Q77 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
The vector equation of plane in parametric form, passing through the points (-1, 2, 0), (2, 2, -1) and parallel to the line $\dfrac{x-1}{1} = \dfrac{2y+1}{2} = \dfrac{z+1}{-1}$ is
- Correct answerA$\bar{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$
- B$\bar{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + 2\hat{j} - \hat{k})$
- C$\bar{r} = (\hat{i} - 2\hat{j}) + \lambda(3\hat{i} + \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$
- D$\bar{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$
Solution
The position vector of the point $A(-1, 2, 0)$ is $\vec{a} = -\hat{i} + 2\hat{j}$.
The plane passes through $A(-1, 2, 0)$ and $B(2, 2, -1)$. A vector parallel to the plane is $\vec{AB} = (2 - (-1))\hat{i} + (2 - 2)\hat{j} + (-1 - 0)\hat{k} = 3\hat{i} - \hat{k}$.
The given line is $\dfrac{x-1}{1} = \dfrac{2y+1}{2} = \dfrac{z+1}{-1}$. Rewriting it in standard form, we get $\dfrac{x-1}{1} = \dfrac{y+1/2}{1} = \dfrac{z+1}{-1}$.
The direction vector of the line is $\vec{v} = \hat{i} + \hat{j} - \hat{k}$. Since the plane is parallel to this line, $\vec{v}$ is also parallel to the plane.
The parametric vector equation of a plane passing through a point with position vector $\vec{a}$ and parallel to two non-collinear vectors $\vec{b}$ and $\vec{c}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.
Substituting $\vec{a} = -\hat{i} + 2\hat{j}$, $\vec{b} = \vec{AB} = 3\hat{i} - \hat{k}$, and $\vec{c} = \vec{v} = \hat{i} + \hat{j} - \hat{k}$, we get:
$\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$
Answer: $\bar{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$
The plane passes through $A(-1, 2, 0)$ and $B(2, 2, -1)$. A vector parallel to the plane is $\vec{AB} = (2 - (-1))\hat{i} + (2 - 2)\hat{j} + (-1 - 0)\hat{k} = 3\hat{i} - \hat{k}$.
The given line is $\dfrac{x-1}{1} = \dfrac{2y+1}{2} = \dfrac{z+1}{-1}$. Rewriting it in standard form, we get $\dfrac{x-1}{1} = \dfrac{y+1/2}{1} = \dfrac{z+1}{-1}$.
The direction vector of the line is $\vec{v} = \hat{i} + \hat{j} - \hat{k}$. Since the plane is parallel to this line, $\vec{v}$ is also parallel to the plane.
The parametric vector equation of a plane passing through a point with position vector $\vec{a}$ and parallel to two non-collinear vectors $\vec{b}$ and $\vec{c}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.
Substituting $\vec{a} = -\hat{i} + 2\hat{j}$, $\vec{b} = \vec{AB} = 3\hat{i} - \hat{k}$, and $\vec{c} = \vec{v} = \hat{i} + \hat{j} - \hat{k}$, we get:
$\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$
Answer: $\bar{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$
Hence, the correct answer is option: 'A'.