Question

Lines $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\lambda {\overrightarrow{b}}_2$ lie on same plane, if $(a_1-a_2)(b_1\times b_2)=0$. Let two lines whose equations are $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda }$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ lie in the same plane.The value of $\lambda$ is equal to

• A. $-4$
• B. $4$
• C. $3$
• D. $2$

Answer 1

Answer: (B)

$4$

$L:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda }$

Vector form, $\overrightarrow{r}=3\hat{i}+2\hat{j}+\hat{k}+\mu (2\hat{i}+3\hat{j}+\lambda \hat{k})$

$L:\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$

Vector form ,$\overrightarrow{r}=2\hat{i}+3\hat{j}+2\hat{k}+\mu (3\hat{i}+2\hat{j}+3\hat{k})$

$\therefore$ For the lines to lie on same phase.

$\left[\right(3\hat{i}+2\hat{j}+2\hat{k})-(2\hat{i}+3\hat{j}+2\hat{k}\left)\right].\left\{\right(2\hat{i}+3\hat{j}+\lambda \hat{k})\times (3\hat{i}+2\hat{j}+3\hat{k}\left)\right\}=0(\hat{i}-\hat{j}-\hat{k})\left[\right(2\hat{i}+3\hat{j}+\lambda \hat{k})\times (3\hat{i}+2\hat{j}+3\hat{k}\left)\right]=0\left[\begin{array}{ccc}\hat{i}+(-1)\hat{j}-\hat{k}& 3\hat{i}+2\hat{j}+3\hat{k}& 2\hat{i}+3\hat{j}+\lambda \hat{k}\end{array}\right]=0\left|\begin{array}{ccc}1& -1& -1\\ 3& 2& 3\\ 2& 3& \lambda \end{array}\right|=01(2\lambda -9)+1(3\lambda -6)-1(9-4)=0\Rightarrow \lambda =4$