbf{v} = 20$ , then $\textbf{v} \cdot \textbf{u} = 20$ .

The statement is true,

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Example Question #33 : Vector Vector Product

$\textbf{u} \times \textbf{v}= (1, 2, 3)$

True or false: It follows that $\textbf{v}\times \textbf{u} = (1, 2, 3)$ .

Possible Answers:

False

True

Correct answer:

False

Explanation:

One property of vector cross products is anticommutativity - that is,

$\textbf{v}\times \textbf{u} = -(\textbf{u} \times \textbf{v})$ .

If $\textbf{u} \times \textbf{v}= (1, 2, 3)$ , it follows that

$\textbf{v}\times \textbf{u} = -(1,2,3) = (-1. -2. -3)$ .

The statement is false.

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Example Question #38 : Vector Vector Product

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

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 $\textbf{w} \times \textbf{x} \in \mathbb{R}^{3}$ . The dot product of two vectors in the same vector space, is a scalar, so $(\textbf{u} \times \textbf{v}) \cdot ( \textbf{w} \times \textbf{x})$ , the dot product of two vectors in $\mathbb{R}^{3}$ , is a scalar in $\mathbb{R}$ .

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Example Question #33 : Vector Vector Product

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

$\textbf{v } = (4, 3, y)$

Find and so that $\textbf{u} \times \textbf{v} =(0 , -15, 15 )$ .

Possible Answers:

x=- 9

y= 1

x= 3

y= -3

There does not exist any such and

x= 9

y= -1

x= -3

y= 3

Correct answer:

x= -3

y= 3

Explanation:

The cross product of two vectors in $\mathbb{R}^{3}$ can be set up and calculated as if it were a determinant of a matrix with its top row comprising $\textbf{i} , \textbf{j}, \textbf{k}$ , the unit vectors of $\mathbb{R}^{3}$ , and the other two comprising the elements of $\textbf{u}$ and $\textbf{v}$ :

$\textbf{u} \times \textbf{v} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 1 & x& -3 \\ 4 & 3 &y \end{vmatrix}$

Calculate as you would a determinant, adding the upper-left to lower-right products and subtracting upper-right to lower-left products:

Cross product

The cross-product is equal to

$(xy)\textbf{ i } -12\textbf{j} + 3\textbf{k } - (4x)\textbf{k} - (-9)\textbf{i} -y\textbf{j}$

$= (xy+9)\textbf{ i } + (-y -12)\textbf{j} +( -4x+ 3)\textbf{k }$

= (xy+9 , -y-12, -4x+3 )

We want this vector to be equal to (0 , -15, 15 ) , so the following must hold:

-4x+ 3 = 15

x= -3

-y - 12 = -15

y = 3

Examining the third equation, xy+9 = 0 , we find that this is consistent with the other equations, since x= -3 and y = 3 make this true. Therefore, x= -3 and y = 3 are the values sought.

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Example Question #41 : Vector Vector Product

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

Which of the following holds for $\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u}$ ?

Note: $\textbf{0}$ (boldface) refers to the vector (0,0,0) ;

Possible Answers:

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u} \in \mathbb{R}^{3}$ , but the result need not be $\textbf{0}$ .

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u} \in \mathbb{R}$ , but the result need not be 0.

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u}$ is an undefined expression.

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u} = 0$ , where 0 refers to the real zero.

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u} = \textbf{0}$ , where $\textbf{0}$ refers to the vector (0,0,0) .

Correct answer:

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u} = \textbf{0}$ , where $\textbf{0}$ refers to the vector (0,0,0) .

Explanation:

One property of vector cross products is anticommutativity - that is,

$\textbf{v}\times \textbf{u} = -(\textbf{u} \times \textbf{v})$ .

It therefore follows that for all $\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ ,

$\textbf{u} \times \textbf{v}+ \textbf{v} \times \textbf{u} = \textbf{0}$ , the zero vector (0.0.0) .

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Example Question #853 : Linear Algebra

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

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

A parallelogram has these two vectors as sides. Find so that the parallelogram is a rhombus.

Possible Answers:

$a = \pm 2\sqrt{7}$

$a = \pm 6$

$a = \pm \frac{9}{4}$

$a = \pm 28$

$a = \pm \frac{3}{2}$

Correct answer:

$a = \pm 2\sqrt{7}$

Explanation:

For the parallelogram formed by $\textbf{u}$ and $\textbf{v}$ to be a rhombus, the vectors must be of equal length, or norm - $\left \| \textbf{u} \right \|= \left \| \textbf{v} \right \|$ . The norm of a vector is equal to the square root of the sum of the squares of its entries; to set the norms equal, it suffices to set the squares of the norms equal, and to solve for :

$\left \| \textbf{u} \right \|^{2}= \left \| \textbf{v} \right \|^{2}$

$4^{2}+ (-1)^{2}+ 4^{2} = 2^{2}+ (-1)^{2}+a^{2}$

$16+1+16 = 4+1 +a^{2}$

$33 = 5 +a^{2}$

$a^{2} = 28$

$a = \pm 2\sqrt{7}$

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Example Question #854 : Linear Algebra

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

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

A parallelogram has these two vectors as sides. Find so that the parallelogram is a rectangle.

Possible Answers:

$a = -\frac{9}{4}$

$a =- \frac{3}{2}$

a = 6

a = 28

$a = 2\sqrt{7}$

Correct answer:

$a = -\frac{9}{4}$

Explanation:

For the parallelogram formed by $\textbf{u}$ and $\textbf{v}$ to be a rectangle, the vectors must be perpendicular - that is, orthogonal, This is true if and only if $\textbf{u} \cdot \textbf{v} = 0$ .

The dot product of the vectors can be found by adding the products of corresponding entries:

$\textbf{u} \cdot \textbf{v} = 4(2)+ (-1)(-1)+4(a) = 4a+9$

Set equal to 0 and solve for :

$\textbf{u} \cdot \textbf{v} = 0$

4a+9 = 0

4a = -9

$a = -\frac{9}{4}$ .

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Example Question #43 : Vector Vector Product

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

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

Find a value of (nearest tenth) so that $\textbf{u}$ and $\textbf{v}$ are the sides of a parallelogram of area 100.

Possible Answers:

26.4

23.8

22.2

22.6

25.9

Correct answer:

26.4

Explanation:

The area of a parallelogram with sides $\textbf{u}$ and $\textbf{v}$ in $\mathbb{R}^{3}$ is the norm of their cross-product:

$A = \left \|\textbf{u} \times \textbf{v} \right \|$

To find the cross-product, take the "determinant" of the matrix formed from the entries of the vectors, as follows

$\textbf{u} \times \textbf{v} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 4 & -1 & 4 \\ 2 & -1 & a \end{vmatrix}$

where $\textbf{i}= (1,0,0), \textbf{j} = (0,1,0), \textbf{k} = (0,0,1)$ .

Find this "determinant" as you would a numeric determinant - add the upper-left to lower-right products, and subtract the upper-right to lower-left products.

$\begin{vmatrix}{\color{Red} \textbf{i}} & {\color{DarkGreen} \textbf{j}} &{\color{Purple} \textbf{k}} \\ 4 & {\color{Red} \textbf{-1}} & {\color{DarkGreen} \textbf{4}}\\ 2 & -1 &{\color{Red} \textbf{a}}\end{vmatrix}\begin{matrix} \textbf{i} & \textbf{j} \\ {\color{Purple} \textbf{4}} & -1 \\ {\color{DarkGreen} \textbf{2}} & {\color{Purple} \textbf{-1}} \end{matrix}$ $\begin{vmatrix} \textbf{i} & \textbf{j} & {\color{Red} \textbf{k} } \\ 4 & {\color{Red}\textbf{-1}} & {\color{DarkGreen}\textbf{4}} \\ {\color{Red}\textbf{2}} & {\color{DarkGreen}\textbf{-1}} & {\color{Purple}\textbf{a}} \end{vmatrix}\begin{matrix} {\color{DarkGreen}\textbf{i} }& {\color{Purple} \textbf{j}} \\ {\color{Purple} \textbf{4}} & -1 \\ 2 & -1 \end{matrix}$

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

$= \textbf{i}(-1)(a)+ \textbf{j}(4)(2)+ \textbf{k}(4)(-1) - \textbf{k}(-1)(1)- \textbf{i}(4)(-1)- \textbf{j}(4)(a)$

$=-a \textbf{i}+8 \textbf{j} -4 \textbf{k} + \textbf{k} +4 \textbf{i} -4a \textbf{j}$

$=(-a+4) \textbf{i}+( -4a+8 ) \textbf{j} -3 \textbf{k}$

=(-a+4, -4a+ 8, -3 )

The norm is the result of adding the squares of the entries, and taking the square root. In terms of , this is

$\left \|\textbf{u} \times \textbf{v} \right \|$

$= \sqrt{(-a+4)^{2}+ ( -4a+ 8)^{2}+( -3 )^{2}}$

$= \sqrt{a^{2}-8a + 16 + 16a^{2} -64a + 64 + 9 }$

$= \sqrt{17a^{2}-72a +25}$

Set this equal to 100:

$\sqrt{17a^{2}-72a +25} = 100$

$17a^{2}-72a +25 = 10,000$

$17a^{2}-72a -9,975 = 0$

Through the quadratic formula:

$a=\frac{ 72 \pm \sqrt{72^{2}- 4(17)(-9,975)}}{2(17)}$

After calculation, we get solutions

$a \approx -22.2, a \approx 26.4$ .

All of the choices are positive, so select 26.4.

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Example Question #855 : Linear Algebra

A parallelogram in $\mathbb{R}^{3}$ has as two of its sides the vectors

$\textbf{u} = (3, 5, 4 )$

and

$\textbf{v}= (5, 0, -5)$ .

Which statement is true of the parallelogram?

Possible Answers:

The parallelogram is a rhombus, but not a rectangle.

The parallelogram is a square.

The parallelogram is neither a rectangle nor a rhombus.

The parallelogram is a rectangle, but not a rhombus.

Correct answer:

The parallelogram is a rhombus, but not a rectangle.

Explanation:

The parallelogram is a rectangle if and only if the two vectors that form the adjacent sides are perpendicular - that is, orthogonal. This happens if the dot product - the sum of the products of elements in corresponding positions - is equal to 0. Test this:

$\textbf{u} \cdot \textbf{v} = 3(5)+5(0)+ 4(-5) = -5 \ne 0$

The parallelogram is not a rectangle.

The parallelogram is a rhombus if and only if the two vectors that form the adjacent sides are of equal lenght - that is, if their norms are equal. The norm of a vector is the square root of the sum of the squares of its elements, so:

$\left \| \textbf{u} \right \| = \sqrt{3^{2}+5^{2}+4^{2}} = \sqrt{9+ 25 + 16}= \sqrt{50}$

$\left \| \textbf{v} \right \| = \sqrt{5^{2}+0^{2}+(-5)^{2}} = \sqrt{ 25 + 0+25}= \sqrt{50}$

$\left \| \textbf{u} \right \| = \left \| \textbf{v} \right \|$ , so the parallelogram is a rhombus.

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Example Question #856 : Linear Algebra

$\textbf{u} = (8, -1,- 3)$

$\textbf{v}= (8, 1, 3)$

Calculate the angle between $\textbf{u}$ and $\textbf{v}$ (nearest degree).

Possible Answers:

$137^{\circ}$

$47^{\circ}$

$90^{\circ}$

$43^{\circ}$

$133^{\circ}$

Correct answer:

$43^{\circ}$

Explanation:

The angle between two vectors $\textbf{u}$ and $\textbf{v}$ is $\theta$ , where

$\cos \theta = \frac{\textbf{u}\cdot \textbf{v}}{\left \| \textbf{u} \right \| \left \| \textbf{v} \right \|}$

The dot product of two vectors is the sum of the products of their corresponding entries:

$\textbf{u} \cdot \textbf{v} = 8(8)+ (-1)(1)+ (-3)3= 54$

The norm of a vector is the square root of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{8^{2}+(-1)^{2}+ (-3)^{2}} = \sqrt{74}$

$\left \| \textbf{v} \right \| = \sqrt{8^{2}+1^{2}+ 3^{2}} = \sqrt{74}$

$\cos \theta = \frac{54}{( \sqrt{74}) (\sqrt{74} )}$

$\cos \theta = \frac{54}{74}$

$\theta = \cos^{-1} \frac{54}{74} \approx 43^{\circ}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #32 : Vector Vector Product

Example Question #33 : Vector Vector Product

Example Question #38 : Vector Vector Product

Example Question #33 : Vector Vector Product

Example Question #41 : Vector Vector Product

Example Question #853 : Linear Algebra

Example Question #854 : Linear Algebra

Example Question #43 : Vector Vector Product

Example Question #855 : Linear Algebra

Example Question #856 : Linear Algebra

All Linear Algebra Resources

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