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Linear Algebra : Norms

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Example Questions

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:

True

False

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 #31 : 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:

Obtuse

Right

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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Example Question #251 : Linear Algebra

$\textbf{u}= (13, -4, 3 )$ and $\textbf{v} = (5, 5, -12)$ .

$\textbf{u}$ and $\textbf{v}$ form two sides of a triangle. Is this triangle scalene, isosceles (but not equilateral), or equilateral?

Possible Answers:

Scalene

Isosceles, but not equilateral

Equilateral

Correct answer:

Isosceles, but not equilateral

Explanation:

First, find the lengths, or norms, of $\textbf{u}$ and $\textbf{v}$ by taking the square roots of the sums of the squares of their entries:

$\textbf{u}= (13, -4, 3 )$

$\left \| \textbf{u} \right \|= \sqrt{13^{2}+(-4)^{2} + 3 ^{2} }= \sqrt{194}$

$\textbf{v} = (5, 5, -12)$

$\left \| \textbf{v} \right \| =\sqrt{ 5^{2}+5^{2}+ (-12)^{2}}= \sqrt{194}$

The length of the third side is $\left \|\textbf{u}-\textbf{v}\right \|$ .

$\textbf{u}-\textbf{v} = (8, -9, 15)$

$\left \|\textbf{u}-\textbf{v}\right \| = \sqrt{8^{2}+(-9)^{2}+ 15^{2}} = \sqrt{370}$

$\left \| \textbf{u} \right \|= \left \| \textbf{v} \right \| \ne \left \| \textbf{u}-\textbf{v} \right \|$ . Exactly two sides are congruent, so the triangle is isosceles, but not equilateral.

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Example Question #34 : Norms

$\textbf{u}= (13, -4, 3 )$ and $\textbf{v} = (5, 5, -12)$ .

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

Possible Answers:

Right

Acute

Obtuse

Correct answer:

Acute

Explanation:

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

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

The lengths, or norms, of $\textbf{u}$ and $\textbf{v}$ , can be found by taking the square roots of the sums of the squares of their entries:

$\left \| \textbf{u} \right \|= \sqrt{13^{2}+(-4)^{2} + 3 ^{2} }= \sqrt{194}$

$\left \| \textbf{v} \right \| =\sqrt{ 5^{2}+5^{2}+ (-12)^{2}}= \sqrt{194}$

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

$\textbf{u}\cdot \textbf{v} = 13(5)+(-4)(5)+3(-12) = 9$

$\cos \theta = \frac{9}{ \sqrt{194} \cdot \sqrt{194} } = \frac{9}{194}$

$\theta = \cos^{-1} \frac{9}{194} \approx 87.3 ^{\circ}$

$\left \| \textbf{u} \right \| = \left \| \textbf{v} \right \|$ , so the triangle is isosceles; their included angle is the vertex angle, so the measures of the other two (base) angles are congruent; they each measure

$\frac{180 - 87.3}{2} ^{\circ} \approx 46.4^{\circ}$

This triangle is acute.

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Linear Algebra : Norms

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

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 #31 : Norms

Example Question #32 : Norms

Example Question #251 : Linear Algebra

Example Question #34 : Norms

All Linear Algebra Resources

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