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Linear Algebra : Linear Algebra

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Example Questions

Example Question #161 : Operations And Properties

$\textbf{u} = \left ( \frac{1}{3}, -\frac{1}{6}, \frac{1}{6} \right )$

Give the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left ( \frac{\sqrt{2 }}{3}, -\frac{\sqrt{2 }}{6}, \frac{\sqrt{2 }}{6} \right )$

$\left ( \frac{\sqrt{3 }}{3}, -\frac{\sqrt{3}}{6}, \frac{\sqrt{3}}{6} \right )$

$\textbf{u}$ is itself a unit vector.

$\left ( \frac{\sqrt{6 }}{3}, -\frac{\sqrt{6 }}{6}, \frac{\sqrt{6 }}{6} \right )$

$\textbf{u} = \left ( \frac{2}{3}, -\frac{1}{3}, \frac{1}{3} \right )$

Correct answer:

$\left ( \frac{\sqrt{6 }}{3}, -\frac{\sqrt{6 }}{6}, \frac{\sqrt{6 }}{6} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is the vector

$\textbf{v}= \frac{\textbf{u}}{\left \| \textbf{u} \right \|}$ ,

where $\left \| \textbf{u} \right \|$ , the norm of $\textbf{u}$ , is the square root of the sum of the squares of its entries.

$\textbf{u} = \left ( \frac{1}{3}, -\frac{1}{6}, \frac{1}{6} \right )$

$\left \| \textbf{u} \right \| = \sqrt{ \left (\frac{1}{3} \right ) ^{2}+ \left (-\frac{1}{6} \right )^{2}+ \left ( \frac{1}{6} \right )^{2}}$

$\left \| \textbf{u} \right \| = \sqrt{ \frac{1}{9} + \frac{1}{36} + \frac{1}{36} }$

$\left \| \textbf{u} \right \| = \sqrt{ \frac{1}{6} }$

$\left \| \textbf{u} \right \| =\frac{1}{ \sqrt{6 }}$

The unit vector is therefore

$\left ( \frac{\frac{1}{3}}{\frac{1}{ \sqrt{6 }}}, \frac{ -\frac{1}{6}}{\frac{1}{ \sqrt{6 }}}, \frac{ \frac{1}{6}}{\frac{1}{ \sqrt{6 }}} \right )$ ,

or, simplified,

$\left ( \frac{\sqrt{6 }}{3}, -\frac{\sqrt{6 }}{6}, \frac{\sqrt{6 }}{6} \right )$ .

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Example Question #161 : Operations And Properties

$\textbf{u} = (2,5, -7, 8,- 1)$

$\textbf{v} = (7, -5, 1, -2, 8)$

Which is a true statement?

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum; $\left \| \textbf{v} \right \|$ can be found similarly.

It is not necessary to actually calculate the norms. It can be observed that the five entries in $\textbf{u}$ and the five entries in $\textbf{v}$ have the same set of absolute values, and, consequently, the same squares. The sum of the squares of the five entries in $\textbf{u}$ is therefore equal to the same sum for $\textbf{v}$ . It immediately follows that

$\left \| \textbf{u} \right \| = \left \| \textbf{v} \right \|$ .

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Example Question #31 : Norms

$\textbf{u} = (a, b)$ , where and are the lengths of the legs of a given right triangle.

True or false: The length of the hypotenuse is $\left \| \textbf{u}\right \|$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum. $\textbf{u} = (a, b)$ , so

$\left \| \textbf{u} \right \| = \sqrt{a\cdot a + b\cdot b} = \sqrt{a^{2}+b^{2}}$

By the Pythagorean Theorem, this is the length of the hypotenuse. The statement is true.

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Example Question #32 : Norms

$\textbf{u} = (2, -4, 1,6)$

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

$\textbf{w} = (4, 1,- 5,- 3)$

$\textbf{x} = (5, 2, 0, 4)$

$\textbf{y} = (5, 5, 0,0)$

Which vector has the greatest norm?

Possible Answers:

$\textbf{v}$

$\textbf{x}$

$\textbf{w}$

$\textbf{y}$

$\textbf{u}$

Correct answer:

$\textbf{u}$

Explanation:

The norm of a vector is equal to the square root of the sum of the squares of its entries. It suffices to compare the sum of the squares, which will be the squares of the norms:

$\left \| \textbf{u} \right \| ^{2}= 2^{2}+(-4)^{2}+1^{2}+ 6^{2} = 4 + 16 + 1 + 36= 57$

$\left \| \textbf{v} \right \|^{2}= (-3)^{2}+ 4^{2} + 2^{2}+ 5^{2} =9 + 16 + 4 + 25 = 54$

$\left \| \textbf{w} \right \|^{2}= 4^{2}+ 1^{2}+ (-5)^{2} + (-3)^{2} = 16+ 1 + 25 + 9 = 51$

$\left \| \textbf{x} \right \|= 5^{2}+ 2^{2}+0^{2}+ 4^{2} = 25+4+ 0 + 16 = 45$

$\left \| \textbf{y} \right \|^{2}= 5^{2} + 5^{2}+ 0^{2}+ 0 ^{2} = 25 + 25+ 0 + 0 = 50$

Of the five squares of the norms, $\left \| \textbf{u} \right \| ^{2}$ is the greatest, so $\left \| \textbf{u} \right \|$ is the greatest norm.

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Example Question #33 : Norms

$\textbf{u}= (1, a, a)$ and $\textbf{v} = (b, -1 , 2)$ for some $a,b \in \mathbb{R}$

Give the relationship between and for $\textbf{u}$ and $\textbf{v}$ to be orthogonal.

Possible Answers:

b = -a

$b = \frac{1}{3}a$

$b = -\frac{1}{3}a$

b = a

$\textbf{u}$ and $\textbf{v}$ are orthogonal regardless of the values of and .

Correct answer:

b = -a

Explanation:

$\textbf{u}$ and $\textbf{v}$ are orthogonal if and only if their dot product is 0. The dot product is equal to the sum of the products of entries in corresponding positions, so

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

Set this equal to 0:

b+ a = 0

b = -a , the correct choice.

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Example Question #32 : Norms

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

True or false: $\left \| \textbf{u} \right \|+ \left \| \textbf{v} \right \|$ is an undefined expression.

Possible Answers:

True

False

Correct answer:

False

Explanation:

$\left \| \textbf{u} \right \|$ refers to the norm of a vector, which is always a scalar quantity regardless of what vector space the vector falls in. It follows that $\left \| \textbf{u} \right \|+ \left \| \textbf{v} \right \|$ , the sum of two scalars, itself a scalar - a defined expression.

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Example Question #35 : Norms

$\textbf{u}$ and $\textbf{v}$ are the sides of a parallelogram in Cartesian space, where $\textbf{u} = (2, 4, -1)$ and $\textbf{v}= (3, 7, 4 )$ .

Give the length of its longer diagonal to the nearest tenth.

Possible Answers:

4.4

13.1

5.9

12.4

8.7

Correct answer:

12.4

Explanation:

The lengths of the diagonals of a parallelogram formed by $\textbf{u}$ and $\textbf{v}$ are the lengths, or norms, of their sum and their difference: $\left \| \textbf{u}+ \textbf{v} \right \|$ and $\left \| \textbf{u}- \textbf{v} \right \|$ .

To find $\textbf{u}+ \textbf{v}$ , add elementwise:

$\textbf{u}+ \textbf{v}$

= (2, 4, -1)+(3, 7, 4 )

= (5, 11, 3)

The norm is equal to the square root of the sum of the squares of the elements:

$\left \| \textbf{u}+ \textbf{v} \right \| = \sqrt{5^{2}+11^{2} +3^{2}} \approx 12.4$ ,

the length of one diagonal.

$\left \| \textbf{u}- \textbf{v} \right \|$ can be found similarly:

$\textbf{u}- \textbf{v}$

= (2, 4, -1)-(3, 7, 4 )

= (-1, -3, -5 )

$\left \| \textbf{u}- \textbf{v} \right \| = \sqrt{(-1)^{2}+(-3)^{2} +(-5)^{2}} \approx 5.9$ ,

the length of the other diagonal.

The longer diagonal has length 12.4.

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Example Question #164 : Operations And Properties

$\textbf{u}$ and $\textbf{v}$ are the sides of a parallelogram in Cartesian space, where $\textbf{u} = (2, 4, -1)$ and $\textbf{v}= (3, 7, 4 )$ .

Give the lengths of its shorter diagonal to the nearest tenth.

Possible Answers:

12.4

4.4

8.7

13.1

5.9

Correct answer:

5.9

Explanation:

To find $\textbf{u}+ \textbf{v}$ , add elementwise:

$\textbf{u}+ \textbf{v}$

= (2, 4, -1)+(3, 7, 4 )

= (5, 11, 3)

The norm is equal to the square root of the sum of the squares of the elements:

$\left \| \textbf{u}+ \textbf{v} \right \| = \sqrt{5^{2}+11^{2} +3^{2}} \approx 12.4$ ,

the length of one diagonal.

$\left \| \textbf{u}- \textbf{v} \right \|$ can be found similarly:

$\textbf{u}- \textbf{v}$

= (2, 4, -1)-(3, 7, 4 )

= (-1, -3, -5 )

$\left \| \textbf{u}- \textbf{v} \right \| = \sqrt{(-1)^{2}+(-3)^{2} +(-5)^{2}} \approx 5.9$ ,

the length of the other diagonal.

The shorter diagonal has length 5.9.

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Example Question #33 : Norms

$\textbf{u} = \left ( \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}, 0 \right )$

$\textbf{v} = \left ( \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3} \right )$

$\textbf{u}$ and $\textbf{v}$ form two sides of a triangle. Is this triangle right, acute, or obtuse?

Possible Answers:

Right

Obtuse

Acute

Correct answer:

Right

Explanation:

The angle between two vectors is $\theta$ , where

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

Their dot product is the sum of the products of their corresponding entries:

$\textbf{u}\cdot \textbf{v} = \left (\frac{\sqrt{2}}{2} \right ) \left ( \frac{\sqrt{3}}{3} \right )+ \left ( - \frac{\sqrt{2}}{2} \right ) \left ( \frac{\sqrt{3}}{3} \right )+ 0 =0$

It follows that $\textbf{u}$ and $\textbf{v}$ are orthogonal, or perpendicular, vectors; this immediately proves that the triangle they form is right, so there is no need to go further.

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Example Question #32 : Norms

$\textbf{u}= (7, 1, -7 )$

and

$\textbf{v}= (-7, -1, -7)$

$\textbf{u}$ and $\textbf{v}$ are the sides of a parallelogram in Cartesian space; which of the following statements describes the parallelogram?

Possible Answers:

The parallelogram is both a rectangle and a rhombus (and, consequently, a square).

The parallelogram is a rectangle, but not a rhombus.

The parallelogram is a rhombus, but not a rectangle.

The parallelogram is neither a rectangle nor a rhombus.

Correct answer:

The parallelogram is a rhombus, but not a rectangle.

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:

$\textbf{u} \cdot \textbf{v} = (7)(-7)+ 1(-1)+(-7)(-7)= -1 \ne 0$

The parallelogram is not a rectangle.

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; it suffices to compare the squares of the norms:

$\left \| \textbf{u} \right \| ^{2} = 7^{2} +1^{2}+ (-7)^{2} = 99$

$\left \| \textbf{v} \right \| ^{2} = (-7)^{2} +(-1)^{2}+ (-7)^{2} = 99$

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

The parallelogram is a rhombus.

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #161 : Operations And Properties

Example Question #161 : Operations And Properties

Example Question #31 : Norms

Example Question #32 : Norms

Example Question #33 : Norms

Example Question #32 : Norms

Example Question #35 : Norms

Example Question #164 : Operations And Properties

Example Question #33 : Norms

Example Question #32 : Norms

All Linear Algebra Resources

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