Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The required plane contains the line $\bar{r}=(2\hat{i}-\hat{k})+\lambda (\hat{i}+\hat{j}+2\hat{k})$, so it passes through the point with position vector $\vec{a}=2\hat{i}-\hat{k}$ and is parallel to the vector $\vec{b}=\hat{i}+\hat{j}+2\hat{k}$.

The plane is perpendicular to the given plane $\bar{r}\cdot (3\hat{i}-\hat{j}+\hat{k})=20$, so its normal vector $\vec{n}$ is perpendicular to the normal vector of the given plane, ${\vec{n}}_1=3\hat{i}-\hat{j}+\hat{k}$.

The normal vector $\vec{n}$ of the required plane is given by the cross product of $\vec{b}$ and ${\vec{n}}_1$:

$\vec{n}=\vec{b}\times {\vec{n}}_1=(\hat{i}+\hat{j}+2\hat{k})\times (3\hat{i}-\hat{j}+\hat{k})$

$\vec{n}=\hat{i}(1-(-2))-\hat{j}(1-6)+\hat{k}(-1-3)$

$\vec{n}=3\hat{i}+5\hat{j}-4\hat{k}$

The equation of the required plane is $(\bar{r}-\vec{a})\cdot \vec{n}=0$, which can be written as $\bar{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$.

Substituting the values of $\vec{a}$ and $\vec{n}$:

$\vec{a}\cdot \vec{n}=(2\hat{i}-\hat{k})\cdot (3\hat{i}+5\hat{j}-4\hat{k})$

$\vec{a}\cdot \vec{n}=(2)(3)+(0)(5)+(-1)(-4)=6+0+4=10$

Therefore, the equation of the plane is $\bar{r}\cdot (3\hat{i}+5\hat{j}-4\hat{k})=10$.

Answer: $\bar{r}\cdot (3\hat{i}+5\hat{j}-4\hat{k})=10$