Revise notes JEE introduction to three dimensional geometry

Revise notes
JEE Maths
Introduction to Three Dimensional Geometry

Topics

1. 3D Geometry
2. Coordinate Axes and Coordinate Planes
3. Coordinates of a Point in Space
4. Distance between 2 points
5. Section Formula
6. Represent a point in Cartesian and Vector form
7. Direction Cosines of a Line
8. Direction Ratios of a Line
9. Equation of a line in Space
10. Equation of a line passing through two given points
11. Angle between two lines
12. Angle between two lines (in terms of Direction cosines)
13. Shortest Distance between two lines
14. Distance between parallel lines

Angle between two lines (in terms of Direction cosines)

Considering 2 line vectorsP and Q and the direction cosines of P (l₁, m₁, n₁) and of

Q(l₂, m₂, n₂). Magnitude of P =r₁ and magnitude of Q = r₂.

Therefore P^->=r₁(l₁î+m₁ĵ +n₁k̂) and Q^-> =r₂(l₂î+m₂ĵ +n₂k̂)

P^->.Q^->= IPIIQIcosϴ

cosϴ= (P^->.Q^->)/(IPIIQI)

=r₁ (l₁î +m₁ĵ +n₁k̂). r₂(l₂î +m₂ĵ +n₂k̂)/(r₁√l₁)²+(m₁)²+(n₁)²) (r₂√((l₂)²+(m₂)²+(n₂)²)

cosϴ =Il₁l₂+m₁m₂+n₁n₂I(using for any line (l)²+ (m)²+ (n)² =1)

Perpendicular and parallel Test between 2 line vectors

Two lines with direction ratios a₁, b₁, c₁and a₂,b₂,c₂.

When Perpendicularϴ=90⁰:-

When direction ratios are given then a₁a₂+b₁b₂+c₁c₂=0,and when direction cosines are given then l₁l₂+m₁m₂+n₁n₂=0.

When Parallelϴ=0:-

(a₁/a₂)=(b₁/b₂)=(c₁/c₂) or (l₁/l₂)/(m₁/m₂)=(n₁/n₂)=0

Angle between lines when line equations are given (Vector)

Consider the equation of a line passing through a₁⃗ (point vector)and parallel to b₁⃗ such that r₁⃗ = a₁⃗ + λ b₁⃗ andanother line passing through a₂⃗ (point vector)and parallel to b₂⃗ such that r₂⃗= a₂⃗ +μ b₂⃗.

From the figure we can say that b₁⃗ and b₂⃗ are line vectors.

Angles between lines r₁⃗ and r₂⃗ are same as angle between line vectors b₁⃗ and b₂⃗

Problem:-

Find the angle between the following pairs of lines:

r ⃗=2î-5ĵ+k̂+λ (3î-2ĵ +6k̂) and r ⃗=7î-6k̂+μ (î+2ĵ +2k̂)?

Answer:-

=3+4+12

=19

=> cosϴ = (19)/ (7x3)

=>ϴ = cos^-1(19/21)

Problem:-

Find the angle between the following pairs of lines:

(x-2)/ (2) =(y-1)/ (5)/ (z+3)/ (-3) and (x+2)/ (-1) =(y-4)/ (8) = (z-5)/ (4).

Answer:-

(x-2)/ (2) =(y-1)/ (5)/ (z+3)/ (-3) and (x+2)/ (-1) =(y-4)/ (8) = (z-5)/ (4)

Therefore b₁⃗=2î+5ĵ -3k̂andb₂⃗=-î +8ĵ +4k̂

Ib₁⃗I=√ ((2)²+ (5)²+ (-3)²) =√38

Ib₂⃗I= √ ((-1)²+ (8)²+ (4)²) =√81 =9

b₁⃗.b₂⃗ = (2î+5ĵ -3k̂). (-î+8ĵ +4k̂)

=2(-1) +5x8+ (-3).4

=-2+40-12

=26

=>cosϴ= (26)/ (9√38)

=>ϴ =cos^-1(26)/ (9√38))

Problem:-

Find the values of p so the line (1-x)/(3)=(7y-14)/(2p)=(z-3)/(2)and (7-7x)/(3p)=(y-5)/(1)=(6-z)/(-5)are atright angles.

Answer:-

The given equations can be written in the standard form as:

(1-x)/(3)=(7y-14)/(2p)=(z-3)/(2)and (7-7x)/(3p)=(y-5)/(1)=(6-z)/(-5)

The direction ratios of the lines are −3, (2p)/ (7), 2 and (-3p)/ (7), 1,-5 respectively.

Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular to each other,

If a₁a₂ + b₁ b₂ + c₁c₂ = 0.

Therefore ((-3)x(-3p)/(7))+((2p)/(1))+((2).(-5))=0

=> (9p)/ (7) + (2p)/ (7) =10

=>11p=70

=>p=(70/11).

Thus, the value of p is (70/11).

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