What's the vector form of the line going through this point?

P = (1, 1, 2), Q = (-1, 0, 3), R = (2, 1, -1).

Line passes through (1, 1, 2) and is perpendicular to the triangle made by these three points. What's the vector form of this line?

2 Answers
May 25, 2018

# vecr=(1,1,2)+k(3,-5,1), k in RR.#

Explanation:

The reqd. line #L bot DeltaPQR#.

Clearly, the direction #vecl# of #L# must be along the normal #vecn# of the

plane #pi# of the #DeltaPQR#.

#P(1,1,2), Q(-1,0,3) and R(2,1,-1) in pi#.

#vec(QP)xxvec(PR),# being #bot vec(QP) and (PR)#, we may take,

# vecn=vec(QP)xxvec(PR)#,

#=(2,1,-1)xx(1,0,-3)#,

#=|(i,j,k),(2,1,-1),(1,0,-3)|#,

#=-3i+5j-k#.

We take, #vecl=vecn=(3,-5,1)#.

Thus, we see that #L# passes through #P(1,1,2)# and has the

direction #vecl=(3,-5,1)#.

Hence, the vector eqn. of #L# is given by,

# L : vecr=(1,1,2)+k(3,-5,1), k in RR.#

Feel the Joy of Maths.!

May 25, 2018

#(x,y,z) = (1,1,2) + t ( 3,-5,1) quad #for real #t#

Explanation:

The plane containing the triangle is

#ax + by + cz = d#

where only the ratio #a:b:c:d# matters.

#1a + 1b + 2c = d#

#-1 a + 0b + 3c = d#

#2a +b -c = d#

Adding first two,

#b+5c = 2d#

Twice the second plus the third,

#b + 5c = 3d #

Subtracting,

#d = 0#

Substituting,

#a = 3c-d = 3c#

#b = d - a-2c= 0 -3c -2c = -5c #

WLOG let #c=1# giving our plane as

#3x - 5y + z = 0#

Check: #(1,1,2) quad sqrt{} quad quad (-1,0,3) quad sqrt{} quad quad (2,1,-1) quad sqrt{}#

So the normal vector to this plane is #(3,-5,1)# and our line through vertex #(1,1,2)# is

#(x,y,z) = (1,1,2) + t ( 3,-5,1) quad #for real #t#

Some teachers like the equivalent form with unit vectors:

#X = (hat{i} + hat{j} + 2hat{k}) + t(3hat{i} -5 hat{j} + hat{k})#