The locus of a point P(x, y, z) at which the | vedclass

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(C) Let the point be $P(x, y, z)$. The points are $A(-3, 1, 2)$ and $B(1, -2, 4)$.Since the line segment $AB$ subtends a right angle at $P$,the vector

Vedclass · Accepted Answer

$x^2+y^2+z^2+2x+y-6z+3=0$

Vedclass · Answer

(C) Let the point be $P(x, y, z)$. The points are $A(-3, 1, 2)$ and $B(1, -2, 4)$.Since the line segment $AB$ subtends a right angle at $P$,the vectors $\vec{PA}$ and $\vec{PB}$ are perpendicular.$\vec{PA} = (-3-x, 1-y, 2-z)$$\vec{PB} = (1-x, -2-y, 4-z)$Since $\vec{PA} \cdot \vec{PB} = 0$,we have:$(-3-x)(1-x) + (1-y)(-2-y) + (2-z)(4-z) = 0$$(x+3)(x-1) + (y-1)(y+2) + (z-2)(z-4) = 0$$(x^2 + 2x - 3) + (y^2 + y - 2) + (z^2 - 6z + 8) = 0$$x^2 + y^2 + z^2 + 2x + y - 6z + (-3 - 2 + 8) = 0$$x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$Thus,the locus is $x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$.

The locus of a point $P(x, y, z)$ at which the line segment joining the points $A(-3, 1, 2)$ and $B(1, -2, 4)$ subtends a right angle is:

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