(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find th distance of point P to the plane OQR, O being the origin.

"PS" is perpendicular to plane "OQR"

P(3,-2,-1)

S(x,y,z)

O(0,0,0)

Q(1,3,4)

R(2,1,-2)

2. Relevant equations

A) The equation in terms of (x,y,z) obtain by triple product of OS,OQ & OR.

OS.(OQXOR)=0......... [all being co-planar therefore =0]

we get,

-10x+10y-5z=0

B)The Equation on which I'm stuck.I didn't get how PS vector is obtained from coordinates of P & S ;PS=(3-x)i+(-2-y)j+(-1-z)k

Then direction ratio is used to get

x+y=1

y+2z=-4

Now we have 3 equations & 3 Unknowns

3. The attempt at a solution

I understood how first equation is obtained that means by using theorem :- three vector being co-planar then

A.(BXC)=0..... (as we get volume from this, being co-planar means no volume)

So equation 1 is obtained that way

Now in Second equation

Vector cross product is done OQ X OR= -10i + 10j -5k

Then,

Where I'm stuck:-

Vector PS is given by (3-X)i + (-2-Y)j+(-1-Z)k,

How??? Vector PS is obtained by coordinate of P & S????

Next two equations are simple it just uses the direction ratio.

Solving 3 equation I'll get x,y& z.

Then find vector PS from above and find magnitude of PS that means it's length.

For reference please check below image :-

https://m.imgur.com/a/dEXAT

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# Homework Help: Finding distance of "P" from a plane OQR using vector

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