Question

Find the distance of the point (−1, −5, −10) from the point of intersection of the line $\overrightarrow{r}=\left(2\hat{i}-\hat{j}+2\hat{k}\right)+\lambda \left(3\hat{i}+4\hat{j}+2\hat{k}\right)$ and the plane $\overrightarrow{r}.\left(\hat{i}-\hat{j}+\hat{k}\right)=5.$

Answer 1

$$\begin{gathered} \text{The given equation of the line is} \\ \overrightarrow{r}=\left(2\hat{i}-\hat{j}+2\hat{k}\right)+\lambda \left(3\hat{i}+4\hat{j}+2\hat{k}\right) \\ \Rightarrow \overrightarrow{r}=\left(2+3\lambda \right)\hat{i}+\left(-1+4\lambda \right)\hat{j}+\left(2+2\lambda \right)\hat{k} \\ \text{T}\mathrm{he\; coordinates\; of\; any\; point\; on\; this\; line\; are\; of\; the\; form}\left(2+3\lambda \right)\hat{i}+\left(-1+4\lambda \right)\hat{j}+\left(2+2\lambda \right)\hat{k}\text{. or}\left(2+3\lambda ,-1+4\lambda ,2+2\lambda \right) \\ \text{Since this point lies on the plane}\overrightarrow{r}\text{.}\left(\hat{i}-\hat{j}+\hat{k}\right)\text{= 5,} \\ \left[\left(2+3\lambda \right)\hat{i}+\left(-1+4\lambda \right)\hat{j}+\left(2+2\lambda \right)\hat{k}\right].\left(\hat{i}-\hat{j}+\hat{k}\right)=5 \\ \Rightarrow 2+3\lambda +1-4\lambda +2+2\lambda -5=0 \\ \Rightarrow \lambda =0 \\ \mathrm{So,\; the\; coordinates\; of\; the\; point}\text{are} \\ \left(2+3\lambda ,-1+4\lambda ,2+2\lambda \right) \\ =\left(2+0,-1+0,2+0\right) \\ =\left(2,-1,2\right) \\ \text{Distance between (2, -1, 2) and (-1, -5, -10)} \\ =\sqrt{{\left(-1-2\right)}^2+{\left(-5+1\right)}^2+{\left(-10-2\right)}^2} \\ =\sqrt{9+16+144} \\ =13\text{units} \end{gathered}$$