Question

The angle between the lines, whose direction cosines are proportional to \((4,√3-1,-√3-1)\) and \((4,-√3-1,√3-1)\) is 

• A. \(\dfrac{\pi}{6}\)
• B. \(\dfrac{\pi}{4}\)
• C. \(\dfrac{\pi}{3}\)
• D. \(\dfrac{\pi}{2}\)
• E. \(\pi\)

Answer 1

The Correct Option is C

Here according to the question, we can use the dot product formula for the angle between two vectors.

Let, the direction cosines of the first line be (l₁, m₁, n₁) and the direction cosines of the second line be (l₂, m₂, n₂).

The dot product of two vectors A and B is given by:\( A · B = |A| * |B| * cos(θ)\)

where \(|A| \) and \(|B|\) are the magnitudes of the vectors A and B, respectively, and \(θ\) is the angle between them.

In this case, the direction cosines of the first line are proportional to \( (4, √3 - 1, -√3 - 1)\) and 

the direction cosines of the second line are proportional to \((4, -√3 - 1, √3 - 1)\).

Now,

 the magnitudes of the two vectors respectively be determine as follows;

\( |A| = √(4^2 + (√3 - 1)^2 + (-√3 - 1)^2) = √(16 + 4 + 4) = √24 = 2√6\)

\( |B| = √(4^2 + (-√3 - 1)^2 + (√3 - 1)^2) = √(16 + 4 + 4) = √24 = 2√6\)

Now, let's calculate the dot product of the two vectors (A · B):

\(A · B = (4 * 4) + ((√3 - 1) * (-√3 - 1)) + ((-√3 - 1) * (√3 - 1)) \)

⇒\(A · B = 16 - (3 - 1) - (3 - 1)\)

⇒ \(A · B = 16 - 2 - 2 A · B = 12\)

Now, we can find the angle θ between the two lines:

\(A · B = |A| * |B| * cos(θ) 12 = (2√6) * (2√6) * cos(θ) 12 = 24 * cos(θ)\)

\(cos(θ) = \dfrac{12 }{ 24 }=cos(θ) = \dfrac{1}{2}\)

\(θ = cos⁻¹(\dfrac{1}{2}) \)

⇒\(θ ≈ 60° =\dfrac{\pi}{3}\)

So, the angle between the lines whose direction cosines are proportional to \((4, √3 - 1, -√3 - 1)\) and \((4, -√3 - 1, √3 - 1)\) is\( \dfrac{\pi}{3}\)  (_Ans)