If the shortest distance between the lines (x-4)/(1)=(y+1)/(2)=(z)/(-3) and (x-)/(2)=(y+1)…

Question

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 :

Accepted Answer

$$\begin{aligned}
& \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \
& \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}
\end{aligned}$$

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|\]

$$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$$

$$\begin{aligned}
& \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \
& 3=|\lambda-4| \
& \lambda-4= \pm 3 \
& \lambda=7,1
\end{aligned}$$

Sum of all possible values of $$\lambda$$ is $=8$

Solutions · 1