Question

Show that the following planes are at right angles.
(i) $\overrightarrow{r}·\left(2\hat{i}-\hat{j}+\hat{k}\right)=5\mathrm{and}\overrightarrow{r}·\left(-\hat{i}-\hat{j}+\hat{k}\right)=3$

(ii) x − 2y + 4z = 10 and 18x + 17y + 4z = 49

Answer 1

$$\begin{gathered} \left(i\right)\text{We know that the planes}\overrightarrow{r}.\overrightarrow{n_1}=d_{\mathit{1}}\mathit{,}\mathit{}\overrightarrow{r}\mathit{.}\mathit{}\overrightarrow{n_{\mathit{2}}}=d_2\text{are perpendicular to each other only if}\overrightarrow{n_1}\text{.}\overrightarrow{n_2}\text{=0.} \\ \text{Here,}\overrightarrow{n_1}=2\hat{i}-\hat{j}+\hat{k};\overrightarrow{n_2}=-\hat{i}-\hat{j}+\hat{k} \\ \text{Now,} \\ \overrightarrow{n_1}.\overrightarrow{n_2}=\left(2\hat{i}-\hat{j}+\hat{k}\right).\left(-\hat{i}-\hat{j}+\hat{k}\right)=-2+1+1=0 \\ \text{So},\text{the given planes are perpendicular.} \end{gathered}$$

$$\begin{gathered} \left(ii\right)\text{We know that the planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{are}\text{perperndicular to each other only if} \\ {a}_1{a}_2+b_1{b}_2+c_1{c}_2=0. \\ \text{The given planes are}x-2y+4z=10\text{and}18x+17y+4z=49. \\ \Rightarrow a_1=1;b_1=-2;c_1=4;a_2=18;b_2=17;c_2=4 \\ \text{Now,} \\ {a}_1{a}_2+b_1{b}_2+c_1{c}_2=\left(1\right)\left(18\right)+\left(-2\right)\left(17\right)+\left(4\right)\left(4\right)=18-34+16=0 \\ \text{So},\text{the given planes are perpendicular.} \end{gathered}$$