Answer :

The distance between points $P(x_1, y_1, z_1)$ and $P(x_2, y_2, z_2)$ is given

(i) Distance between points $(2,3,5)$ and $(4,3,1)$

$=\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}$

$=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}$

$=\sqrt{4+16}$

$=\sqrt{20}$

$=2 \sqrt{5}$

(ii) Distance between points (-3, 7, 2) and (2, 4, -1) $=\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}$

$=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}$

$=\sqrt{25+9+9}$

$=\sqrt{43}$

(iii) Distance between points (-1, 3, -4) and (1, -3, 4)

$=\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}$

$=\sqrt{(2)^{2}+(-6)^{3}+(8)^{2}}$

$=\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}$

(iv) Distance between points (2, -1,3) and (-2, 1, 3)

$=\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}$

$=\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}$

$=\sqrt{16+4}$

$=\sqrt{20}$

$=2 \sqrt{5}$

Answer :

Let points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) be denoted by P, Q, and R respectively.

Points $P, Q$, and $R$ are collinear if they lie on a line.

$ \begin{aligned} PQ & =\sqrt{(1+2)^{2}+(2-3)^{2}+(3-5)^{2}} \\ & =\sqrt{(3)^{2}+(-1)^{2}+(-2)^{2}} \\ & =\sqrt{9+1+4} \\ & =\sqrt{14} \\ QR & =\sqrt{(7-1)^{2}+(0-2)^{2}+(-1-3)^{2}} \\ & =\sqrt{(6)^{2}+(-2)^{2}+(-4)^{2}} \\ & =\sqrt{36+4+16} \\ & =\sqrt{56} \\ & =2 \sqrt{14} \end{aligned} $

$ \begin{aligned} P R= & \sqrt{(7+2)^{2}+(0-3)^{2}+(-1-5)^{2}} \\ & =\sqrt{(9)^{2}+(-3)^{2}+(-6)^{2}} \\ & =\sqrt{81+9+36} \\ & =\sqrt{126} \\ & =3 \sqrt{14} \end{aligned} $

Here, $PQ+QR=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=PR$

Hence, points $P(- 2,3,5), Q(1,2,3)$, and $R(7,0$, - 1 $)$ are collinear.

Answer :

(i) Let points $(0,7$, -10), (1, 6, - 6 ), and (4, 9, -6) be denoted by A, B, and C respectively.

$ \begin{aligned} AB & =\sqrt{(1-0)^{2}+(6-7)^{2}+(-6+10)^{2}} \\ & =\sqrt{(1)^{2}+(-1)^{2}+(4)^{2}} \\ & =\sqrt{1+1+16} \\ & =\sqrt{18} \\ & =3 \sqrt{2} \\ BC & =\sqrt{(4-1)^{2}+(9-6)^{2}+(-6+6)^{2}} \\ & =\sqrt{(3)^{2}+(3)^{2}} \\ & =\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\ CA & =\sqrt{(0-4)^{2}+(7-9)^{2}+(-10+6)^{2}} \\ & =\sqrt{(-4)^{2}+(-2)^{2}+(-4)^{2}} \\ & =\sqrt{16+4+16}=\sqrt{36}=6 \end{aligned} $

Here, $A B=B C \neq C A$

Thus, the given points are the vertices of an isosceles triangle.

(ii) Let $(0,7,10)$, (-1, 6, 6), and (- $ 4,9,6 )$ be denoted by A, B, and C respectively.

$ \begin{aligned} AB & =\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}} \\ & =\sqrt{(-1)^{2}+(-1)^{2}+(-4)^{2}} \\ & =\sqrt{1+1+16}=\sqrt{18} \\ & =3 \sqrt{2} \\ BC & =\sqrt{(-4+1)^{2}+(9-6)^{2}+(6-6)^{2}} \\ & =\sqrt{(-3)^{2}+(3)^{2}+(0)^{2}} \\ & =\sqrt{9+9}=\sqrt{18} \\ & =3 \sqrt{2} \end{aligned} $

$ \begin{aligned} CA & =\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}} \\ & =\sqrt{(4)^{2}+(-2)^{2}+(4)^{2}} \\ & =\sqrt{16+4+16} \\ & =\sqrt{36} \\ & =6 \end{aligned} $

Now, $AB^{2}+BC^{2}=(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=18+18=36=AC^{2}$

Therefore, by Pythagoras theorem, $A B C$ is a right triangle.

Hence, the given points are the vertices of a right-angled triangle.

(iii) Let $(-1,2,1),(1, - 2,5),(4, - 7,8)$, and $(2, - 3,4)$ be denoted by $A, B, C$, and $D$ respectively.

$ \begin{aligned} AB & =\sqrt{(1+1)^{2}+(-2-2)^{2}+(5-1)^{2}} \\ & =\sqrt{4+16+16} \\ & =\sqrt{36} \\ & =6 \end{aligned} $

$ \begin{aligned} BC & =\sqrt{(4-1)^{2}+(-7+2)^{2}+(8-5)^{2}} \\ & =\sqrt{9+25+9}=\sqrt{43} \\ CD & =\sqrt{(2-4)^{2}+(-3+7)^{2}+(4-8)^{2}} \\ & =\sqrt{4+16+16} \\ & =\sqrt{36} \\ & =6 \end{aligned} $

$ \begin{aligned} DA & =\sqrt{(-1-2)^{2}+(2+3)^{2}+(1-4)^{2}} \\ & =\sqrt{9+25+9}=\sqrt{43} \end{aligned} $

Here, $A B=C D=6, B C=A D=\sqrt{43}$

Hence, the opposite sides of quadrilateral $A B C D$, whose vertices are taken in order, are equal.

Therefore, $A B C D$ is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

Answer :

Let $P(x, y, z)$ be the point that is equidistant from points $A(1,2,3)$ and $B(3,2, - 1)$.

Accordingly, PA $=$ PB

$\Rightarrow PA^{2}=PB^{2}$

$\Rightarrow(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=(x-3)^{2}+(y-2)^{2}+(z+1)^{2}$

$ x^2-2x+1 +y^2 - 4y +4 + z^2-6z+9 = x^2 -6x+9 + y^2 - 4y +4 +z^2 +2z+1 $

$\Rightarrow$ - $2 x$ - $4 y - 6 z+14=- 6 x$ - $4 y+2 z+14$

$\Rightarrow$ - $2 x - 6 z+6 x - 2 z=0$

$\Rightarrow 4 x - 8 z=0$

$\Rightarrow x - 2 z=0$

Thus, the required equation is $x - 2 z=0$.

Answer :

Let the coordinates of $P$ be $(x, y, z)$.

The coordinates of points $A$ and $B$ are $(4,0,0)$ and $(-4,0,0) $ respectively.

It is given that $PA+PB=10$.

$ \begin{aligned} & \Rightarrow \sqrt{(x-4)^{2}+y^{2}+z^{2}}+\sqrt{(x+4)^{2}+y^{2}+z^{2}}=10 \\ & \Rightarrow \sqrt{(x-4)^{2}+y^{2}+z^{2}}=10-\sqrt{(x+4)^{2}+y^{2}+z^{2}} \end{aligned} $

On squaring both sides, we obtain

$ \begin{aligned} & \Rightarrow(x-4)^{2}+y^{2}+z^{2}=100-20 \sqrt{(x+4)^{2}+y^{2}+z^{2}}+(x+4)^{2}+y^{2}+z^{2} \\ & \Rightarrow x^{2}-8 x+16+y^{2}+z^{2}=100-20 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}+x^{2}+8 x+16+y^{2}+z^{2} \\ & \Rightarrow 20 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}=100+16 x \\ & \Rightarrow 5 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}=(25+4 x) \end{aligned} $

On squaring both sides again, we obtain

$25(x^{2}+8 x+16+y^{2}+z^{2})=625+16 x^{2}+200 x$

$\Rightarrow 25 x^{2}+200 x+400+25 y^{2}+25 z^{2}=625+16 x^{2}+200 x$

$\Rightarrow 9 x^{2}+25 y^{2}+25 z^{2} - 225=0$

Thus, the required equation is $9 x^{2}+25 y^{2}+25 z^{2} - 225=0$.