Question

The acute angle between the lines $x = -2 + 2t, y = 3 - 4t, z = -4 + t$ and $x = -2 - t, y = 3 + 2t, z = -4 + 3t$ is

• A. $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$
• B. $\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• C. $\sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$
• D. $\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$

Hint:

Parametric equations $x = x_0 + at$ reveal direction ratios directly as $(a, b, c)$.

Answer 1

The Correct Option is A

Step 1: Concept
Direction ratios are the coefficients of the parameter $t$.

Step 2: Meaning
$\vec{b}_1 = (2, -4, 1)$ and $\vec{b}_2 = (-1, 2, 3)$.

Step 3: Analysis
$\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|} = \frac{|(2)(-1) + (-4)(2) + (1)(3)|}{\sqrt{4+16+1} \sqrt{1+4+9}}$. $\cos \theta = \frac{|-2 - 8 + 3|}{\sqrt{21} \sqrt{14}} = \frac{7}{\sqrt{294}}$. $\sqrt{294} = \sqrt{49 \times 6} = 7\sqrt{6}$. $\cos \theta = \frac{7}{7\sqrt{6}} = \frac{1}{\sqrt{6}}$.

Step 4: Conclusion
$\theta = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$. Final Answer: (A)