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Linear Algebra : Norms

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Example Question #41 : 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 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:

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

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

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 #42 : Norms

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

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

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

Possible Answers:

Equilateral

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:

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

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

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 #171 : 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 #44 : 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:

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

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

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

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

Correct answer:

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

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 #45 : 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:

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

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

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

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

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 #41 : 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:

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

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

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

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

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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Example Question #47 : Norms

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

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

These two vectors form two sides of a triangle. In which range does the area of the triangle fall?

Possible Answers:

$A \in (20 , 25)$

$A \in (10 , 15)$

$A \in (5,10)$

$A \in (15,20)$

$A \in (25,30)$

Correct answer:

$A \in (10 , 15)$

Explanation:

The area of a triangle formed by two vectors in $\mathbb{R}^{3}$ is half the norm of their cross-product - that is

$A =\frac{1}{2} \left \| \textbf{u} \times \textbf{v} \right \|$

The cross-product is equal to the "determinant" of the matrix

$\begin{bmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ 4 & 2 & -3\\ 6 & 2& 2 \end{bmatrix}$ ,

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

which can be calculated by adding the upper-left to lower-right products and subtracting the upper-right to lower-left products:

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

$=\textbf{i} (2)(2)+ \textbf{j} (-3)(6)+ \textbf{k} (4)(2)- \textbf{k}(2)(6)- \textbf{i} (-3)(2) - \textbf{j} (4) (2)$

$=4 \textbf{i} -18 \textbf{j} +8 \textbf{k} - 12 \textbf{k}+6 \textbf{i} -8 \textbf{j}$

$=10 \textbf{i} -26\textbf{j} - 4 \textbf{k}$

= (10, -26, -4)

The norm of this vector is the square root of the sum of the squares of the entries:

$\left \| \textbf{u} \times \textbf{v} \right \| = \sqrt{10^{2}+ (-26)^{2}+ (-4)^{2}} \approx 28.1$

$A \approx \frac{1}{2} \cdot 28.1 \approx 14.1$

The area of the triangle falls in the range $A \in (10 , 15)$ .

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Example Question #181 : 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 acute, right, or obtuse?

Possible Answers:

Acute

Right

Obtuse

Correct answer:

Acute

Explanation:

First, find the angle $\theta$ between the vectors using the formula

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

Find the dot product $\textbf{u} \cdot \textbf{v}$ by adding the products of corresponding entries:

$\textbf{u} \cdot \textbf{v}= 4(6)+ 2(2)+ (-3)(2) = 22$

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:

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

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

Therefore,

$\cos \theta = \frac{22}{ \sqrt{29} \sqrt{44}} \approx 0.6159$

$\theta \approx 52.0^{\circ }$

To find the measures of the other angles, it is necessary to find the length of the third side, which is equal to $\left \|\textbf{v}-\textbf{u}\right \|$ .

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

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

It follows that the triangle is isosceles. This third side, which is one of the two congruent sides, is opposite the $52.0^{\circ }$ angle; by the Isosceles Triangle Theorem the angle opposite the other congruent side is also $52.0^{\circ }$ . The measure of the third angle is

$180^{\circ } -( 52.0 ^{\circ }+ 52.0^{\circ } ) =76^{\circ }$ .

All three angles are acute, so the triangle is an acute triangle.

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Example Question #49 : Norms

$\textbf{u}= (-2, 4, 7)$

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

Possible Answers:

$\left ( -\frac{2}{\sqrt{53}}, \frac{4}{\sqrt{53}}, \frac{7}{\sqrt{53}} \right )$

$\left ( -\frac{2}{\sqrt{69}}, \frac{4}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

None of the other choices gives the correct response.

$\left ( -\frac{2}{\sqrt{13}}, \frac{4}{\sqrt{13}}, \frac{7}{\sqrt{13}} \right )$

$\left ( -\frac{2}{\sqrt{61}}, \frac{4}{\sqrt{61}}, \frac{7}{\sqrt{61}} \right )$

Correct answer:

$\left ( -\frac{2}{\sqrt{69}}, \frac{4}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

Explanation:

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

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{(-2)^{2}+4^{2}+7^{2}} = \sqrt{4+16+49} = \sqrt{69}$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} = \frac{ (-2, 4, 7)}{\sqrt{69}} = \left ( -\frac{2}{\sqrt{69}}, \frac{4}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

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Example Question #50 : Norms

$\textbf{u}= (-4 , -2, 7)$

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

Possible Answers:

$\left ( -\frac{4}{\sqrt{53}},-\frac{2}{\sqrt{53}}, \frac{7}{\sqrt{53}} \right )$

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

$\left ( -\frac{4}{\sqrt{69}},-\frac{2}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

$\left ( -\frac{4}{\sqrt{61}},-\frac{2}{\sqrt{61}}, \frac{7}{\sqrt{61}} \right )$

$\left ( -\frac{4}{\sqrt{13}},-\frac{2}{\sqrt{13}}, \frac{7}{\sqrt{13}} \right )$

Correct answer:

$\left ( -\frac{4}{\sqrt{69}},-\frac{2}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

Explanation:

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

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{(-4)^{2}+ (-2)^{2} +7^{2}} = \sqrt{16+4+ 49} = \sqrt{69}$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} = \frac{ (-4 , -2, 7) }{\sqrt{69}}$

$= \left ( -\frac{4}{\sqrt{69}},-\frac{2}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

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Linear Algebra : Norms

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #41 : Norms

Example Question #42 : Norms

Example Question #171 : Operations And Properties

Example Question #44 : Norms

Example Question #45 : Norms

Example Question #41 : Norms

Example Question #47 : Norms

Example Question #181 : Operations And Properties

Example Question #49 : Norms

Example Question #50 : Norms

All Linear Algebra Resources

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