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RD Sharma Class 12 Exercise 27.3 Straight line in space Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 27.3 Straight line in space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 09:47 AM IST

The RD Sharma class 12 solution of Straight line in space exercise 27.3 is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12th exercise 27.3 consists of 10 questions including subparts which explores vector and cartesian form, finding the point of intersection between the two lines.

RD Sharma Class 12 Solutions Chapter27 Straight line in space - Other Exercise

Straight Line in Space Excercise: 27.3

Straight Line in Space exercise 27.3 question 1

Answer: The given line intersect at a point (2,6,3).

Hint: By putting the value of γ.

Given:x1=y22=z+33 …(i)
and x22=y63=z34 …(ii)

Solution:x1=y22=z+33=γ (say)
Then x=γ,y=2γ+2,z=3γ3
Substituting x,y,z in eqn (ii)
γ22=2γ+263=3γ334
Taking first and second ratios, we get
γ22=2γ433γ6=4γ8γ=2
Now taking second and third ratios, we get
2γ43=3γ648γ16=9γ18γ=2
The line intersects at a point (γ,2γ+2,3γ3)
taking γ=2
Points will be (2,6,3).

Straight Line in Space exercise 27.3 question 2

Answer: The given line don’t intersect each other.

Hint: By substituting λ=17 and μ=12

Given:x13=y+12=z15 and x+24=y13=z+12

Solution: The co- ordinates of any point on the first line are given by
x13=y+12=z15=λ( say )
x=3λ+1y=2λ1z=5λ+1
The co-ordinates of first line is (3λ+1,2λ1,5λ+1)
The co- ordinates of any point on the second line are given by
x+24=y13=z+12=μ(say)
 x=4μ2 y=3μ+1 z=2μ1
The co-ordinates of second line is (4μ2,3μ+1,2μ1)
If the line intersect, then they have a common point. So, for some value of λ and μ.
We must have,
3λ+1=4μ2,2λ1=3μ+1,5λ+1=2μ1
3λ4μ=3(i)2λ3μ=2(ii)5λ+2μ=2(iii)
Solving (i)and (ii), we get
λ=17μ=12
Substituting λ=17 and μ=12 in (iii)
L.H.S
=5λ+2μ=5(17)+2(12)
=8524=1092 L.H.S R.H.S

This shows that the given line will not intersect each other.

Straight Line in Space exercise 27.3 question 3

Answer: The answer is (12,12,32)

Hint: x+13=y+35=z+57=λ and x21=y43=z65=μ( say )

Given:x+13=y+35=z+57 and x21=y43=z65

Solution: we have,
x+13=y+35=z+57=λ(say)x=3λ1,y=5λ3,z=7λ5
So, the co-ordinates of a general point on this line are (3λ1,5λ3,7λ5)
The equation of second line is given
x21=y43=z65=μ( say )x=μ+2,y=3μ+4,z=5μ+6
So, the co-ordinates of general point on this line are (μ+2,3μ+4,5μ+6)
If the line intersect, then they have a common point. So, for some value of λ and μ,we must have
3λ1=μ+2,5λ3=3μ+4,7λ5=5μ+63λμ=3(i)5λ3μ=7(ii)7λ5μ=11(iii)
Solving the first two equations (i) and (ii), we get
λ=12μ=32.
Since,λ=12 and μ=32 satisfying the eqn(iii) 7λ5μ=11

Hence,the given lines intersect each other.
When λ=12 in (3λ1,5λ3,7λ5)
the co-ordinates of the required point of intersection are (12,12,32)

Straight Line in Space exercise 27.3 question 4

Answer: The answer of the given equation is (10,14,4).

Hint: By substituting the value of λ and μ.

Given:A(0,1,1),B(4,5,1),C(3,9,4)andD(4,4,4).

Solution: The co-ordinates of any point on the line AB are given by
x040=y+15+1=z+11+1=λx=4λy=6λ1z=2λ1
The co-ordinates of general point on AB are
(4λ,6λ1,2λ1)
The co-ordinates of any point on the line CD are given by
x343=y949=z444=μx=7μ+3y=5μ+9z=4
The co-ordinates of general point on CD are
(7μ+3,5μ+9,4)
If the line intersect, then they have a common point. So, for some value of λ and μ.
We must have,
4λ=7μ+3,6λ1=5μ+9,2λ1=44λ+7μ=3(i)6λ+5μ=10(ii)λ=52(iii)
Solving (ii) and (iii), we get
λ=52μ=1
Substitutingλ=52 and μ=1 in (i),we get
 L.H.S =4λ7μ=4(52)7(1)=3 R.H.S 
Since λ=52 and μ=1 satisfying eqn(iii),
Hence, the given line intersect.
Substituting the value of λ and μ in the Co- ordinates of a general point on the line AB and CD, we get
x=10y=14z=4

Hence, AB and CD intersect at point (10,14,4).

Straight Line in Space exercise 27.3 question 5

Answer: The answer of the given question is (4,0,1).

Hint: Equate the coefficient of vector equation of line.

Given: vector r=(ı^+ȷ^k^)+λ(3ı^ȷ^) and vector r=(4ı^k^)+μ(2ı^+3k^)

Solution: General points on the lines are
(1+3λ)ı^+(1λ)ȷ^k^ and (4+2μ)ı^+(3μ1)k^
Lines intersect if
(1+3λ)=(4+2μ)1λ=0 and 3μ1=1
From (ii) and (iii), λ=1,μ=0
Substituting in equation (i)
Since,1+3(1)=4+2(0) is true.
Hence, the lines intersect.
Point of intersection is :4ı^k^ or (4,0,1)

Straight Line in Space exercise 27.3 question 6(i)

Answer: The answer of the given question is that the line do not intersect each other.

Hint: Assume λ=λ1 and μ=μ1

Given: r=(ı^ȷ^)+λ(2ı^+k^) and r=(2ı^ȷ^)+μ(ı^+ȷ^k^)

Solution: let’s first assume that the line intersect for λ=λ1 and μ=μ1
Then, (2λ1+1)ı^ȷ^+λ1k^=(2+μ1)ı^+(μ11)ȷ^μ1k^
Equations we get,
(2λ1+1)=(2+μ1)(i)1=μ11(ii)λ1=μ1(iii)
Solving equation (ii) and (iii), we get λ1=μ1=0
Which doesn’t satisfy the equation (i) which is a contradiction.
Thus, the above lines are skew lines i.e. they neither intersect nor parallel to each other.

Straight Line in Space exercise 27.3 question 6(ii)

Answer: The given line do not intersect.

Hint: If (a2a1)(b1×b2)=0 then the lines intersect each other.

Given:x12=y+13=z and x+15=y21;z=2

Solution:x12=y+13=z01 and x+15=y21=z20
The first line passes through the point (1,-1,0) and has direction ratio proportional to 2,3,1 and its vector equation is
r=a1+λb1 …(i)
Here,a1=i^j^+0k^b1=2i^+3j^+k^
Also, the second line passes through the point (1,2,2) and has direction ratios proportional to 5,1,0.
Its vector equation is
r=a2+μb2 ...........(ii)
Here,
a2=i^+2j^+2k^b2=5i^+j^+0k^
Now,
a2a1=2i^+3j^+2k^
And
b1×b2=|i^j^k^231510|=ı^+513k^(a2a1)(b1×b2)=(2i^+3j^+2k^)(i^+5j^13k^)=2+1526=9.
We observe , (a2a1)(b1×b2)0
Thus,the given lines don't intersect.

Straight Line in Space exercise 27.3 question 6(iii)

Answer: The line intersect at the point of intersection is (4,0,1).

Hint: Equate the coefficient of vector equation of line.

Given:x13=y11=z+10 and x42=y00=z+13

Solution: co-ordinate of first equation of line
x13=y11=z+10=λ( let )(i)x=3λ+1,y=λ+1,z=1
Co-ordinate of second equation of line
x42=y00=z+13=μ( let )(ii)x=2μ+4,y=0,z=3μ1
If the lines intersect, then they have a common point for some value of λ and μ.
 So, 3λ+1=2μ+4(iii)λ+1=0λ=1(iv)3μ1=1μ=0(v)λ=1 and μ=0
Satisfy equation (iii). So, the given line intersect and the point of intersection is (4,0,1).

Straight Line in Space exercise 27.3 question 6(iv)

Answer: The line intersect each other and the point of intersection is (1,3,2).

Hint: Equate the coefficient of vector equation of line.

Given: Two equation of line.

Solution: The equation of the first line,
x54=y74=z+35=λ(let)x=4λ+5,y=4λ+7,z=5λ3
So, the co-ordinates of a general point on this line are (4λ+5,4λ+7,5λ3)
The equation of the second line,
x87=y41=z53=μ( let )x=7μ+8,y=μ+4,z=3μ+5
So, the co-ordinates of a general point on this line are ((7μ+8,μ+4,3μ+5)
If the lines intersect, then they have a common point for some value of λ and μ
We have,
4λ+5=7μ+8,4λ+7=μ+4,5λ3=3μ+54λ7μ=3(i)μ=4λ+3(ii) 5λ3μ=8(iii)
Putting the value of μ from equation (ii) to (i), we get
4λ7μ=34λ7(4λ+3)=34λ28λ21=3
24λ=24λ=1
Now putting the value of λ in equation (ii), we get
μ=4λ+3μ=4(1)+3μ=1
As we can see by putting the value of λ and μ in equation (iii), that it satisfy the equation
Check,  5λ3μ=8
5(1)3(1)=8
$5+3=8 $
8=8
L.H.S.=R.H.S.
Hence intersection point exists or line of intersects.
Thus, point of intersection (4λ+5,4λ+7,5λ3)
=(4(1)+5,4(1)+7,5(1)3)=(1,3,2)

Straight Line in Space exercise 27.3 question 7

Answer: The point of intersection is (1,6,12)

Hint: Equate the coefficient of vector equation of line.

Given: r=3ı^+2ȷ^4k^+λ(ı^+2ȷ^+2k^) and r=5ı^2ȷ^+μ(3ı^+2ȷ^+6k^)

Solution: let the lines are
M:r=3t+2J4k+λ(ı+2ȷ+2k) N:r=5ı2ȷ+μ(3l+2J+6k)
Co-ordinate of any random point on M are P(3+λ,2+2λ,4+2λ) and on N are Q(5+3μ,2+2μ,6μ)
If the lines Mand N intersect then ,they must have a common point on them i.e. P and Q must coincide for some values of λ and μ.
Now,
3+λ=5+3μ... (i)2+2λ=2+2μ... (ii)4+2λ=6μ... (iii)
Solving (i) and (ii) ,we get
λ=4μ=2
Substitute the value in equation (iii),
4+2(4)=6(2)4=12+84=4
So, the given line intersect each other.
Now, point of intersection isP(1,6,12).


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