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

Answer: (C)

$\dfrac{\pi}{3}$

Answer 2

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)