If the points (1, 1, lambda) and (-3, 0, 1) a | vedclass

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

Question

(B) The distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 +

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Given the points $A(1, 1, \lambda)$ and $B(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$.Distance from $A$: $d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\lambda|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.Distance from $B$: $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since $d_1 = d_2$,we have $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$,which implies $|20 - 12\lambda| = 8$.Case $1$: $20 - 12\lambda = 8 \Rightarrow 12\lambda = 12 \Rightarrow \lambda = 1$.Case $2$: $20 - 12\lambda = -8 \Rightarrow 12\lambda = 28 \Rightarrow \lambda = \frac{28}{12} = \frac{7}{3}$.Since $\lambda$ must be an integer,the value is $\lambda = 1$.

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$,then the integer value of $\lambda$ is:

Similar Questions

Find the length of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ and the coordinates of the foot of the perpendicular.

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $............$.

The vector equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$,which is perpendicular to the plane $x-y+z=0$,is

The line $\ell$ passes through the point $P(2, 1, 1)$ and is parallel to the plane $x + y + 2z = 18$. If line $\ell$ intersects the line $L_2: \frac{x + 2}{3} = \frac{y + 1}{-1} = \frac{z - 2}{1}$, then the equation of the line $\ell$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module