Direction vectors: b1=(1,2,2), b2=(3,2,6). b1⋅b2=3+4+12=19. ∣b1∣=3, ∣b2∣=7. cosθ=2119. So θ=cos−1(2119).
The angle between the pairs of lines
r=(3i^+2j^−4k^)+λ(i^+2j^+2k^)
r=(5i^−2j^)+μ(3i^+2j^+6k^) is :
Held on 23 Aug 2022 · Verified 13 Jul 2026.
θ=cos−1(2119)
θ=cos−1(2119)
θ=cos−1(2119)
θ=cos−1(219)
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
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
Work through every CUET UG Geometry PYQ, year by year.