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# Show that the distance between the parallel planes $ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$ is$$D = \frac{\mid d_1 - d_2 \mid}{\sqrt{a^2 + b^2 + c^2}}$$

## $\frac{\left|d_{2}-d_{1}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

in the question they're asking too. So that the distance between the panel planes given by the equation X plus B. Y proceeded LSD one is equal to zero and expressly where he proceeded to Z two is equal to zero. And the distance between these planes that a panel to each other is equal to model the value of the one minus D. Two divided by route under a script. This discrepancy square, okay. Therefore in order to find the distance between these palette lines first we assume a point that lies on a plane one that is let P one B a point that this X one, Y one, Z one be a point that lies on plane one. Therefore the distance of perpendicular from plane one point X one x 1, 01 two plane two which are parlance planes is equal to capital D, which is a distance from this point too plain too, which are parallel to each other is equal to modular value of a X one plus B Y one plus C. Z one plus the two divided by route under a square plus b square plus c square. So if no, we put the value of the point That we have just assumed to be on the plane. P one That is X one, Y 1 and Z one in equation of plain one, we get the equation written as a X one plus B Y one plus cesid one. In place of X Y Z report the value of X one by one answered one Is equal to -1. Mhm. Therefore in this situation that is the value of the Distance between the points on playing one and the plane to is denoted by capital D. So this is equation one. So if we put the value of this equation to in putting value of equation too in equation one we get capital D. Is equal to modular value of in place of X one, B by one plus. Is that when we put the value of equation two that is minus the one. Therefore in place of equation one we right the substituted value and we get the tu minus D. One in the numerator, the scalar value in the numerator and divided by the value. That is in equation one route under a squared plus B squared plus C. Square. So therefore this is the distance between the Two Parallel Planes P one and P two. Mhm does. This is the answer. I think you're in question.

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp