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Linear Algebra : Operations and Properties

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Example Questions

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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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

Obtuse

Acute

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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Example Question #171 : Operations And Properties

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

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

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

Possible Answers:

Equilateral

Isosceles, but not equilateral

Scalene

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:

$\left \| \textbf{u} \right \| = \sqrt{\left ( \frac{\sqrt{2}}{2}\right ) ^{2} + \left (- \frac{\sqrt{2}}{2} \right )^{2}+0 ^{2}} = \sqrt{1} = 1$

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

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

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

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

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

$=\sqrt{2}$

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

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Example Question #171 : Operations And Properties

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

$\textbf{v} = (6,2,2)$

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

Possible Answers:

Equilateral

Scalene

Isosceles, but not 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:

$\left \| \textbf{u} \right \|= \sqrt{4^{2}+2^{2}+ (-3)^{2} } = \sqrt{29}$

$\left \| \textbf{v} \right \| =\sqrt{ 6^{2}+ 2^{2}+ 2^{2} }=\sqrt{44}$

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

$\textbf{v}-\textbf{u} = (-2, 0, -5 )$

$\left \|\textbf{v}-\textbf{u}\right \| = \sqrt{(-2)^{2}+ 0^{2}+ (-5 )^{2}} = \sqrt{29}$ .

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

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Example Question #172 : Operations And Properties

$\textbf{v} = (\sqrt{\pi}, r )$ , where is the radius of a circle. Let be the area of the circle.

True or false:

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

Possible Answers:

True

False

Correct answer:

False

Explanation:

The norm of a vector is equal to the square root of the sum of the squares of its entries; the square of the norm is equal to this sum itself. Thus,

$\left \| \textbf{v} \right \|^{2} =( \sqrt{\pi })^{2} + r^{2} = \pi + r^{2}$

The area of a circle, however, is

$A = \pi r^{2}$ .

The statement is false.

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Example Question #42 : Norms

Let the center of an ellipse on the coordinate plane be the point (h, k ) . Let the lengths of its horizontal axis and vertical axis be and , respectively.

The equation of the ellipse can be written as

$\left \| \textbf{u} \right \| ^{2} = 1$ ,

where $\textbf{u}$ is the vector:

Possible Answers:

$\left (\left ( \frac{x-h }{a} \right )i, \frac{y-k }{b } \right )$

$\left (\left ( \frac{x-h }{a} \right )i, \left ( \frac{y-k }{b } \right )i \right )$

$\left ( \frac{x-h }{a} ,\left ( \frac{y-k }{b } \right )i \right )$

$\left ( \frac{x-h }{a} , \frac{y-k }{b } \right )$

Correct answer:

$\left ( \frac{x-h }{a} , \frac{y-k }{b } \right )$

Explanation:

The square of a norm of a vector $\textbf{u} = (u_{1}, u_{2})$ is equal to

$\left \| \textbf{u} \right \|^{2}= u_{1}^{2} + u_{2}^{2}$

The equation for an ellipse with the given characteristics is

$\frac{(x-h)^{2} }{a^{2}} + \frac{(y-k)^{2} }{b^{2}} = 1$ ,

or, equivalently,

$\left (\frac{x-h }{a} \right )^{2} +\left ( \frac{y-k }{b } \right )^{2} = 1$

It follows that if we set $u_{1} = \frac{x-h }{a}$ and $u_{2} =\frac{y-k }{b }$ , the equation of an ellipse can be restated as $\left \| \textbf{u} \right \|^{2}= 1$ .

The correct choice is $\textbf{u} = \left ( \frac{x-h }{a} , \frac{y-k }{b } \right )$ .

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Example Question #42 : Norms

Let the center of a horizontal hyperbola on the coordinate plane be the point (h, k ) . Let the width and height of its central rectangle be and , respectively.

The equation of the hyperbola can be written as

$\left \| \textbf{u} \right \| ^{2} = 1$ ,

where $\textbf{u}$ is the vector:

Possible Answers:

$\left ( \frac{x-h }{a} , \frac{y-k }{b } \right )$

$\left ( \frac{x-h }{a} ,\left ( \frac{y-k }{b } \right )i \right )$

$\left (\left ( \frac{x-h }{a} \right )i, \frac{y-k }{b } \right )$

$\left (\left ( \frac{x-h }{a} \right )i, \left ( \frac{y-k }{b } \right )i \right )$

Correct answer:

$\left ( \frac{x-h }{a} ,\left ( \frac{y-k }{b } \right )i \right )$

Explanation:

The square of a norm of a vector $\textbf{u} = (u_{1}, u_{2})$ is equal to

$\left \| \textbf{u} \right \|^{2}= u_{1}^{2} + u_{2}^{2}$

The equation for a horizontal hyperbola with the given characteristics is

$\frac{(x-h)^{2} }{a^{2}} - \frac{(y-k)^{2} }{b^{2}} = 1$ ,

or, equivalently,

$\left (\frac{x-h }{a} \right )^{2} +\left ( \left ( \frac{y-k }{b } \right ) i \right ) ^{2} = 1$

It follows that if we set $u_{1} = \frac{x-h }{a}$ and $u_{2} =\left (\frac{y-k }{b } \right ) i$ , the equation of a horizontal hyperbola can be restated as $\left \| \textbf{u} \right \|^{2}= 1$ .

The correct choice is $\textbf{u} = \left ( \frac{x-h }{a} ,\left ( \frac{y-k }{b } \right )i \right )$ .

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Example Question #43 : Norms

Let the center of a vertical hyperbola on the coordinate plane be the point (h, k ) . Let the width and height of its central rectangle be and , respectively.

The equation of the hyperbola can be written as

$\left \| \textbf{u} \right \| ^{2} = 1$ ,

where $\textbf{u}$ is what vector?

Possible Answers:

$\left (\left ( \frac{x-h }{a} \right )i, \frac{y-k }{b } \right )$

$\left (\left ( \frac{x-h }{a} \right )i, \left ( \frac{y-k }{b } \right )i \right )$

$\left ( \frac{x-h }{a} ,\left ( \frac{y-k }{b } \right )i \right )$

$\left ( \frac{x-h }{a} , \frac{y-k }{b } \right )$

Correct answer:

$\left (\left ( \frac{x-h }{a} \right )i, \frac{y-k }{b } \right )$

Explanation:

The square of a norm of a vector $\textbf{u} = (u_{1}, u_{2})$ is equal to

$\left \| \textbf{u} \right \|^{2}= u_{1}^{2} + u_{2}^{2}$

The equation for a vertical hyperbola with the given characteristics is

$- \frac{(x-h)^{2} }{a^{2}} + \frac{(y-k)^{2} }{b^{2}} = 1$ ,

or, equivalently,

$\left (\left (\frac{x-h }{a} \right )i \right )^{2} +\left ( \frac{y-k }{b } \right )^{2} = 1$

It follows that if we set $u_{1} =\left ( \frac{x-h }{a} \right )i$ and $u_{2} = \frac{y-k }{b }$ , the equation of a vertical hyperbola can be restated as $\left \| \textbf{u} \right \|^{2}= 1$ .

The correct choice is $\textbf{u}= \left (\left ( \frac{x-h }{a} \right )i, \frac{y-k }{b } \right )$ .

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #33 : Norms

Example Question #32 : Norms

Example Question #251 : Linear Algebra

Example Question #34 : Norms

Example Question #171 : Operations And Properties

Example Question #171 : Operations And Properties

Example Question #172 : Operations And Properties

Example Question #42 : Norms

Example Question #42 : Norms

Example Question #43 : Norms

All Linear Algebra Resources

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