Find the Cartesian equation of the following | vedclass

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

Question

(A) The given vector equation of the plane is $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$ ............$(1)$For any arbitrary point $P(x, y, z)$ on the

Vedclass · Accepted Answer

$x+y-z=2$

Vedclass · Answer

(A) The given vector equation of the plane is $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$ ............$(1)$For any arbitrary point $P(x, y, z)$ on the plane,the position vector $\vec{r}$ is given by $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.Substituting the value of $\vec{r}$ in equation $(1)$,we get:$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 2$Taking the dot product of the vectors:$x(1) + y(1) + z(-1) = 2$$x + y - z = 2$Thus,the Cartesian equation of the plane is $x + y - z = 2$.

Find the Cartesian equation of the following plane: $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$

Similar Questions

If $l, m, n$ are the direction cosines of a normal to the plane passing through the points $(0, 1, 2)$, $(3, 0, 2)$, and $(4, 5, 0)$, then $|l| + |m| + |n| = $

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

Find the Cartesian equation of the plane passing through the point with position vector $2\hat{i} + 3\hat{j} - 4\hat{k}$ and perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$.

In the following case,determine the direction cosines of the normal to the plane and the distance from the origin: $x+y+z=1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module