MCQ
If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :
- A$5$
- ✓$8$
- C$7$
- D$10$
$ \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$
the shortest distance between the lines
$ =\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|$
$=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right|$
$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$
$\frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right|$
$ 3=|\lambda-4|$
$ \lambda-4= \pm 3 $
$\lambda=7,1$
Sum of all possible values of $\lambda$ is $=8$
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$\frac{{dy}}{{dx}} + \frac{1}{x}\sin 2y = {x^3}\,{\cos ^2}\,y$ represented by family of curves which is is givey by
| $X = x_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X = X_i)$ | $0$ | $2p$ | $2p$ | $3p$ | $p^2$ | $2p^2$ | $7p^2$ | $2p$ |