39:24

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

Q is the midpoint of segment pr. pq=2x+1 and qr= 3x-5. what is the length of segment pr

viva [34]

I’m not positive this is right but try 26. I set 2x+1 and 3x-5 equal to each other and solved for x. I got x=6 then plugged it into both of the equations then added the equations together.

8 0

3 years ago

Which inequality's solution is graphed here?

blagie [28]

The answer to this inequality solution is 6x+42<0

7 0

3 years ago

State whether the lines are parallel, perpendicular,or neither.

cluponka [151]

Answer:

Please check the explanation.

Step-by-step explanation:

Two lines are parallel if their slopes are equal.

Two lines are perpendicular if the product of their slope is -1

We also know that the slope-intercept form of the line equation is

$y=mx+b$

where m is the slope and b is the y-intercept

Given the lines

1)

y = 6х - 3

Comparing with y=mx+b, the slope of y = 6х - 3:

m₁=6

y = - 1/6x + 7

Comparing with y=mx+b, the slope of y = - 1/6x + 7:

m₂=-1/6

As

m₁ × m₂ = -1

6 × - 1/6 = -1

-1 = -1

Thus, the lines y = 6х - 3 and y = - 1/6x + 7 are perpendicular.

2)

y = 3x + 2

Comparing with y=mx+b, the slope of y = 3x + 2:

m₁=3

2y = 6x - 6

simplifying to write in slope-intercept form

y=3x-3

Comparing with y=mx+b, the slope of y=3x-3:

m₂=3

As the slopes of y = 3x + 2 and 2y = 6x - 6 are equal.

i.e. m₁ = m₂ → 3 = 3

Thus, the lines y = 3x + 2 and 2y = 6x - 6 are paralle.

3)

8x - 2y = 3

simplifying to write in slope-intercept form

y = 4x - 3/2

Comparing with y=mx+b, the slope of y = 4x - 3/2:

m₁=4

x + 4y = - 1

simplifying to write in slope-intercept form

y=-1/4x-1/4

Comparing with y=mx+b, the slope of y=-1/4x-1/4:

m₂=-1/4

As

m₁ × m₂ = -1

4 × - 1/4 = -1

-1 = -1

Thus, the lines 8x - 2y = 3 and x + 4y = - 1 are perpendicular.

4)

3x+2y = 5

simplifying to write in slope-intercept form

y = -3/2x + 5/2

Comparing with y=mx+b, the slope of y = -3/2x + 5/2:

m₁=-3/2

3y + 2x = - 3

simplifying to write in slope-intercept form

y = -2/3x - 1

Comparing with y=mx+b, the slope of y = -2/3x - 1:

m₂=-2/3

As m₁ and m₂ are neither equal nor their product is -1, hence the lines neither perpendicular nor parallel.

5)

y - 5 = 6x

simplifying to write in slope-intercept form

y=6x+5

Comparing with y=mx+b, the slope of y=6x+5:

m₁=6

y - 6x = - 1

simplifying to write in slope-intercept form

y=6x-1

Comparing with y=mx+b, the slope of y=6x-1:

m₂=6

As the slopes of y - 5 = 6x and y - 6x = -1 are equal.

i.e. m₁ = m₂ → 6 = 6

Thus, the lines y - 5 = 6x and y - 6x = -1 are paralle.

6)

y = 3х + 9

Comparing with y=mx+b, the slope of y = 3х + 9:

m₁=3

y = -1/3x - 4

Comparing with y=mx+b, the slope of y = 1/3x - 4:

m₂=1/3

As m₁ and m₂ are neither equal nor their product is -1, hence the lines neither perpendicular nor parallel.

8 0

3 years ago

How much is half of 2.25?

Mashcka [7]

The answer to the question is 1.125

3 0

4 years ago

Read 2 more answers