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Chapter 10: Q .42. (page 849)

Show that the planes given by y 鈭� 7z = 16 and 2y 鈭� 14z = 5 are parallel, and find the distance between the planes.

Short Answer

Expert verified

$Therefore, the distance from the point to the plane is \frac{11 \sqrt{26}}{13} units.$

Step by step solution

01

Step 1:Given information

The planes given by y 鈭� 7z = 16 and 2y 鈭� 14z = 5

02

Step 2:Calculation

$Consider the point P (1, 2, - 3) and the line with the equation r (t) = 鉄� 2 + 5 t, - 3 + t, - 4 t 鉄� and the$

$plane with the equation x - 3 y + 4 z = 5 .$

$The objective is to find the distance from the point P (1, 2, - 3) to the plane with the equation$

$x - 3 y + 4 z = 5$

$The normal vector of the plane x - 3 y + 4 z = 5 is:$

$N = 鉄� 1, - 3, 4 鉄�$

$The point R = (5, 0, 0) is on the plane x - 3 y + 4 z = 5 .$

$The vector \overset{鈫�}{R P} is given by:$

$\overset{鈫�}{R P} = 鉄� 1 - 5, 2 - 0, - 3 - 0 鉄�$

$= 鉄� - 4, 2, - 3 鉄�$

The distance from P to the plane is:

$|{comp}_{N} \overset{鈫�}{R P}| = \frac{| N 路 R P |}{鈥� N 鈥�}$

$= \frac{| 鉄� 1, - 3, 4 鉄� 路鉄� - 4, 2, - 3 鉄� |}{鈥� 鉄� 1, - 3, 4 鉄� 鈥�}$

$= \frac{| - 4 - 6 - 12 |}{\sqrt{1 + 9 + 16}}$

$= \frac{| - 22 |}{\sqrt{26}}$

$= \frac{22}{\sqrt{26}}$

$= \frac{11 \sqrt{26}}{13}$

$Therefore, the distance from the point to the plane is \frac{11 \sqrt{26}}{13} units.$

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