If the points (1, 1, k) and (-3, 0, 1) are eq | 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 plane $3x + 4y - 12z + 13 = 0$,the distance of point $P(1, 1, k)$ is $d_1 = \frac{|3(1) + 4(1) - 12(k) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12k|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12k|}{13}$.The distance of point $Q(-3, 0, 1)$ 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 $\frac{|20 - 12k|}{13} = \frac{8}{13}$.This implies $|20 - 12k| = 8$,which gives two cases:Case $1$: $20 - 12k = 8 \Rightarrow 12k = 12 \Rightarrow k = 1$.Case $2$: $20 - 12k = -8 \Rightarrow 12k = 28 \Rightarrow k = \frac{28}{12} = \frac{7}{3}$.Since $k=1$ is provided in the options,the correct answer is $1$.

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

Similar Questions

The equation of a plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is

The plane which bisects the line segment joining the points $(-3, -3, 4)$ and $(3, 7, 6)$ at right angles,passes through which one of the following points?

$A$ 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 $\cos \theta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module