The vector equation of line passing through (−1,3,−2) and perpendicular to the lines 1x+4=2y=3z−3 and −3x+2=2y+5=5z−6 is
Held on 3 Jun 2025 · Verified 13 Jul 2026.
r=(−3i^+4j^+15k^)+λ(−i+3j^−2k^)
r=(−i^+3j^−2k^)+λ(−3i^+4j^+15k^)
r=2i^−7j^+4k^+λ(−i+3j^−2k^)
r=(−i+3j^−2k^)+λ(2i^−7j^+4k^)
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The shortest distance between the following lines: $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + s(2\hat{i} + \hat{j} + \hat{k})$ $\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t(4\hat{i} + 2\hat{j} + 2\hat{k})$, where s and t are scalars, is:
The angle between the line $2x = 3y = z$ and $x$- axis is:
The angle at which the line, $\frac{x-1}{0} = \frac{2-y}{-1} = \frac{2z-3}{-2}$ is inclined with the positive direction of z-axis is
$\sin^{-1}(\cos\frac{3\pi}{5})$ equals
If lines $\frac{x+5}{5\lambda+2} = \frac{4-2y}{10} = \frac{1-3z}{-3}$ and $\frac{x-2}{1} = \frac{1+2y}{4\lambda} = \frac{2+z}{3}$ are perpendicular, than value of '$\lambda$' is
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