Let the shortest distance between the lines L: x-5-2=y-0=z+1, 0, and L_1: x+1=y-1=4-z be 2…

Question

Let the shortest distance between the lines $L:\ \dfrac{x-5}{-2}=\dfrac{y-\lambda}{0}=\dfrac{z+\lambda}{1}$, $\lambda\geq0$, and $L_1:\ x+1=y-1=4-z$ be $2\sqrt6$. If $(\alpha,\beta,\gamma)$ lies on $L$, then which of the following is NOT possible?

Accepted Answer

$\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$

$$
\begin{aligned}
& \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \\
& \vec{a}_{1}=5 \hat{i}+\lambda \hat{j}-\lambda \hat{k}_{,} \vec{a}_{2}=-\hat{i}+\hat{j}+4 \hat{k} \\
& \vec{a}_{1}-\vec{a}_{2}=6 \hat{i}+(\lambda-1) \hat{j}-(\lambda+4) \hat{k} \\
& \vec{b}_{1}=-2 \hat{i}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-\hat{k} \\
& \vec{b}_{1} \times \vec{b}_{2}=\left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ -2 & 0 & 1 \ 1 & 1 & -1 \end{array}\end{aligned}$$\right| \
\& =-\hat{i}-\hat{j}-2 \hat{k}

\]

$\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot \vec{b}_{1} \times \vec{b}_{2}=-6+1-\lambda+2 \lambda+8=\lambda+3$

and $\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{6}$

$\because \frac{|\lambda+3|}{\sqrt{6}}=2 \sqrt{6}$

$\therefore \lambda=9, \because \lambda \geq 0$

$\therefore \quad L: \frac{x-5}{-2}=\frac{y-9}{0}=\frac{z+9}{1}=k$

$\therefore \quad \alpha=-2 k+5, \beta=9, \gamma=k-9$

Here $k$ is real then

$\alpha+2 \gamma=-13 \neq 24$.

But all other are in terms of $k$ hence possible.

Solutions · 1