Illustration: Find angle b/w lines whose direction cosines are & .
Ans: Let be angle
cos = l1l2 + m1m2 + n1n2
cos = - = 1200
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.
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
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 line
6x + 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 =
Condition for perpendicularity:
. = 0 a1a2 + b1b2 + c1c2
Condition for parallelism:
Dumb Question: Why angle between pair of lines
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