The distance between the planes x + 2y + 3z + | vedclass

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

Question

(A) The given equations of the planes are $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$. First,we rewrite the second equation to have the same coef

Vedclass · Accepted Answer

$\frac{\sqrt{7}}{2\sqrt{2}}$

Vedclass · Answer

(A) The given equations of the planes are $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$. First,we rewrite the second equation to have the same coefficients for $x, y, z$ as the first plane by dividing it by $2$: $x + 2y + 3z + \frac{7}{2} = 0$. The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by the formula $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A = 1, B = 2, C = 3, D_1 = 7$,and $D_2 = \frac{7}{2}$. Substituting these values into the formula: $d = \frac{|7 - \frac{7}{2}|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|\frac{14 - 7}{2}|}{\sqrt{1 + 4 + 9}} = \frac{\frac{7}{2}}{\sqrt{14}}$. Simplifying the expression: $d = \frac{7}{2\sqrt{14}} = \frac{7}{2\sqrt{2} 	imes \sqrt{7}} = \frac{\sqrt{7}}{2\sqrt{2}}$. Thus,the correct option is $A$.

The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is

Similar Questions

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin.
$Z=2$

In three-dimensional space,if $f(x, y, z) = xy + xz$,then the locus of all points which satisfy the equation $f(x, y, z) = 0$ is -

The equation of the plane passing through the origin and perpendicular to the planes $x+2y-z=1$ and $3x-4y+z=5$ is:

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$, then the equation of the plane is

$A$ vector $n$ of magnitude $8$ units is inclined to $x$-axis at $45^\circ$,$y$-axis at $60^\circ$ and an acute angle with $z$-axis. If a plane passes through a point $(\sqrt{2}, -1, 1)$ and is normal to $n$,then its equation in vector form is

Vedclass Test Series

Exam Paper Generator

Online Exam Module