Practicing Success
The vector equation of the plane $\vec{r}=(\hat{i}-\hat{j}) + λ(\hat{i}+\hat{j}+\hat{k}) + \mu (\hat{i}-2\hat{j}+3\hat{k})$ in a scalar product form, is |
$\vec{r}. (5\hat{i}-2\hat{j}-3\hat{k})=7$ $\vec{r}. (5\hat{i}+2\hat{j}+3\hat{k})=7$ $\vec{r}. (5\hat{i}-2\hat{j}-3\hat{k})=-7$ none of these |
$\vec{r}. (5\hat{i}-2\hat{j}-3\hat{k})=7$ |
Clearly, given plane passes through the point $a = \hat{i} - \hat{j}$ and is parallel to the vectors $\vec{b}= \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}= \hat{i}-2\hat{j}+3\hat{k}$. So, it is perpendicular to the vector $\vec{n}$ given by $\vec{n} = \vec{b} × \vec{c}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 1 & 1\\1 & -2 & 3\end{vmatrix}= 5\hat{i}-2\hat{j}-3\hat{k}$ The vector equation of the plane in scalar product form is $\vec{r}.\vec{n}=\vec{a}.\vec{n}$ $⇒\vec{r}.(5\hat{i}-2\hat{j}-3\hat{k})=(\hat{i}-\hat{j}).(5\hat{i}-2\hat{j}-3\hat{k})$ $⇒\vec{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7 $ |