Q.8 Find the angle between the lines r = (i + 2j + 3k) + λ(2i - j + k) r = (i + 2j + 3k) + μ(i + 2j + k)

Answer: θ = cos⁻¹(1/6) Direction vectors: b₁ = 2i - j + k , b₂ = i + 2j + k . b₁·b₂ = 2 - 2 + 1 = 1 , |b₁| = √(6) , |b₂| = √(6) . cosθ = |b₁·b₂|/|b₁||b₂| = 1/6 , so θ = cos⁻¹(1/6) .

Find the angle between the lines\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - \hat{j} + \hat{k})\)\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + \hat{k})\)

Q.8 Find the angle between the lines
\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - \hat{j} + \hat{k})\)
\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + \hat{k})\)

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: \(\theta = \cos^{-1}\left(\frac{1}{6}\right)\)

Direction vectors: \(\bar{b}_1 = 2\hat{i} - \hat{j} + \hat{k}\), \(\bar{b}_2 = \hat{i} + 2\hat{j} + \hat{k}\).

\(\bar{b}_1\cdot\bar{b}_2 = 2 - 2 + 1 = 1\), \(|\bar{b}_1| = \sqrt{6}\), \(|\bar{b}_2| = \sqrt{6}\).

\(\cos\theta = \frac{|\bar{b}_1\cdot\bar{b}_2|}{|\bar{b}_1||\bar{b}_2|} = \frac{1}{6}\), so \(\theta = \cos^{-1}\left(\frac{1}{6}\right)\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Find the angle between the lines\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - \hat{j} + \hat{k})\)\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + \hat{k})\)

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