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algol [13]
3 years ago
15

For the lines x=3t, y=1-2t, z=2-3t and x (3, 1, 4) +s(-9, 6, 9) (a) Show that the lines are parallel. (b) Calculate the distance

between the paralle lines.
Mathematics
1 answer:
son4ous [18]3 years ago
6 0

Answer:

Step-by-step explanation:

Given lines in parametric form

line L_1

\frac{x}{3}=\frac{y-1}{-2} =\frac{z-2}{-3}

direction vector of L_1 v_1=

Line L_2

direction vector of L_2 v_2=

therefore

v_2=-3v_1

thus lines are parallel.

(ii)distance between two lines is

L_2 is given by

\frac{x-3}{-9}=\frac{y-1}{6} =\frac{z-4}{9}=s

\frac{x-3}{-3}=\frac{y-1}{2} =\frac{z-4}{3}=3s

\left | \frac{\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1 \\ a_1&b_1&c_1 \\ a_2&b_2&c_2\end{vmatrix}}{\sqrt{\left ( a_1b_2-a_2b_1 \right )^2+\left ( b_1c_2-b_2c_1 \right )^2+\left ( c_1a_2-c_2a_1 \right )^2}}\right |

where a_1=3

b_1=-2

c_1=-3

a_2=-9

b_2=6

c_2=9

distance(d)=0 units since value of the matrix

\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1\\ a_1&b_1&c_1 \\a_2&b_2 &c_2 \end{vmatrix}

is zero

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<u>Complete Question:</u>

Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.

<u>Correct Answer:</u>

A) (10-x)(12-x)=15

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<u>Step-by-step explanation:</u>

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Let the reduced dimensions is by x , So the new dimensions are

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According to question , Area of new image is :

⇒ Area = \frac{1}{8}Length(breadth)

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So the equation will be :

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⇒ x =7

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