Chapter 10: Q .41. (page 849)

Show that the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8 are parallel, and find the distance between the planes

Short Answer

Expert verified

$Therefore, the distance from the point P (1, 2, - 3) to the line r (t) = ⟨ 2 + 5 t, - 3 + t, - 4 t ⟩ is 6.10 u n i t s$

Step by step solution

Step 1:Given information

the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8

Step 2:Calculation

Consider the point $P (1, 2, - 3)$ and the line with the equation $r (t) = ⟨ 2 + 5 t, - 3 + t, - 4 t ⟩$ and the plane with the equation $x - 3 y + 4 z = 5$ .

$The point P_{0} (2, - 3, 0) corresponds to r (0) and the direction vector of the line is d = ⟨ 5, 1, - 4 ⟩ .$

$The norm of the direction vector d = ⟨ 5, 1, - 4 ⟩ is:$

$‖ d ‖ = \sqrt{5^{2} + 1^{2} + (- 4)^{2}}$

$= \sqrt{25 + 1 + 16} = \sqrt{42}$

$The vector \vec{P_{0} P} is given by:$

$\vec{P_{0} P} = ⟨ 1 - 2, 2 - (- 3), - 3 - 0 ⟩$

$= ⟨ - 1, 5, - 3 ⟩$

$The value of d \times \vec{P_{0} P} is:$

$d \times {\vec{P}}_{0} P = |\begin{matrix} i & j & k \\ 5 & 1 & - 4 \\ - 1 & 5 & - 3 \end{matrix}|$

$= - 23 i - 19 j + 26 k$

$= ⟨ - 23, - 19, 26 ⟩$

The distance from point P (1, 2, - 3) to the line r (t) = ⟨ 2 + 5 t, - 3 + t, - 4 t ⟩ is:

$\frac{||d \times \vec{P_{0} P}||}{‖ d ‖} = \frac{‖ ⟨ - 23, - 19, 26 ⟩ ‖}{\sqrt{42}}$

$= \frac{\sqrt{1566}}{\sqrt{42}}$

$\approx 6.10 units$

$Therefore, the distance from the point P (1, 2, - 3) to the line r (t) = ⟨ 2 + 5 t, - 3 + t, - 4 t ⟩ is 6.10 u n i t s$

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Show that the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8 are parallel, and find the distance between the planes

Short Answer

Step by step solution

Step 1:Given information

Step 2:Calculation

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