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The planes are: $$2x-3y+z=4 \quad \text{and}\\ 4x-6y+2z=3,$$ my answer that I got as I tried so far is $\dfrac{5}{\sqrt{56}}$

Qurultay
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Antonius Mateus I
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    Can you show us what have you tried? – openspace Mar 06 '20 at 09:17
  • If you show what steps you took to reach your answer of $\frac{5}{\sqrt{56}}$ then we will be able to see whether you have gone wrong somewhere or whether there is something about this question you are misunderstanding. – lioness99a Mar 06 '20 at 09:27
  • When I tried I got 5÷√(56) as the final answer. I find a point in the first equation (0,0,4) by letting X and y equal to zeros. Then used the second equation in the distance formula, and substitute that point . – Antonius Mateus I Mar 06 '20 at 10:04

2 Answers2

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Hint:

Take any arbitrary point on any plane (lets say plane $1$)

Can you use that point's position vector and the normal vector of the plane to find when the normal vector from this point touches plane $2$?

Can you reason why this is the shortest distance i.e perpendicular distance between the $2$ planes

h-squared
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Let $\Pi_1$ be the plane $$2x-3y+z=4$$

Let $\Pi_2$ be the plane $$4x-6y+2z=3$$ The vector [2,-3,1] is normal to both planes. The point (1,1,5) is on the plane $\Pi_1.$ Let $L$ be the line through (1,1,5) in the direction of the normal vector. Parametric equations of $L$ are $$x=1+2t, y=1-3t, z=5+t$$ Substitute these expressions for $x,y$ and $z$ into the equation of $\Pi_2$ and solve for $t.$ Substituting this value of $t$ into the equations of $L$ gives you the point $P$ where the line $L$ pierces the plane $\Pi_2.$ The distance between the two planes is the distance beween the points (1,1,5) and $P.$

P. Lawrence
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