Let L_1:r=(i-j+2k)+ (i-j+2k), R, L_2:r=(j-k)+ (3i+j+pk), RL_3:r= (li+mj+nk), R be three li…

Question

Let ${L}_{1}:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda (\hat{i}-\hat{j}+2\hat{k}),\lambda \in R$,

${L}_{2}:\vec{r}=(\hat{j}-\hat{k})+\mu (3\hat{i}+\hat{j}+p\hat{k}),\mu \in R{L}_{3}:\vec{r}=\delta (l\hat{i}+m\hat{j}+n\hat{k}),\delta \in R$

be three lines such that ${L}_{1}$ is perpendicular to ${L}_{2}$ and ${L}_{3}$ is perpendicular to both ${L}_{1}$ and ${L}_{2}$. Then, the point which lies on ${L}_{3}$ is

Accepted Answer

$$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$$

$$\begin{aligned}
& 3-1+2 \mathrm{P}=0 \
& \mathrm{P}=-1 \
& \left|\[\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 1 & -1 & 2 \ 3 & 1 & -1 \end{array}\end{aligned}$$\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \
\& \therefore(-\delta, 7 \delta, 4 \delta) \text { will lie on } \mathrm{L}\_3
\]

For $$\delta=1$$ the point will be $(-1,7,4)$

Solutions · 1