The vector equation of a plane in parametric | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: Identify the position vector of a point on the plane. Let $\vec{a} = -\hat{i} + 2\hat{j}$.Step $2$: Find the first direction vector $\ve

Vedclass · Accepted Answer

$\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$

Vedclass · Answer

(A) Step $1$: Identify the position vector of a point on the plane. Let $\vec{a} = -\hat{i} + 2\hat{j}$.Step $2$: Find the first direction vector $\vec{b}$ by the vector joining points $A(-1, 2, 0)$ and $B(2, 2, -1)$. $\vec{b} = (2 - (-1))\hat{i} + (2 - 2)\hat{j} + (-1 - 0)\hat{k} = 3\hat{i} - \hat{k}$.Step $3$: Find the second direction vector $\vec{c}$ from the line $\frac{x - 1}{1} = \frac{y + 1/2}{1} = \frac{z + 1}{-1}$. The direction ratios are $(1, 1, -1)$, so $\vec{c} = \hat{i} + \hat{j} - \hat{k}$.Step $4$: The parametric form of the plane is $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.Substituting the vectors: $\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$.This matches option $A$.

The vector equation of a plane in parametric form, passing through the points $A(-1, 2, 0)$ and $B(2, 2, -1)$ and parallel to the line $\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{-1}$ is:

Similar Questions

The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ meets the line $\frac{x - 3}{-1} = \frac{y + 2}{5} = \frac{z - 2}{-4}$ at $A$ and the plane $2 x - y + 2 z + 7 = 0$ at $B$. Then $AB = $

The distance of the point $P(1, 0, -3)$ from the plane $x - y - z = 9$ measured parallel to the line $\frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{-6}$ is (in $\text{ units}$)

If a line $L$ is the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive $X$-axis,then the value of $\sec \alpha$ is

The equation of the line passing through the intersection of the plane $x+2y+3z=4$ and the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}$ and parallel to the vector $(2\hat{i}-3\hat{j}) \times (\hat{i}+2\hat{j}-\hat{k})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module