Round 1.086 to the nearest whole number

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,

Dahasolnce [82]

Answer:

The direction cosines are:

$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = $\frac{a . i}{|a| . |i|}$ ---------------------(i)

cos β = $\frac{a.j}{|a||j|}$ ---------------------(ii)

cos γ = $\frac{a.k}{|a|.|k|}$ ----------------------(iii)

And from these we can get the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i = (5i + j + 4k) . (i)

a . i = 5 [a.i is just the x component of the vector]

a . j = 1 [the y component of the vector]

a . k = 4 [the z component of the vector]

Also

|a|. |i| = |a|. |j| = |a|. |k| = |a| [since |i| = |j| = |k| = 1]

|a| = $\sqrt{5^2 + 1^2 + 4^2}$

|a| = $\sqrt{25 + 1 + 16}$

|a| = $\sqrt{42}$

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = $\frac{5}{\sqrt{42} }$

cos β = $\frac{1}{\sqrt{42} }$

cos γ = $\frac{4}{\sqrt{42} }$

From the value, now find the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

α = cos⁻¹ ( $\frac{5}{\sqrt{42} }$ )

α = cos⁻¹ ( $\frac{5}{6.481}$ )

α = cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

β = cos⁻¹ ( $\frac{1}{\sqrt{42} }$ )

β = cos⁻¹ ( $\frac{1}{6.481 }$ )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

γ = cos⁻¹ ( $\frac{4}{\sqrt{42} }$ )

γ = cos⁻¹ ( $\frac{4}{6.481}$ )

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

Conclusion:

The direction cosines are:

$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

3 0

3 years ago

Please help!!!!!!!!!!

Degger [83]

I hope this helps you

4 0

3 years ago

Please Help, this is math problems

MariettaO [177]

Answer:

1/8 for the first one then 1/6 for the second one

Step-by-step explanation:

5 0

3 years ago

An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t

Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .