Direction of L2: (3,6,−k) since k1−z=−kz−1.
Vector joining points: (1,1,1).
Coplanarity: scalar triple product = 0.
12315614−k=−3k−15=0, so k=−5.
If two lines 2x−3=5y−4=4z and 3x−4=6y−5=k1−z, are coplanar, then k is equal to
Held on 6 Aug 2022 · Verified 13 Jul 2026.
5
−5
9
−3
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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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