Use the Gram-Schmidt process to find an orthonormal basis for the subspace of R4 spanned by the vectors (1, 0, 1, 1), (1, 0, 1, 0), (0, 0, 1, 1).

Answers

Answer 1

Answer:

Answer:

$$ e_1 = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}$ $$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}$ $$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{2}}}{\textbf{2}}\n\n0\end{pmatrix}$

Step-by-step explanation:

we have to orthonormalize the vectors:

$v_1 = \begin{pmatrix} 1 \n 0 \n 1 \n 1 \end{pmatrix}$ $v_2 = \begin{pmatrix} 1 \n 0 \n 1 \n 0 \end{pmatrix}$ $$ v_3 = \begin{pmatrix} 0 \n 0 \n 1 \n 1 \end{pmatrix}$

According to Gram - Schmidt process, we have:

$u_k = v_k - \sum_(j = 1) ^ {k - 1} proj_(uj) (v_k)$ where, $$ proj_u (v) = (u . v)/(u . u)u$

The normalized vector is: $e_k = (u_k)/(√(u_k.u_k))$

Now, the first step.

$v_1 = \begin{pmatrix} 1 \n 0 \n 1 \n 1 \end{pmatrix}$ = u₁

Therefore, e₁ = $(u_1)/(√(u_1.u_1))$

$$ = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}$

Now, we find e₂.

$u_2 = v_2 - (u_1.v_2)/(u_1.u_1)u_1$

$$ = \begin{pmatrix} (1)/(3)\n \n 0 \n\n (1)/(3)\n\n (-2)/(3) \end{pmatrix}$

Therefore, $e_2 = (u_2)/(√(u_2.u_2))$

$$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}$

To find e₃:

$u_3 = v_3 - (u_1. v_3)/(u_1.u_1)u_1 - (u_2. v_3)/(u_2.u_2) u_2$

$$ = \begin{pmatrix} (-1)/(2) \n\n 0\n \n (1)/(2) \n\n 0 \n\end{pmatrix}$

$e_3 = (u_3)/(√(u_3.u_3))$

$$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{2}}}{\textbf{2}}\n\n0\end{pmatrix}$

So, we have the orthonormalized vectors $e_1, e_2, e_3$ .

Hence, the answer.

Answer 2

Answer:

Final answer:

To find the orthonormal basis using the Gram-Schmidt process, we calculate the first vector by dividing the first given vector by its magnitude and normalize it. Then, we subtract the projection of each subsequent vector onto the previously found orthonormal vectors and normalize the resulting vector.

Explanation:

To find an orthonormal basis for the subspace of R4 spanned by the given vectors using the Gram-Schmidt process, we will start by finding the first vector of the orthonormal basis. Let's call the given vectors v1, v2, and v3, respectively. The first vector of the orthonormal basis, u1, is equal to v1 divided by its magnitude, which is ||v1||. So, u1 = v1 / ||v1||. We can calculate ||v1|| as √(1^2 + 0^2 + 1^2 + 1^2) = √3.

Therefore, u1 = (1/√3, 0/√3, 1/√3, 1/√3).

Now, we need to find u2, the second vector of the orthonormal basis. To do this, we subtract the projection of v2 onto u1 from v2, then divide the result by its magnitude. We calculate the projection of v2 onto u1 as proj_u1(v2) = u1 * dot(u1, v2), where dot(u1, v2) represents the dot product of u1 and v2.

Finally, we subtract proj_u1(v2) from v2 to get v2' = v2 - proj_u1(v2), and then normalize v2' to get u2 = v2' / ||v2'||.

We can repeat this process to find u3, the third vector of the orthonormal basis. Subtract proj_u1(v3) and proj_u2(v3) from v3, then normalize the result to get u3 = v3' / ||v3'||.

Therefore, the orthonormal basis for the subspace spanned by the given vectors is (1/√3, 0/√3, 1/√3, 1/√3), (0, 0, 0, 1), and (-1/√3, 0/√3, 1/√3, 1/√3).

Learn more about Gram-Schmidt process here:

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Billy finished his history assignment in 1/4 hours. Then he completed he’s chemistry assignment in 1/3 hours. What’s the total amount of time bill spent doing these 2 assignments

Answers

Answer:

7/12 of an hour

Step-by-step explanation:

you add them, by finding the common denominator.

1/3=4/12

1/4=3/12

3/12+4/12=7/12

Which of the following is equivalent to 18 minus StartRoot negative 25 EndRoot?

Answers

Answer:

18-5i

Step-by-step explanation:

Answer: 12-i

12-(√-1)

Step-by-step explanation:

$18-√(-25)$ Original Question

$18-(√(25) * √(-1) )$ Split

$18-5*(√(-1) )$ Solve for square root

$12-√(-1)$ Subtract

You can substitute $√(-1)$ for i

$12-i$ Substitute

Line A y= 2x + 3 is parallel to another line B, what is the slope of the line B?

Answers

Answer:

Step-by-step explanation:

Since the lines are parallel, then the slope of line B would be the same as line A.

In one full day, a kudzu vine can grow 15 inches in length. How many inches per hour is this?

Answers

1.66666666667 inches per hour

Answer:

0.625

Step-by-step explanation:

15 divided by 24 which equals 0.625

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the cost to make the can if the metal for the sides will cost $1.25 per 2 cm and the metal for the bottom will cost $2.00 per 2 cm ?

Answers

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=(25)/(\pi r^2)$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=(25)/(\pi r^2)$

$\therefore C=2\pi r^2+2.5 \pi r * (25)/(\pi r^2)$

$\Rightarrow C=2\pi r^2+ (62.5)/( r)$

Differentiating with respect to r

$C'=4\pi r- (62.5)/( r^2)$

Again differentiating with respect to r

$C''=4\pi + (125)/( r^3)$

To find the minimize cost, we set C'=0

$4\pi r- (62.5)/( r^2)=0$

$\Rightarrow 4\pi r=(62.5)/( r^2)$

$\Rightarrow r^3=(62.5)/( 4\pi)$

⇒r=1.71

Now,

$\left C''\right|_(x=1.71)=4\pi +(125)/(1.71^3)>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=(25)/(\pi* 1.71^2)$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

The probability is 1 in 4,011,000 that a single auto trip in the United States will result in a fatality. Over a lifetime, an average U.S. driver takes 46,000 trips.(a) What is the probability of a

Answers

Answer:

0.0114

Step-by-step explanation:

(a) What is the probability of a fatal accident over a lifetime?

Suppose A be the event of a fatal accident occurring in a single trip.

Given that:

P(1 single auto trip in the United States result in a fatality) = P(A)

Then;

P(A) = 1/4011000

P(A) = 2.493 × 10⁻⁷

Now;

P(1 single auto trip in the United States NOT resulting in a fatality) is:

P( $\mathbf{\overline A}$ ) = 1 - P(A)

P( $\mathbf{\overline A}$ ) = 1 - 2.493 × 10⁻⁷

P( $\mathbf{\overline A}$ ) = 0.9999997507

However, P(fatal accident over a lifetime) = P(at least 1 fatal accident in lifetime i.e. 46000 trips)

= 1 - P(NO fatal accidents in 46000 trips)

Similarly,

P(No fatal accidents over a lifetime) = P(No fatal accident in the 46000 trips) = P(No fatality on the 1st trip and No fatality on the 2nd trip ... and no fatality on the 45999 trip and no fatality on the 46000 trip)

= $[P(\overline A)] ^(46000) \ \ \ (since \ trips \ are \ independent \ events)$

= $[0.9999997507]^(46000)$

= 0.9885977032

Finally;

P(fatal accident over a lifetime) = 1 - 0.9885977032

P(fatal accident over a lifetime) = 0.0114022968

P(fatal accident over a lifetime) ≅ 0.0114

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