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Find the distance between the given parallel planes.

$ 2x - 3y + z = 4 , 4x - 6y + 2z = 3 $

$$5 /(2 \sqrt{14})$$

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you have a question and then we need to find, we need to find the distance between the panel planes. This is two weeks minus D. Y. Blessed 24 and four x -6 way Plus. Do that equal to three. We know that distance of a quiet and a plane. If this point is excellent, comma Y. One commas that one. And this is a plane X. B Y. Let's see that there is equal to zero. So this distance becomes a quick too Excellent. Just b. y. one. This these Edwin, LSD modelers. He went by a squared plus B squared plus C square and it all. So basically we need to find points on is a single point on added on this or other on this line this plane. So it is considered this plane that is supposed acceptable to zero And why call to 0? Which means that will be equal to four at this point definitely lies on this. Okay. According to this formula, distance will be equal to with respect to this plane. four in 2, zero -6 in 2, 0 Plus two into 4 minus three months went by four square plus -60 square. Let's do a square under this battle. So this is 008 -3 which is five, bye 16 plus 36 plus four. And the road. Hmm five Banks 36 and the root. So five by to win the road 14. So if you multiply with 14 and divide with 14 fresh lays, we'll be getting distance equal to right under reported by 28 units. Thank you