# 3D Geometry - 3

Introduction

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**Vector Form**Since L lies on line AB

P.V. of L = P.V. of line AB

= +

dr's of

Since

P.V. of L is +

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**Reflection of point in straight line**__:__

**Castesian Form**From above, we get coordinate of L(foot of )

But L is mid point of PQ

= x

_{1}+ a, = y

_{1}+ b, = z

_{1}+ c

' = 2(x

_{1}+ a) - , ' = 2(y

_{1}+ b) - , ' = 2(z

_{1}+ c) -

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**Vector Form**From above, we get

P.V. of L, +

Let P.V. of Q is

Since L is mid point of PQ

__: Find reflection of point P(2, 3, 1) in line__

**Illustration**Ans:

Since L lies on line AB

coordinate of L (3 + 2, 2 + 1, 4 - 3)

DR's of PL are

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

= (3, 2 - 2, 4 - 4)

DR's of AB are (3, 2, 4)

Since PL AB

3(3) + 2(2 - 2) + 4(4 - 4) = 0

9 + 4 - 4 + 16 - 16 = 0

29 = 20 =

Since L is mid point of PQ

So, = 3 + 2, = 2 + 1, = 4 - 3

= (6 + 4 - 2), = (4 + 2 - 3), = 8 - 6 - 1

= 6 + 2, = 4 - 1, = 8 - 7

where =

__: Those lines which do not lies in same plane.__

**Skew lines**__:__

**Shortest distence b/w two skew lines**The line which is to both line l

_{1}& l

_{2}are c/d line of shortest distance.

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**Vector Form**Let l

_{1}& l

_{2}are:

& respectively.

Since is to both l

_{1}& l

_{2}which are parallel to &

is || to x

Let be unit vector along , then = ±

PQ = Projection of on

= .

__: How PQ = Projection on ?__

**Dumb Question**Ans: Form fig., it is clear that PQ is projection of on .

PQ =

__: Two skew lines__

**Cartesian form**&

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**Shortest distance**__:__

**Condition for lines to intersect**Two lines intersects if shortest distance = 0

or

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**Shortest distance b/w parallel line**Let l

_{1}& l

_{2}are

respectively

& BM is shortest distance b/w l

_{1}& l

_{2}

sin = BM = AB sin = || sin

| x | = || || sin( - )

= || || sin = (|| sin) || =

__: Why we have taken sin( - ) ?__

**Dumb Question**Ans:

Since direction of vector is opposite to || lines. So, we have taken ( - ) instead of

BM =

__: Find shortest distance b/w lines:__

**Illustration**Ans:

On comparing

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**Plane**(i) Eq. of plane passing through a given point (x

_{1}, y

_{1}, z

_{1}) is:

a(x - x

_{1}) + b(y - y

_{1}) + c(z - z

_{1}) = 0 where a, b, c constants.

Proof: General eq. of plane is ax + by + cz + d = 0 ....................................... (i)

It is passes through (x

_{1}, y

_{1}, z

_{1})

ax

_{1}+ by

_{1}+ cz

_{1}+ d = 0 ............................................. (ii)

By (i) - (ii), we get

a(x - x

_{1}) + b(y - y

_{1}) + c(z - z

_{1}) = 0

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**Intercept form of a plane**eq. of plane of intercepting lengths a, b & c with x, y & z-axis respectively is,

__: A variable plane moves in such a way that sum of reciprocals of its intercepts on 3 coordinate axes is constant. Show that plane passes through fixed point.__

**Illustration**Ans: Let eq. of plane is . Then, intercepts of plane with axes are:

A(a, 0, 0), B(0, b, 0), c(0, 0, c)

= constant (k) (given)

= 1 & comparing with fixed point

x = , y = , z =

This shows plane passes through fixed point (, , )

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**Vector eq. of plane passing through a given point & normal to given vector**VEctor eq. of plane passing through u point of P.V. & normal to vector is ( - ). = 0

__: What is normal to vector ?__

**Dumb Question**Ans: Plane normal to vector means the every line in plane is to that given vector.

Proof: Let plane passes through A() & normal to vector & be P.V. of every point 'P' on palne.

Since lies in plane & is normal to plane.

. = 0

( - ). = 0 ( = - )

eq. of plane ( - ). = 0

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**Eq. of plane in normal form**Vector eq. of plane normal to unit vector & at O distance d from origin is . = d.

Proof:

ON is to plane such that & =

Since . = 0

( - ). = 0

( - d).d = 0 .d - d

^{2}. = 0

d(.) - d

^{2}= 0 . = d

eq. of plane is . = d

__: Let l, m, n be d.r.'s of normal to given plane & P is length of from origin to plane, then eq. of plane is lx + my + nz = P.__

**Cartesian Form**