Find the Cartesian equation of the plane pass | vedclass

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

Question

(A) The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:$(\vec{r} - \vec{

Vedclass · Accepted Answer

$x + 5y - 6z + 18 = 0$

Vedclass · Answer

(A) The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ or $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \dots (i)$Given that the plane passes through the point $P(3, -3, 1)$,we have $\vec{a} = 3\hat{i} - 3\hat{j} + \hat{k}$.The plane is normal to the line joining $A(3, 4, -1)$ and $B(2, -1, 5)$. Thus,the normal vector $\vec{n}$ is the vector $\vec{AB}$:$\vec{n} = \vec{AB} = (2 - 3)\hat{i} + (-1 - 4)\hat{j} + (5 - (-1))\hat{k} = -\hat{i} - 5\hat{j} + 6\hat{k}$.Substituting $\vec{a}$ and $\vec{n}$ into equation $(i)$:$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = (3\hat{i} - 3\hat{j} + \hat{k}) \cdot (-\hat{i} - 5\hat{j} + 6\hat{k})$$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = (3)(-1) + (-3)(-5) + (1)(6)$$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = -3 + 15 + 6 = 18$.To find the Cartesian equation,substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = 18$$-x - 5y + 6z = 18$$x + 5y - 6z + 18 = 0$.

Find the Cartesian equation of the plane passing through the point $(3, -3, 1)$ and normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$.

Similar Questions

Find the distance of a point $(2, 5, -3)$ from the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$. (in $/7$)

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to

$A$ vector $\vec{n}$ is inclined to $X$-axis at $45^{\circ}$,$Y$-axis at $60^{\circ}$ and at an acute angle to $Z$-axis. If $\vec{n}$ is normal to a plane passing through the point $(-\sqrt{2}, 1, 1)$,then the equation of the plane is

$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

The equation of the plane,passing through the point $(1, 1, 1)$ and perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module