Question
Find the vector equation of the plane passing through points $3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$ and $7\hat{\text{i}}+6\hat{\text{k}}.$
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Answer
Let A(3, 4, 2), B(2, -2, -1) and C(7, 0, 6) be the points respresented by the given position vectors. The required plane passes through the point A(3, 4, 2) whose
position vector is $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by $\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$ Clearly, $\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$ $=(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})-(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$ $=-\hat{\text{i}}-6\hat{\text{j}}-3\hat{\text{k}}$ $\overrightarrow{\text{AC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$ $=(7\hat{\text{i}}+0\hat{\text{j}}+6\hat{\text{k}})-(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$ $=4\hat{\text{i}}-4\hat{\text{j}}+4\hat{\text{k}}$ $\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$ $=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&-6&-3\\4&-4&4\end{vmatrix}$ $=-36\hat{\text{i}}-8\hat{\text{j}}+28\hat{\text{k}}$ The vector equation of the required plane is $\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$ $\Rightarrow\vec{\text{r}}\cdot(36\hat{\text{i}}-8\hat{\text{j}}+28\hat{\text{k}})=(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})\cdot(36\hat{\text{i}}-8\hat{\text{j}}+28\hat{\text{k}})$ $\Rightarrow\vec{\text{r}}\cdot\Big[-4(9\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}})\Big]=-108-32+56$ $\Rightarrow\vec{\text{r}}\cdot\Big[-4(9\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}})\Big]=-84$ $\Rightarrow\vec{\text{r}}\cdot(9\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}})=21$Need a full question paper?
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