If the plane 2x - y + 2z + 3 = 0 has distance | vedclass

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

Question

(B) The given plane is $2x - y + 2z + 3 = 0$. First,rewrite the plane $4x - 2y + 4z + \lambda = 0$ as $2x - y + 2z + \frac{\lambda}{2} = 0$. The dista

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) The given plane is $2x - y + 2z + 3 = 0$. First,rewrite the plane $4x - 2y + 4z + \lambda = 0$ as $2x - y + 2z + \frac{\lambda}{2} = 0$. The distance between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. For the first plane,the distance is $\frac{|\frac{\lambda}{2} - 3|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{1}{3}$. $\frac{|\frac{\lambda - 6}{2}|}{3} = \frac{1}{3} \implies |\lambda - 6| = 2$. Thus,$\lambda - 6 = 2$ or $\lambda - 6 = -2$,which gives $\lambda = 8$ or $\lambda = 4$. For the second plane $2x - y + 2z + \mu = 0$,the distance is $\frac{|\mu - 3|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{2}{3}$. $\frac{|\mu - 3|}{3} = \frac{2}{3} \implies |\mu - 3| = 2$. Thus,$\mu - 3 = 2$ or $\mu - 3 = -2$,which gives $\mu = 5$ or $\mu = 1$. The maximum value of $\lambda + \mu$ is $8 + 5 = 13$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

If the plane $2x - y + 2z + 3 = 0$ has distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x - 2y + 4z + \lambda = 0$ and $2x - y + 2z + \mu = 0$ respectively,then the maximum value of $\lambda + \mu$ is equal to:

Similar Questions

In the following case,find the distance of the given point from the corresponding given plane.

Point Plane

$(-6, 0, 0)$ $2x - 3y + 6z - 2 = 0$

If $\bar{r}=(2-\lambda+\mu) \hat{i}+(1-\mu) \hat{j}+(2-3 \lambda+2 \mu) \hat{k}$ is the vector equation of a plane,then the equivalent cartesian equation of the plane is

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x + y + 3z - 2 = 0$ and $x - 2y + 5 = 0$

$A$ tetrahedron has vertices $O(0,0,0)$,$A(1,2,1)$,$B(2,1,3)$,and $C(-1,1,2)$. If $\theta$ is the angle between the faces $OAB$ and $ABC$,then $\cos \theta =$

Find the Cartesian equation of the following plane: $\vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module