Solution

Given, coordinates are

(i) $(1,-1,3)$

(ii) $B(-1,2,4)$

(iii) $C(-2,-4,-7)$

(iv) $D(-4,2,-5)$

X-increment $=Y$-increment $=Z$-increment $=1$

Solution

(i) Point $(1,2,3)$ lies in first quadrant.

(ii) $(4,-2,3)$ in fourth octant.

(iii) $(4,-2,-5)$ in eight octant.

(iv) $(4,2,-5)$ in fifth octant.

(v) $(-4,2,5)$ in second octant.

(vi) $(-3,-1,6)$ in third octant.

(vii) $(2,-4,-7)$ in eight octant.

(viii) $(-4,2,-5)$ in sixth octant.

Solution

The coordinates of $A, B$ and $C$ are the following

(i) $A(3,0,0), B(0,4,0), C(0,0,2)$

(ii) $A(-5,0,0), B(0,3,0), C(0,0,7)$

(iii) $A(4,0,0), B(0,-3,0), C(0,0,-5)$

Solution

We know that, on $X Y$-plane $z=0$, on YZ-plane, $x=0$ and on ZX-plane, $y=0$. Thus, the coordinate of $A, B$ and $C$ are following

(i) $A(3,4,0), B(0,4,5), C(3,0,5)$

(ii) $A(-5,3,0), B(0,3,7), C(-5,0,7)$

(iii) $A(4,-3,0), B(0,-3,-5), C(4,0,-5)$

Solution

Given points, $A(2,0,0)$ and $B(-3,0,0)$

$ A B=\sqrt{(2+3)^{2}+0^{2}+0^{2}}=5 $

Solution

Distance from origin to the point $(6,6,7)$

$ \begin{aligned} & =\sqrt{(0-6)^{2}+(0-6)^{2}+(0-7)^{2}} \quad[\because d=\sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}+(z_1-z_2)^{2}}] \\ & =\sqrt{36+36+49} \\ & =\sqrt{121}=11 \end{aligned} $

Solution

Given that, $x^{2}+y^{2}=1$

$\therefore \quad$ Distance of the point $(x, y, \sqrt{1-x^{2}-y^{2}})$ from origin is given as

$ \begin{aligned} d & =|\sqrt{x^{2}+y^{2}+(\sqrt{1-x^{2}-y^{2}})^{2}}| \\ & =|\sqrt{x^{2}+y^{2}+1-x^{2}-y^{2}}|=1 \end{aligned} $

Hence proved.

Solution

Given points, $A(1,-1,3), B(2,-4,5)$ and $C(5,-13,11)$.

$ \begin{aligned} A B & =\sqrt{(1-2)^{2}+(-1+4)^{2}+(3-5)^{2}} \\ & =\sqrt{1+9+4}=\sqrt{14} \\ B C & =\sqrt{(2-5)^{2}+(-4+13)^{2}+(5-11)^{2}} \\ & =\sqrt{9+81+36}=\sqrt{126} \\ A C & =\sqrt{(1-5)^{2}+(-1+13)^{2}+(3-11)^{2}} \\ & =\sqrt{16+144+64}=\sqrt{224} \\ \because \quad A B+B C & =A C \\ \Rightarrow \quad \sqrt{14}+\sqrt{126} & =\sqrt{224} \\ \Rightarrow \quad \sqrt{14}+3 \sqrt{14} & =4 \sqrt{14} \end{aligned} $

So, the points $A, B$ and $C$ are collinear.

Solution

Let the coordinates of the fourth vertices $D(x, y, z)$.

Mid-points of diagonal $A C$,

and

$ \begin{aligned} & x=\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}, z=\frac{z_1+z_2}{2} \\ & x=\frac{6-2}{2}=2, y=\frac{-2+2}{2}=0, z=\frac{4+4}{2}=4 \end{aligned} $

Since, the mid-point of $A C$ are $(2,0,4)$.

Now, $\quad$ mid-point of $B D, 2=\frac{x+2}{2} \Rightarrow x=2$

$\Rightarrow \quad 0=\frac{y+4}{2} \Rightarrow y=-4$

$\Rightarrow \quad 4=\frac{z-8}{2} \Rightarrow z=16$

So, the coordinates of fourth vertex $D$ is $(2,-4,16)$.

Solution

Given that, the vertices of the $\triangle A B C$ are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$. Now,

$ \begin{aligned} A B & =\sqrt{(0-2)^{2}+(4-3)^{3}+(1+1)^{2}} \\ & =\sqrt{4+1+4}=3 \\ B C & =\sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2}} \\ & =\sqrt{4+4+1}=3 \\ A C & =\sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2}} \\ & =\sqrt{16+1+1}=\sqrt{18} \end{aligned} $

$ \begin{matrix} \because & A C^{2}=A B^{2}+B C^{2} \\ \Rightarrow & 18=9+9 \end{matrix} $

Hence, vertices $\triangle A B C$ is a right angled triangle.

Solution

Let third vertex of $\triangle A B C$ i.e., is $A(x, y, z)$.

$(2,4,6)$

Given that, the coordinate of centroid $G$ are $(0,0,0)$.

$\because \quad \begin{aligned} 0 & =\frac{x+2+0}{3} \Rightarrow x=-2 \\ 0 & =\frac{y+4-2}{3} \Rightarrow y=-2 \\ 0 & =\frac{z+6-5}{2} \Rightarrow z=-1\end{aligned}$

Hence, the third vertex of triangle is $(-2,-2,-1)$.

Solution

Given that, mid-points of sides are $D(1,2,-3), E(3,0,1)$ and $F(-1,1,-4)$.

Let the vertices of the $\triangle A B C$ are $A(x_1, y_1, z_1), B(x_2, y_2, z_3)$ and $C(x_3, y_3, z_3)$.

Then, mid-point of $B C$ are $(1,2,-3)$.

$ \begin{aligned} 1 & =\frac{x_2+x_3}{2} \Rightarrow x_2+x_3=2 \\ 2 & =\frac{y_2+y_3}{2} \Rightarrow y_2+y_3=4 \\ -3 & =\frac{z_2+z_3}{2} \Rightarrow z_2+z_3=-6 \end{aligned} $

Similarly for the sides $A B$ and $A C$,

$ \begin{matrix} \Rightarrow & -1=\frac{x_1+x_2}{2} \Rightarrow x_1+x_2=-2 \\ \Rightarrow & 1=\frac{y_1+y_2}{2} \Rightarrow y_1+y_2=2 \\ \Rightarrow & -4=\frac{z_1+z_2}{2} \Rightarrow z_1+z_2=-8 \\ \Rightarrow & 3=\frac{x_1+x_3}{2} \Rightarrow x_1+x_3=6 \\ \Rightarrow & 0=\frac{y_1+y_3}{2} \Rightarrow y_1+y_3=0 \\ \Rightarrow & 1=\frac{z_1+z_3}{2} \Rightarrow z_1+z_3=2 . \end{matrix} $

On adding Eqs. (i) and (iv), we get

On adding Eqs. (ii) and (v), we get

$ x_1+2 x_2+x_3=0 $

On adding Eqs. (iii) and (vi), we get

$ y_1+2 y_2+y_3=6 $

From Eqs. (vii) and (x),

$ z_1+2 z_2+z_3=-14 $

If $x_2=-3$, then $x_3=5$

$ 2 x_2=-6 \Rightarrow x_2=-3 $

If $x_3=5$, then $x_1=1, x_2=-3, x_3=5$

From Eqs. (xi) and (viii),

$ 2 y_2=6 \Rightarrow y_2=3 $

If $y_2=3$, then $y_1=-1 \quad$ If $y_1=-1$, then $y_3=1, \quad y_2=3, y_3=1$

From Eqs. (xii) and (ix),

$ \begin{gathered} 2 z_2=-16 \Rightarrow z_2=-8 \\ z_2=-8, \text { then } z_1=0 \\ z_1=0, \text { then } z_3=2 \\ z_1=0, z_2=-8, z_3=2 \end{gathered} $

So, the points are $A(1-1,0), B(-3,3,-8)$ and $C(5,1,2)$.

$\therefore$ Centroid of the triangle $=G \frac{1-3+5}{3}, \frac{-1+3+1}{3}, \frac{0-8+2}{3}$ i.e., $G(1,1,-2)$

Solution

Let vertices of the $\triangle A B C$ are $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$, then the mid-point of $B C(5,7,11)$.

$ \begin{aligned} & 5=\frac{x_2+x_3}{2} \Rightarrow x_2+x_3=10 \\ & 7=\frac{y_2+y_3}{2} \Rightarrow y_2+y_3=14 \\ & 11=\frac{z_2+z_3}{2} \Rightarrow z_2+z_3=22 \end{aligned} $

Similarly for the sides $A B$ and $A C$,

$ \begin{gathered} 2=\frac{x_1+x_2}{2} \Rightarrow x_1+x_2=4 \\ 3=\frac{y_1+y_2}{2} \Rightarrow y_1+y_2=6 \\ -1=\frac{z_1+z_2}{2} \Rightarrow z_1+z_2=-2 \\ 0=\frac{x_1+x_3}{2} \Rightarrow x_1+x_3=0 \\ 8=\frac{y_1+y_3}{2} \Rightarrow y_1+y_3=16 \\ 5=\frac{z_1+z_3}{2} \Rightarrow z_1+z_3=10 \end{gathered} $

From Eqs. (i) and (iv),

$ x_1+2 x_2+x_3=14 $

From Eqs. (ii) and (v),

From Eqs. (iii) and (vi),

$ y_1+2 y_2+y_3=20 $

From Eqs. (vii) and (x),

From Eqs. (viii) and (xi),

$ z_1+2 z_2+z_3=20 $

$ \begin{aligned} 2 x_2 & =14 \Rightarrow x_2=7 \\ x_2 & =7, \text { then } x_3=10-7=3 \\ x_3 & =3, \text { then } x_1=-3 \\ x_1 & =-3, x_2=7, x_3=3 \end{aligned} $

$ \begin{aligned} 2 y_2 & =4 \Rightarrow y_2=2 \\ y_2 & =2, \text { then } y_1=4 \\ y_1 & =4, \text { then } y_3=12 \\ y_1 & =4, y_2=2, y_3=12 \end{aligned} $

From Eqs. (ix) and (xii),

$ \begin{aligned} 2 z_2 & =10 \Rightarrow z_2=5 \\ z_2 & =5, \text { then } z_1=-7 \\ z_1 & =-7, \text { then } z_3=17 \\ z_1 & =-7, z_2=5, z_3=17 \end{aligned} $

So, the vertices are $A(-3,4,-7), B(7,2,5)$ and $C(3,12,17)$.

Solution

Let the fourth vertex of the parallelogram $A B C D$ is $D(x, y, z)$. Then, the mid-point of $A C$ are

$ P \frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2} \text { i.e., } P \frac{3}{2}, \frac{5}{2}, \frac{5}{2} \text {. } $

Now, mid-point of $B D$,

$ \begin{aligned} & \frac{3}{2}=\frac{-1+x}{2} \Rightarrow x=4 \\ & \frac{5}{2}=\frac{-2+y}{2} \Rightarrow y=7 \\ & \frac{5}{2}=\frac{-1+z}{2} \Rightarrow z=6 \end{aligned} $

So, the coordinates of fourth vertex is $(4,7,6)$.

Solution

Let the $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ trisect line segment $A B$.

Since, the point $P$ divided line $A B$ in $1: 2$ internally, then

$ \begin{aligned} & x_1=\frac{2 \times 2+1 \times 5}{1+2}=\frac{9}{3}=3 \\ & y_1=\frac{2 \times 1+1 \times(-8)}{3}=\frac{-6}{3}=-2 \\ & z_1=\frac{2 \times(-3)+1 \times 3}{3}=\frac{-6+3}{3}=\frac{-3}{3}=-1 \end{aligned} $

Since, the point $Q$ divide the line segment $A B$ in $2: 1$, then

$ \begin{aligned} & x_2=\frac{1 \times 2+2 \times 5}{3}=4, \\ & y_2=\frac{1 \times 1+(-8 \times 2)}{3}=-5 \\ & z_2=\frac{1 \times(-3)+2 \times 3}{3}=-1 \end{aligned} $

So, the coordinates of $P$ are $(3,-2,-1)$ and the coordinates of $Q$ are $(4,-5,1)$.

Solution

Given that origin is the centroid of the $\triangle A B C$ i.e., $G(0,0,0)$.

$\begin{array}{ll} \because \quad 0=\frac{a-2+4}{3} \Rightarrow a=-2 \\ 0=\frac{1+b+7}{3} \Rightarrow b=-8 \\ \quad 0=\frac{3-5+c}{3} \Rightarrow c=+2 \\ \therefore \quad a=-2, b=-8 \text { and } c=2 \end{array}$

Solution

Let the coordinates of $D$ are $(x, y, z)$.

$ \begin{aligned} A B & =\sqrt{9+16+144}=\sqrt{169}=13 \\ A C & =\sqrt{0+25+144}=\sqrt{169}=\sqrt{13} \\ \frac{A B}{A C} & =\frac{13}{13} \Rightarrow A B=A C \\ \frac{B D}{D C} & =\frac{1}{1} \Rightarrow B D=D C \end{aligned} $

$ \Rightarrow \quad \frac{A B}{A C}=\frac{13}{13} \Rightarrow A B=A C $

Since, $D$ is divide the line $B C$ in two equal parts. So, $D$ is the mid-point of $B C$.

$ \begin{matrix} \therefore & x=\frac{5+2}{2}=7 / 2 \\ \Rightarrow & y=\frac{6+7}{2}=13 / 2 \\ \Rightarrow & z=\frac{9+9}{2}=9 \end{matrix} $

So, the coordinates of $D$ are $\frac{7}{2}, \frac{13}{2}, 9$.

Solution

Given points are $A(2,3,4), B(-1,2,-3)$ and $C(-4,1,-10)$.

$ \begin{aligned} & \therefore \quad A B=\sqrt{(2+1)^{2}+(3-2)^{2}+(4+3)^{2}} \\ & =\sqrt{9+1+49}=\sqrt{59} \\ & B C=\sqrt{(-1+4)^{2}+(2-1)^{2}+(-3+10)^{2}} \\ & =\sqrt{9+1+49}=\sqrt{59} \\ & A C=\sqrt{(2+4)^{2}+(3-1)^{2}+(4+10)^{2}} \\ & =\sqrt{36+4+196} \\ & =\sqrt{236}=2 \sqrt{59} \\ & A B+B C=\sqrt{59}+\sqrt{59}=2 \sqrt{59} \end{aligned} $

Solution

Let the vertices of $\triangle A B C$ are $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.

Since, the mid-point of side $B C$ is $D(1,5,-1)$.

Then,

$ \begin{aligned} & \frac{x_2+x_3}{2}=1 \Rightarrow x_2+x_3=2 \\ & \frac{y_2+y_3}{2}=5 \Rightarrow y_2+y_3=10 \\ & \frac{z_2+z_3}{2}=-1 \Rightarrow z_2+z_3=-2 \end{aligned} $

Similarly, the mid-points of $A B$ and $A C$ are $F(2,3,4)$ and $E(0,4,-2)$,

and

$ \begin{aligned} & \frac{x_1+x_2}{2}=2 \Rightarrow x_1+x_2=4 \\ & \frac{y_1+y_2}{2}=3 \Rightarrow y_1+y_2=6 \\ & \frac{z_1+z_2}{2}=4 \Rightarrow z_1+z_2=8 \end{aligned} $

Now,

$ \begin{gathered} \frac{x_1+x_3}{2}=0 \Rightarrow x_1+x_3=0 \\ \frac{y_1+y_3}{2}=4 \Rightarrow y_1+y_3=8 \\ \frac{z_1+z_3}{2}=-2 \Rightarrow z_1+z_3=-4 \end{gathered} $

From Eqs. (i) and (iv),

From Eqs. (ii) and (v),

$ x_1+2 x_2+x_3=6 $

From Eqs. (iii) and (vi),

$ y_1+2 y_2+y_3=16 $

From Eqs. (vii) and (x),

Then,

$ z_1+2 z_2+z_3=6 $

From Eqs. (viii) and (xi),

Then,

$ \begin{aligned} 2 x_2 & =6 \Rightarrow x_2=3 \\ x_2 & =3, \text { then } x_3=-1 \\ x_3 & =-1 \\ x_1 & =1 \Rightarrow x_1=1, x_2=3, x_2=-1 \end{aligned} $

$ \begin{aligned} 2 y_2 & =8 \Rightarrow y_2=4 \\ y_2 & =4 \\ y_1 & =2 \\ y_1 & =2 \\ y_3 & =6 \\ y_1 & =2, y_2=4, y_3=6 \end{aligned} $

$ \Rightarrow $

From Eqs. (ix) and (xii),

Then,

$ \begin{aligned} 2 z_2 & =10 \Rightarrow z_2=5 \\ z_2 & =5 \end{aligned} $

$ z_1=3 $

$ z_1=3, $

Then,

$ z_3=-7 $

$\Rightarrow$

$ z_1=3, z_2=5, z_3=-7 $

So, the vertices of the triangle $A(1,2,3), B(3,4,5)$ and $C(-1,6,-7)$.

Hence, centroid of the triangle $G \frac{1+3-1}{3}, \frac{2+4+6}{3}, \frac{3+5-7}{3}$ i.e., $G(1,4,1 / 3)$.

Solution

Given points are $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$

$ \begin{aligned} & A B=\sqrt{(0-2)^{2}+(-1-1)^{2}+(-7+9)^{2}}=\sqrt{4+4+4}=2 \sqrt{3} \\ & B C=\sqrt{(2-6)^{2}+(1-5)^{2}+(-9+13)^{2}}=\sqrt{16+16+16}=4 \sqrt{3} \\ & A C=\sqrt{(0-6)^{2}+(-1-5)^{2}+(-7+13)^{2}}=\sqrt{36+36+36}=6 \sqrt{3} \end{aligned} $

$ \begin{matrix} \because & A B+B C=2 \sqrt{3}+4 \sqrt{3}=6 \sqrt{3} \\ \text { So, } & A B+B C=A C \\ \text { Hence, the points } A, B \text { and } C \text { are collinear. } \\ & A \quad B \quad B \\ & A B: A C=2 \sqrt{3}: 6 \sqrt{3}=1: 3 \end{matrix} $

So, point $A$ divide $B$ and $C$ in $1: 3$ externally,

Solution

The coordinates of the cube which edge is 2 units, are $(2,0,0),(2,2,0),(0,2,0)$, $(0,2,2),(0,0,2),(2,0,2),(0,0,0)$ and $(2,2,2)$.

Solution

(a) Given, point is $P(3,4,5)$.

Distance of $P$ from YZ-plane, $\quad[\because Y Z$-plane, $x=0]$

$ d=\sqrt{(0-3)^{2}+(4-4)^{2}+(5-5)^{2}}=3 $

Solution

(b) We know that, on the $Y$-axis, $x=0$ and $z=0$.

$\therefore$ Point $A(0,4,0)$,

$ \begin{aligned} P A & =\sqrt{(0-3)^{2}+(4-4)^{2}+(0-5)^{2}} \\ & =\sqrt{9+0+25}=\sqrt{34} \end{aligned} $

Solution

(a) Given, points $P(3,4,5)$ and $O(0,0,0)$,

$ \begin{aligned} P O & =\sqrt{(0-3)^{2}+(0-4)^{2}+(0-5)^{2}} \\ & =\sqrt{9+16+25}=\sqrt{50} \end{aligned} $

Solution

(b) Given, the points are $A(a, 0,1)$ and $B(0,1,2)$.

$ \begin{matrix} \therefore & A B=\sqrt{(a-0)^{2}+(0-1)^{2}+(1-2)^{2}} \\ \Rightarrow & \sqrt{27}=\sqrt{a^{2}+1+1} \\ \Rightarrow & 27=a^{2}+2 \\ \Rightarrow & a^{2}=25 \\ \Rightarrow & a= \pm 5 \end{matrix} $

Solution

(a) We know that, on the $X Y$ and $X Z$-planes, the line of intersection is $X$-axis.

Solution

(c) On the Y-axis, $x=0$ and $z=0$.

Solution

(b) The point $(-2,-3,-4)$ lies in seventh octant.

Solution

(a) A plane is parallel to YZ-plane, so it is perpendicular to $X$-axis.

Solution

(a) We know that, equation on the $X$-axis, $y=0, z=0$. So, the locus of the point is equation of $X$-axis.

Solution

(b) On the YZ-plane, $x=0$, hence the locus of the point is $Y Z$-plane.

Solution

(a) Given points of the parallelopiped are $A(5,8,10)$ and $B(3,6,8)$.

$ \begin{aligned} \therefore \quad A B & =\sqrt{(5-3)^{2}+(6-8)^{2}+(10-8)^{2}} \\ & =\sqrt{4+4+4}=2 \sqrt{3} \end{aligned} $

Solution

(d) We know that, on the $X Y$-plane $z=0$. Hence, the coordinates of the points $L$ are $(3,4,0)$.

Solution

(a) On the X-axis, $y=0$ and $z=0$

Hence, the required coordinates are $(3,0,0)$.

Solution

The three axes $O X, O Y$ and $O Z$ determine three coordinates planes.

Solution

Three points

Solution

Given points

Solution

Eight parts

Solution

We know that, on YZ-plane, $x=0$. So, the coordinates of the required point is $(0, y, z)$.

Solution

The equation of YZ-plane is $x=0$.

Solution

On the $Z$-axis, $x=0$ and $y=0$. So, the required coordinates are $(0,0, z)$.

Solution

The equation of $Z$-axis, $x=0$ and $y=0$.

Solution

z-coordinates.

Solution

$y$ and $z$-coordinates.

Solution

$x=$ a represent a plane parallel to $Y Z$-plane.

Solution

The plane parallel to $Y Z$-plane is perpendicular to $X$-axis.

Solution

Given dimensions are $a=10, b=13$ and $c=8$.

$ \begin{aligned} \therefore \quad \text { Required length } & =\sqrt{a^{2}+b^{2}+c^{2}} \\ & =\sqrt{100+169+64}=\sqrt{333} \end{aligned} $

Solution

Given points are $(a, 2,1)$ and $(1,-1,1)$.

$ \begin{matrix} \therefore & \sqrt{(a-1)^{2}+(2+1)^{2}+(1-1)^{2}} & =5 \\ \Rightarrow & (a-1)^{2}+9+0 & =25 \\ \Rightarrow & a^{2}-2 a+1+9 & =25 \\ \Rightarrow & a^{2}-2 a-15 & =0 \\ \Rightarrow & a^{2}-5 a+3 a-15 & =0 \\ \Rightarrow & a(a-5)+3(a-5) & =0 \\ \Rightarrow & (a-5)(a+3) & =0 \\ \Rightarrow & a-5=0 \text { or } a+3 & =0 \\ \therefore & a & =+5 \text { or }-3 \end{matrix} $

Solution

Let the vertices of $\triangle A B C$ is $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.

Since, $D$ is the mid-point of $A B$, then

$ \begin{aligned} & \frac{x_1+x_2}{2}=1 \Rightarrow x_1+x_2=2 \\ & \frac{y_1+y_2}{2}=2 \Rightarrow y_1+y_2=4 \\ & \frac{z_1+z_2}{2}=-3 \Rightarrow z_1+z_2=-6 \end{aligned} $

Similarly, $E$ and $F$ are the mid-points of sides $B C$ and $A C$, respectively.

$ \begin{aligned} & \frac{x_2+x_3}{2}=3 \Rightarrow x_2+x_3=6 \\ & \frac{y_2+y_3}{2}=0 \Rightarrow y_2+y_3=0 \\ & \frac{z_2+z_3}{2}=1 \Rightarrow z_2+z_3=2 \\ & \frac{x_1+x_3}{2}=-1 \Rightarrow x_1+x_3=-2 \\ & \frac{y_1+y_3}{2}=1 \Rightarrow y_1+y_3=2 \\ & \frac{z_1+z_3}{2}=-4 \Rightarrow z_1+z_3=-8 \end{aligned} $

From Eqs. (i) and (iv),

$ x_1+2 x_2+x_3=8 $

From Eqs. (ii) and (v),

From Eqs. (iii) and (vi),

$ y_1+2 y_2+y_3=4 $

From Eqs. (vii) and (x),

$ z_1+2 z_2+z_3=-4 $

$\Rightarrow$

$ 2 x_2=10 \Rightarrow x_2=5 $

If $x_3=1$, then $x_1=-3$

$x_2=5$, then $x_3=1$

$\because$ From Eqs. (viii) and (xi),

If

$ x_1=-3, x_2=5, x_3=1 $

$ \begin{aligned} 2 y_2 & =2 \Rightarrow y_2=1 \\ y_2 & =1, \text { then } y_3=-1 \\ y_3 & =-1, \text { then } y_1=3 \\ y_1 & =3, y_2=1, y_3=-1 \end{aligned} $

From Eqs. (ix) and (xii),

$ \begin{aligned} 2 z_2 & =4 \Rightarrow z_2=2 \\ z_2 & =2, \text { then } z_3=0 \\ z_3 & =0, \text { then } z_1=-8 \\ z_1 & =-8, z_2=2, z_3=0 \end{aligned} $

So, the vertices of $\triangle A B C$ are $A(-3,3,-8), B(5,1,2)$ and $C(1,-1,0)$. Hence, coordinates of centroid of $\triangle A B C, G \frac{-3+5+1}{3}, \frac{3+1-1}{3}, \frac{-8+2+0}{3}$

i.e.,

$G(1,1,-2)$.

Solution

(i) In XY-plane, z-coordinates is zero.

(ii) The point $(2,3,4)$ lies in 1 st octant .

(iii) Locus of the points having $x$-coordinate is zero is $Y Z$-plane.

(iv) $A$ line is parallel to $X$-axis if and only if all the points on the line have equal $y$ and $z$-coordinates.

(v) $x=0, y=0$ represent Z-axis.

(vi) $z=c$ represent the plane parallel to $X Y$-plane.

(vii) The planes $x=a, y=b$ represent the line parallel to $Z$-axis.

(viii) Coordinates of a point are the distances from the origin to the feet of perpendicular from the point on the respective.

(ix) A ball is the solid region in the space enclosed by a sphere.

(x) The region in the plane enclosed by a circle is known as a disc.