Class 12 Maths Three Dimensional Geometry | Angle between two lines (in terms of Direction cosines) |

**Angle between two lines (in terms of Direction cosines)**

Considering 2 line vectorsP and Q and the direction cosines of P (l_{1}, m_{1}, n_{1}) and of

Q(l_{2}, m_{2}, n_{2}). Magnitude of P =r_{1} and magnitude of Q = r_{2}.

Therefore P^{->}=r_{1}(l_{1}î+m_{1}ĵ +n_{1}k̂) and Q^{->} =r_{2}(l_{2}î+m_{2}ĵ +n_{2}k̂)

P^{->}.Q^{->}= IPIIQIcosϴ

cosϴ= (P^{->}.Q^{->})/(IPIIQI)

=r_{1} (l_{1}î +m_{1}ĵ +n_{1}k̂). r_{2}(l_{2}î +m_{2}ĵ +n_{2}k̂)/(r_{1}√l_{1})^{2}+(m_{1})^{2}+(n_{1})^{2}) (r_{2}√((l_{2})^{2}+(m_{2})^{2}+(n_{2})^{2})

**cosϴ ****=Il _{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}I**(using for any line (l)

__Perpendicular and parallel Test between 2 line vectors__

Two lines with direction ratios a_{1}, b_{1}, c_{1}and a_{2},b_{2},c_{2}.

__When Perpendicularϴ=90 ^{0}:-__

When direction ratios are given then a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0,and when direction cosines are given then **l _{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0**.

__When Parallelϴ=0:-__

**(a _{1}/a_{2})=(b_{1}/b_{2})=(c_{1}/c_{2}) or (l_{1}/l_{2})/(m_{1}/m_{2})=(n_{1}/n_{2})=0**

**Angle between lines when line equations are given (Vector)**

Consider the equation of a line passing through a_{1}⃗ (point vector)and parallel to b_{1}⃗ such that r_{1}⃗ = a_{1}⃗** + **λ b_{1}⃗ andanother line passing through a_{2}⃗ (point vector)and parallel to b_{2}⃗ such that r_{2}⃗= a_{2}⃗ +μ b_{2}⃗.

From the figure we can say that b_{1}⃗ and b_{2}⃗ are line vectors.

Angles between lines r_{1}⃗ and r_{2}⃗ are same as angle between line vectors b_{1}⃗ and b_{2}⃗

__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 b_{1}⃗=2î+5ĵ -3k̂andb_{2}⃗=-î +8ĵ +4k̂

Ib_{1}⃗I=√ ((2)^{2}+ (5)^{2}+ (-3)^{2}) =√38

Ib_{2}⃗I= √ ((-1)^{2}+ (8)^{2}+ (4)^{2}) =√81 =9

b_{1}⃗.b_{2}⃗ = (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, a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2}, are perpendicular to each other,

If a_{1}a_{2} + b_{1} b_{2} + c_{1}c_{2} = 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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