If \( \theta \) is the angle between the lines \( \mathrm{L}_1 \) and \( \mathrm{L}_2 \) , whose direction cosines are \( l_1,m_1,n_1 \) and \( l_2,m_2,n_2 \) respectively, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \cos\theta \) | (I) | \( l_1l_2+m_1m_2+n_1n_2 \) |
| (B) | \( \sin\theta \) | (II) | \( l_1l_2+m_1m_2+n_1n_2=0 \) |
| (C) | \( \mathrm{L}_1 \) and \( \mathrm{L}_2 \) are perpendicular | (III) | \( \displaystyle\frac{l_1}{l_2}=\displaystyle\frac{m_1}{m_2}=\displaystyle\frac{n_1}{n_2} \) |
| (D) | \( \mathrm{L}_1 \) and \( \mathrm{L}_2 \) are parallel | (IV) | \( \sqrt{\sum(m_1n_2-m_2n_1)^2} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (I), (B) - (IV), (C) - (II), (D) - (III)
(A) \( \cos\theta \)
This matches (I) \( l_1l_2+m_1m_2+n_1n_2 \). For unit direction vectors, \( \cos\theta \) is their dot product.
(B) \( \sin\theta \)
This matches (IV) \( \sqrt{\sum(m_1n_2-m_2n_1)^2} \). Since \( l_1^2+m_1^2+n_1^2=1 \) and \( l_2^2+m_2^2+n_2^2=1 \),
\( \displaystyle \sin^2\theta=1-\cos^2\theta \)
\( \displaystyle =(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)-(l_1l_2+m_1m_2+n_1n_2)^2 \)
By Lagrange’s identity, the right side equals \( \sum(m_1n_2-m_2n_1)^2 \). (It is also \( |\hat{b}_1\times\hat{b}_2|^2 \), the squared length of the cross product of the unit vectors.)
(C) \( \mathrm{L}_1 \) and \( \mathrm{L}_2 \) are perpendicular
This matches (II) \( l_1l_2+m_1m_2+n_1n_2=0 \). Perpendicular means \( \theta=90^\circ \), so \( \cos\theta=0 \).
(D) \( \mathrm{L}_1 \) and \( \mathrm{L}_2 \) are parallel
This matches (III) \( \frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2} \). Parallel lines have proportional direction cosines (\( \sin\theta=0 \)).
So the matching is (A)-(I), (B)-(IV), (C)-(II), (D)-(III).