Problem

(a) Find the point at which the given lines inter…

08:27

Problem 63 Hard Difficulty

Find an equation of the plane with $ x $-intercept $ a $, $ y $-intercept $ b $, and $ z $-intercept $ c $.

Name: find an equation of the plane with x intercept a y intercept b and z intercept c
Uploaded: 2021-08-20
Duration: 6 min 51 s
Channel: Numerade
Description: Find an equation of the plane with $ x $-intercept $ a $, $ y $-intercept $ b $, and $ z $-intercept $ c $.

Answer

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

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Video Transcript

the question they are asking to find the equation of the plane with X intercept a, Y intercept B. And they're intercept. See therefore the access can be drawn as this, this is the Y axis, this is the X axis and this is the the nexus. These negatives that access negative Y axis and is the negative X. Axis. And or is the origin. So from this glove according to the question, the intercept for each of the excess are for X axis. It is A. For why access it is me. And for that access it is see so up top connecting this points. Therefore I'm stopped playing. Who's the equation? We have to be found out. So uh for finding out the situation of the coordinates for these points are for B. It is zero B. Zero For a this is a 00 but see it is 00 C. So for in order to find the aggression we can find any two lines in this plane. That is Any two victors of this. Um The insides that is supposed the victim S. C. Is equal to okay zero minus a cap less. 0 0 Jacob plus C zero K. Cup. It is equal to minus a. I. Cap. Let's see K. Cap. And this is equal to minus a comma zero comma C. This is a better A C. And another vector for the plane is uh B. C. So B. C. Is equal to 0 0 IPs last zero minus be Jacob plus zero minus Plus C -0 0 Jacob. Therefore this is equal do I? This is equal to -7. Jacob plus he Jacob equal to zero minus Bc. So in order to find the normal vector to the plane it is equal to the cross product of these two vectors. That is a C vector. Cross bc vector. This is equal to the cross product rule I J K. And the values of our- is zero C. And the value of B C. R minus B. Yeah zero minus Bc. So this is equal to icap, zero plus busy minus Jacob minus a c minus zero. Let's kick up a B minus zero. That is equal to B. C. Icap plus a C. Jacob plus a B k. Cup. So the normal vector to the plane is equal to B C, a c a b. And we consider any point in the plane, I suppose the point C. Is equal to 00 C. To find the equation of the plane. Therefore equation of the len is BC into X -0 0 Plus AC into Y zero. Less baby into zed minus C Is equal to zero. This is equal to B, C. X plus S E Y Plus AB, Z- ABC. Equal to zero. So now dividing both sides of the equation by abc, we get X by a plus by baby present by sequel to one. And this is the required equation on the plane

Problem

(a) Find the point at which the given lines inter…

Problem 63 Hard Difficulty

Find an equation of the plane with $ x $-intercept $ a $, $ y $-intercept $ b $, and $ z $-intercept $ c $.

Answer

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

More Answers

Topics

Calculus: Early Transcendentals