Question

Calculate the value of P, such that the line
x-2/6=y-1/P=z-5/4 is perpendicular to the plane 3x-y-2z=0

Answer 1

Do you mean \[\frac{x-2}{6}=\frac{y-1}{P}=\frac{z-5}{4}\]

Answer 2

yes

Answer 3

There is one clumsy way.

The equation for the line implies there exists a parametric equation in the form \[\vec{r}(t)=t(4,P-1,-1)+(2,1,5)\] and the equation of the surface satisfies the equation\[\vec{X}(t,s)=(t,s,\frac{3t-s}{2})\]Find the direction of the normal of the surface by\[\vec{n}=\frac{\partial \vec{X}}{\partial t}\times\frac{\partial\vec{X}}{\partial s}\]This means \[(4,P-1,-1)=\lambda\vec{n},\lambda\in R\]Find λ and solve for P.

This is definitely not the best way to do it, maybe someone else will give you a simpler version. :P

Answer 4

let the point on the line be...
\[x=6t+2\quad y=Pt+\quad z=4t+5\]
eq of normal to the plane:
\[n⃗ =<3,−1,−2>\]
this is the direction of the given line.

Answer 5

Upon some Google-ling, for plane with equation ax+by+cz+d=0 passing through (x1,y1,z1), i.e. (0,0,0) in this case, have the equation (x-x1)/a=(y-y1)/b=(z-z1)/c.

Then we have b=-1=P

Answer 6

do we know that the given line passes throught (0,0,0)?

Answer 7

We know the plane passes through (0,0,0)

Answer 8

@electrokid
what to do next

Answer 9

@msingh It is done already, P=-1

Answer 10

@agostino oh yea.. duh

Answer 11

@electrokid lol

Answer 12

i didn't get it from where @electrokid
has left

Answer 13

@msingh forget me.. use agostino's method mentioned in the most recent post

Answer 14

after this step, plse explain
let the point on the line be...
x=6t+2y=Pt+z=4t+5

eq of normal to the plane:
n⃗ =<3,−1,−2>

this is the direction of the given line.

Answer 15

i m not able to understand , plse anyone explain

Answer 16

what do you know about planes?

Answer 17

it is infinte set of poinnts

Answer 18

forming a flat surface which is extended infinitely far in alll directions

Answer 19

good, this flat surface can be defined by a vector that is perpendicular to all the vectors in the plane. This perp vector is called a normal.

Answer 20

there is a property of vectors such that there dot product equals zero if the vectors are per to each other.

Answer 21

lets define the a vector in the plane as a directed line from point(x,y,z) to point(xo,yo,zo)

all the vectors then take the form <x-xo, y-yo, z-zo>

the vector that is perp to all these vectors in the plane is the normal <a,b,c>

the dot product of the plane and normal vectors is then

a(x-xo) + b(y-yo) + c(z-zo) = 0