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Question 1.

Given position vectors OA : i+2j-k and OB: 2i+j+3k in a 3D Cartesian space with origin O of the points A and B.

a) Find the scalar equation of the plane which contains A and which is perpendicular to vector AB.

b) Find the shortest distance from the point (1,-1,1) to the plane obtained in a (a)

2. Relevant equations

3. The attempt at a solution

Here is what i did

(a) Equation of line AB

r= (2,1,3)+ t(1,-1,4)

AP.n = 0

OP.n = OA.n

(x,y,z).(1,-1,4) = (2,1,3).(1,-1,4)

x-y+4z = 13 [equation of plane]?

(b) vector form of plane => r.(1,-1,4) = 13

line equation => r= (1,-1,1) + t(1,-1,4)

Using r=r for both equations i get t= 2/9

Substitute t=2/9 into line equation giving me (11/9, -2/9, 17/9)

I'm stuck here then for part (b)

I'm pretty weak in this chapter, couldn't seem to grab the concept here.

1. The problem statement, all variables and given/known data

Question 2.

Obtain the scalar equation of the plane which passes through point P(1,2,3) and contain a straight line

x(t)= 3t

y(t)= 1+t

z(t)= 2-t

t is the parameter of the line

2. Relevant equations

3. The attempt at a solution

Equation of the line= (0,1,2) + t(3,1,-1)

Let Q and R be the points on the line

t=0 Q=(0,1,2)

t=1 R= (3,2,1)

Therefore PQ= (-1,-1,-1) PR= (2,0,2)

PQxPR= (-2,0,-2)

Thus the equation of the plane

a(x-x1) + b(y-y1) + c(z-z1)= 0

-2(x-1) + 0 -2(z-3) = 0

-2x - 2z + 8 = 0

is this correct?

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# Homework Help: Vectors and Planes

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