3D Geometry - 2
Illustration: Find angle b/w lines whose direction cosines are 
& 
.
Ans: Let 
be angle
cos = l1l2 + m1m2 + n1n2
cos = - 
= 1200
Straight line:
Vectorial eq. of line:
Let a line passing through a point of P.V.
& || to given line EF(
).Then, eq. of line
Proof: Let P be any point on line AP & its P.V. is
Then,
= 
= 
(By 
law)Dumb Question: Why
= 
?Ans: Since
is || to EF of P.V.So,
= Note: Eq. of line through origin & || to
is = Cartesion eq. of straight line:
Passing through point & parallel to given vector .
Let line is passing through A(x1, y1, z1) & || to EF whose d.r.'s are a, b, c.
d.r.'s of AP = (x - x1, y - y1, z - z1)
d.r.'s of EF = (a, b, c)
Since EF || AP, then,
b Eq. of line in parametric form.Illustration: Find eq. of line || to
& passing through pt.(1, 2, 3) ?Ans: A(1, 2, 3)
= 
eq. of line passing through & || to = () + 
()Vector eq. of line passing through two given points:
Vector eq. of line passing through two points whose P.V. S
& is: = + ( - )Proof:
is collinear with 
= 
Cartesian form:
Eq. of line passing through (x1, y1, z1) & (x2, y2, z2)
D.R.'s of AB = (x2 - x1, y2 - y1, z2 - z1)
D.R.'s of AP = (x - x1, y - y1, z - z1)
Since AB || AP
Illustration: Find cartesian eq. of line are 6x + 2 = 3y - 1 = 2z + 2. Find its direction ratios.
Ans:
is cartesion eq. of line6x + 2 = 3y - 1 = 2z + 2
6(x +
) = 3(y - ) = 2(z + 1)
on comparing a =
, b = 
, c = 
Angle b/w Two lines:
= 
Angle b/w two lines cos
= 
If
=
Condition for perpendicularity:
.
= 0 a1a2 + b1b2 + c1c2Condition for parallelism:
Dumb Question: Why angle between pair of lines
Ans:
& 
Perpendicular distance of point froma line:
(a) Castesian form:
Suppose 'L' is foot of line of
.Since L lies on line AB so,
coordinate of L is
= i.e. L(x1 +
a, y1 + b, z1 + c)direction ratios of PL are:
(x1 +
a - 
, y1 + b - 
, z1 + c - 
)also direction ratios of AB are (a, b, c)
Since PL
AB a(x1 + a - ) + b(b - ) + c(z1 + c - ) = 0