Mathematics, 08.09.2020 14:01, Frenchfries13

Let L be the line with parametric equations x=5+t, y=6,z=−2−3t. Find the vector equation for a line that passes through the point P=(−5,7,−8) and intersects L at a point that is distance 3 from the point Q=(5,6,−2). Note that there are two possible correct answers. Use the square root symbol √ where needed to give an exact value for your answer.

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answered: liljett

The required equations are

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$ and

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$ .

Step-by-step explanation:

The given parametric equation of the line, , is x=5+t, y=6, z=-2-3t, so, an arbitrary point on the line is R(x,y,z)=R(5+t, 6, -2-3t)

The vector equation of the line passing through the points P(-5,7,-8) and R(5+t, 6, -2-3t) is

$\vec P + \lambda \vec{(PR)}=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)$

The given equation can also be written as

$\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)$

The standard equation of the line passes through the point P_0(x_0,y_0,z_0) and having direction $\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k$ is

$\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)$

Here, The value of the parameter,, of any point at a distance from the point, P_0 , can be determined by

$|t \vec v|=d\;\cdots (iv)$

Comparing equations (ii) and (iii)

The line is passing through the point P_0 (5,6,-2) having direction $\vec v=\hat i -3 \hat k$ .

Note that the point Q(5,6,-2) is the same as P_0 obtained above.

Now, the value of the parameter, , for point at distance d=3 from the point Q(5,6,-2) can be determined by equation (iv) , we have

$|t(\hat i -3 \hat k)|=3$

$\Rightarrow t^2|(\hat i -3 \hat k)|^2=9$

$\Rightarrow 10t^2=9$

$\Rightarrow t^2=\frac {9}{10}$

$\Rightarrow t=\pm \frac {3}{\sqrt {10}}$

Put the value of in equation (i) , the required equations are as follows:

For $t= \frac {3}{\sqrt {10}}$

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$

and for $t= -\frac {3}{\sqrt {10}}$ ,

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$

answered: Guest

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200. after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes 1800 - 6400 which gets you 11,600.

answered: Guest

the most boring thing ever written

step-by-step explanation:

i am writing something. yes, i plan to make it the most boring thing ever written. i go to the store. a car is parked. many cars are parked or moving. some are blue. some are tan. they have windows. in the store, there are items for sale. these include such things as soap, detergent, magazines, and lettuce. you can enhance your life with these products. soap can be used for bathing, be it in a bathtub or in a shower. apply the soap to your body and rinse. detergent is used to wash clothes. place your dirty clothes into a washing machine and add some detergent as directed on the box. select the appropriate settings on your washing machine and you should be ready to begin. magazines are stapled reading material made with glossy paper, and they cover a wide variety of topics, ranging from news and politics to business and stock market information. some magazines are concerned with more recreational topics, like sports card collecting or different kinds of hairstyles. lettuce is a vegetable.