Let A (1, 3, 5) and B (-2, 3, -4) be two poin | vedclass

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

Question

(B) Let the coordinates of point $P$ be $(x, y, z)$.Given $PA^2 - PB^2 = 6c$.Using the distance formula,$PA^2 = (x - 1)^2 + (y - 3)^2 + (z - 5)^2$ and

Vedclass · Accepted Answer

$x + 3z - 1 + c = 0$

Vedclass · Answer

(B) Let the coordinates of point $P$ be $(x, y, z)$.Given $PA^2 - PB^2 = 6c$.Using the distance formula,$PA^2 = (x - 1)^2 + (y - 3)^2 + (z - 5)^2$ and $PB^2 = (x + 2)^2 + (y - 3)^2 + (z + 4)^2$.Substituting these into the equation:$[(x - 1)^2 + (y - 3)^2 + (z - 5)^2] - [(x + 2)^2 + (y - 3)^2 + (z + 4)^2] = 6c$.The $(y - 3)^2$ terms cancel out:$(x - 1)^2 - (x + 2)^2 + (z - 5)^2 - (z + 4)^2 = 6c$.Expanding the squares:$(x^2 - 2x + 1) - (x^2 + 4x + 4) + (z^2 - 10z + 25) - (z^2 + 8z + 16) = 6c$.Simplifying the expression:$-6x - 3 - 18z + 9 = 6c$.$-6x - 18z + 6 = 6c$.Dividing by $-6$:$x + 3z - 1 = -c$,which rearranges to $x + 3z - 1 + c = 0$.

Let $A (1, 3, 5)$ and $B (-2, 3, -4)$ be two points. If a point $P(x, y, z)$ moves such that $PA^2 - PB^2 = 6c$,find the locus of $P$.

Similar Questions

The vector equation of the plane passing through the point $A(1, 2, -1)$ and parallel to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} - \hat{j} + 3 \hat{k}$ is

The equation of the plane passing through $(-1, 1, 2)$,whose normal makes equal acute angles with the coordinate axes is

Find the equation of a plane,given that the foot of the perpendicular drawn to the plane from the origin is $(2, 1, 2)$.

Find the equation of the plane passing through the points $A(2, 2, -1)$,$B(3, 4, 2)$,and $C(7, 0, 6)$.

In three-dimensional space,the equation $3y + 4z = 0$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module