Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at March 22, 2023 by Teachoo

Assertion (A):
The acute angle between the line
r = i ̂+j ̂+2k ̂+λ(i ̂-j ̂ ) and the x-axis is π/4
Reason(R):
The acute angle 𝜃 between the lines
r = x
_{
1
}
i ̂+y
_{
1
}
j ̂+z
_{
1
}
k ̂+λ(a
_{
1
}
i ̂+b
_{
1
}
j ̂+c
_{
1
}
k ̂ ) and
r = x
_{
1
}
i ̂+y
_{
1
}
j ̂+z
_{
1
}
k ̂+μ(a
_{
1
}
i ̂+b
_{
1
}
j ̂+c
_{
1
}
k ̂ )is given by
𝑐𝑜𝑠𝜃 = |a
_{
1
}
a
_{
2
}
+ b
_{
1
}
b
_{
2
}
+ c
_{
1
}
c
_{
2
}
|/(√a
_{
1
}
^{
2
}
+ b
_{
1
}
^{
2
}
+ c
_{
1
}
^{
2
}
√a
_{
2
}
^{
2
}
+ b2
^{
2
}
. + c
_{
2
}
^{
2
}

Get live Maths 1-on-1 Classs - Class 6 to 12

Transcript
Question 20 Assertion (A): The acute angle between the line 𝑟 ⃗ = 𝑖 ̂+𝑗 ̂+2𝑘 ̂+𝜆(𝑖 ̂−𝑗 ̂ ) and the x-axis is 𝜋/4 Reason(R): The acute angle 𝜃 between the lines 𝑟 ⃗ = 𝑥_1 𝑖 ̂+𝑦_1 𝑗 ̂+𝑧_1 𝑘 ̂+𝜆(𝑎_1 𝑖 ̂+𝑏_1 𝑗 ̂+𝑐_1 𝑘 ̂ ) and 𝑟 ⃗ = 𝑥_2 𝑖 ̂+𝑦_2 𝑗 ̂+𝑧_2 𝑘 ̂+𝜇(𝑎_2 𝑖 ̂+𝑏_2 𝑗 ̂+𝑐_2 𝑘 ̂ )is given by 𝑐𝑜𝑠𝜃 = |𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 + 〖 𝑐〗_1 𝑐_2 |/(√(〖𝑎1〗^2 + 〖𝑏1〗^2 + 〖𝑐1〗^2 ) √(〖𝑎2〗^2 + 〖𝑏2〗^2. + 〖𝑐2〗^2 ))
Checking Assertion
Assertion (A): The acute angle between the line 𝑟 ⃗ = 𝑖 ̂+𝑗 ̂+2𝑘 ̂+𝜆(𝑖 ̂−𝑗 ̂ ) and the x-axis is 𝜋/4
Equation of x-axis
Let’s consider two points on x-axis – (a, 0, 0), and (0, 0, 0)
Vector equation of a line passing though two points with position vectors 𝑎 ⃗ and 𝑏 ⃗ is
𝒓 ⃗ = (𝒂 ) ⃗ + 𝜆 (𝒃 ⃗ − 𝒂 ⃗)
Here,
(a, 0, 0)
𝒂 ⃗ = a𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂
(0, 0, 0)
𝒃 ⃗ = 0𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂
Thus, equation of line is
𝒓 ⃗ = (a𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂) + 𝜆 [(0𝑖 ̂+0𝑗+0𝑘 ̂ ) − (𝑎𝑖 ̂ +0𝑗 ̂ + 20)]
= a𝑖 ̂ + 𝜆 [−𝑎𝑖 ̂ ]
= (a − 𝜆a)𝒊 ̂
Since (a − 𝜆a) is a constant, let (a − 𝜆a) = k
= k𝒊 ̂
Now, we need to find angle between the line 𝑟 ⃗ = 𝑖 ̂+𝑗 ̂+2𝑘 ̂+𝜆(𝑖 ̂−𝑗 ̂ ) and the x-axis
i.e. Angle between 𝒓 ⃗ = 𝒊 ̂+𝒋 ̂+𝟐𝒌 ̂+𝝀(𝒊 ̂−𝒋 ̂ ) and 𝒓 ⃗ = k𝒊 ̂
Using formula from Reasoning
𝑐𝑜𝑠 𝜃 = |𝒂_𝟏 𝒂_𝟐 + 𝒃_𝟏 𝒃_𝟐 + 〖 𝒄〗_𝟏 𝒄_𝟐 |/(√(〖𝒂𝟏〗^𝟐 + 〖𝒃𝟏〗^𝟐 + 〖𝒄𝟏〗^𝟐 ) √(〖𝒂𝟐〗^𝟐 + 〖𝒃𝟐〗^𝟐. + 〖𝒄𝟐〗^𝟐 ))
Thus, equation of line is
𝒓 ⃗ = (a𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂) + 𝜆 [(0𝑖 ̂+0𝑗+0𝑘 ̂ ) − (𝑎𝑖 ̂ +0𝑗 ̂ + 20)]
= a𝑖 ̂ + 𝜆 [−𝑎𝑖 ̂ ]
= (a − 𝜆a)𝒊 ̂
Since (a − 𝜆a) is a constant, let (a − 𝜆a) = k
= k𝒊 ̂
Now, we need to find angle between the line 𝑟 ⃗ = 𝑖 ̂+𝑗 ̂+2𝑘 ̂+𝜆(𝑖 ̂−𝑗 ̂ ) and the x-axis
i.e. Angle between 𝒓 ⃗ = 𝒊 ̂+𝒋 ̂+𝟐𝒌 ̂+𝝀(𝒊 ̂−𝒋 ̂ ) and 𝒓 ⃗ = k𝒊 ̂
Using formula from Reasoning
𝑐𝑜𝑠 𝜃 = |𝒂_𝟏 𝒂_𝟐 + 𝒃_𝟏 𝒃_𝟐 + 〖 𝒄〗_𝟏 𝒄_𝟐 |/(√(〖𝒂𝟏〗^𝟐 + 〖𝒃𝟏〗^𝟐 + 〖𝒄𝟏〗^𝟐 ) √(〖𝒂𝟐〗^𝟐 + 〖𝒃𝟐〗^𝟐. + 〖𝒄𝟐〗^𝟐 ))
𝒓 ⃗ = 𝒊 ̂+𝒋 ̂+𝟐𝒌 ̂+𝝀(𝒊 ̂−𝒋 ̂ )
Comparing with
𝑟 ⃗ = 𝑥_1 𝑖 ̂+𝑦_1 𝑗 ̂+𝑧_1 𝑘 ̂+𝜆(𝑎_1 𝑖 ̂+𝑏_1 𝑗 ̂+𝑐_1 𝑘 ̂ )
𝒂1 = 1, b1 = −1, c1 = 0
𝒓 ⃗" = k" 𝒊 ̂
Comparing with
𝑟 ⃗ = 𝑥_2 𝑖 ̂+𝑦_2 𝑗 ̂+𝑧_2 𝑘 ̂+𝜆(𝑎_2 𝑖 ̂+𝑏_2 𝑗 ̂+𝑐_2 𝑘 ̂ )
𝒂2 = 1, 𝒃2 = 0, 𝒄2 = 0
Now, cos θ = |(𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 +〖 𝑐〗_1 𝑐_2)/(√(〖𝑎_1〗^2 + 〖𝑏_1〗^2+ 〖𝑐_1〗^2 ) √(〖𝑎_2〗^2 +〖〖 𝑏〗_2〗^2+ 〖𝑐_2〗^2 ))|
= |((1 × 1) + (−1 × 0) + (0 × 0))/(√(1^2 +(−1)^2 + 0^2 ) × √(1^2 + 0^2 + 0^2 ))|
= |1/(√(1 + 1) × √1)|
= |1/√2|
= 𝟏/√𝟐
So, cos θ = 1/√2
∴ θ = 𝜋/𝟒
Therefore, the angle between the given pair of line is 𝜋/𝟒
So, Assertion is true
Checking Reason
Reason(R): The acute angle 𝜃 between the lines 𝑟 ⃗ = 𝑥_1 𝑖 ̂+𝑦_1 𝑗 ̂+𝑧_1 𝑘 ̂+𝜆(𝑎_1 𝑖 ̂+𝑏_1 𝑗 ̂+𝑐_1 𝑘 ̂ ) and 𝑟 ⃗ = 𝑥_2 𝑖 ̂+𝑦_2 𝑗 ̂+𝑧_2 𝑘 ̂+𝜇(𝑎_2 𝑖 ̂+𝑏_2 𝑗 ̂+𝑐_2 𝑘 ̂ )is given by 𝑐𝑜𝑠𝜃 = |𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 + 〖 𝑐〗_1 𝑐_2 |/(√(〖𝑎1〗^2 + 〖𝑏1〗^2 + 〖𝑐1〗^2 ) √(〖𝑎2〗^2 + 〖𝑏2〗^2. + 〖𝑐2〗^2 ))
This is a formula and it is a correct formula
Thus, Reasoning is true
Is Reason a Correct explanation for Assertion?
Sine we used formula mentioned in Reasoning to find Assertion
Therefore, Reasoning is a correct explanation for Assertion
So,
Assertion is true
Reasoning is true
And, Reasoning is a correct explanation for Assertion
So, the correct answer is (a)

Show More