Bionomial Theoram |
| Introduction
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 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 = 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 |




&
.
be angle

& || to given line EF(
).
=
(By
law)
=
? 
b Eq. of line in parametric form.
& passing through pt.(1, 2, 3) ?
=
eq. of line passing through
(



is cartesion eq. of line
) = 3(y -
, b =
, c =

= 0 
& 


.
, y1 +
, z1 +
)


