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

(D) The distance $d$ 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

Vedclass · Accepted Answer

$1, \frac{7}{3}$

Vedclass · Answer

(D) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For the point $(1, 1, \lambda)$,the distance $d_1$ from the plane $3x + 4y - 12z + 13 = 0$ is:$d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|3 + 4 - 12\lambda + 13|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.For the point $(-3, 0, 1)$,the distance $d_2$ from the plane is:$d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since the points are equidistant,$d_1 = d_2$,so $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$.This implies $|20 - 12\lambda| = 8$,which gives two cases:Case $1$: $20 - 12\lambda = 8 \implies 12\lambda = 12 \implies \lambda = 1$.Case $2$: $20 - 12\lambda = -8 \implies 12\lambda = 28 \implies \lambda = \frac{28}{12} = \frac{7}{3}$.Thus,the values of $\lambda$ are $1$ and $\frac{7}{3}$.

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

Similar Questions

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The distance between the line $\bar{r} = 3\hat{i} - 2\hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and the plane $\bar{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 4$ is

$A$ straight line joining the points $(1, 1, 1)$ and $(0, 0, 0)$ intersects the plane $2x + 2y + z = 10$ at

For non-coplanar vectors $a, b$ and $c$,if the point of intersection of the line $r=a+t(b-c)$ and the plane $r=b+c+x(a-b)+y(c+a)$ is $l a+m b+n c$,then $3 l+4 m+2 n=$

Find the point of intersection of the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{2}$ and the plane $2x + y + z = 6$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module