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 $XY$-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 $B(-3,0,0)$

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

Solution

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

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

Solution

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

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

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

Hence proved.

Solution

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

$\begin{array}{rl}AB& =\sqrt{(1-2{)}^2+(-1+4{)}^2+(3-5{)}^2}\\ & =\sqrt{1+9+4}=\sqrt{14}\\ BC& =\sqrt{(2-5{)}^2+(-4+13{)}^2+(5-11{)}^2}\\ & =\sqrt{9+81+36}=\sqrt{126}\\ AC& =\sqrt{(1-5{)}^2+(-1+13{)}^2+(3-11{)}^2}\\ & =\sqrt{16+144+64}=\sqrt{224}\\ \because AB+BC& =AC\\ \Rightarrow \sqrt{14}+\sqrt{126}& =\sqrt{224}\\ \Rightarrow \sqrt{14}+3\sqrt{14}& =4\sqrt{14}\end{array}$

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 $AC$,

and

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

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

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

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

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

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

Solution

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

$\begin{array}{rl}AB& =\sqrt{(0-2{)}^2+(4-3{)}^3+(1+1{)}^2}\\ & =\sqrt{4+1+4}=3\\ BC& =\sqrt{(2-4{)}^2+(3-5{)}^2+(-1-0{)}^2}\\ & =\sqrt{4+4+1}=3\\ AC& =\sqrt{(0-4{)}^2+(4-5{)}^2+(1-0{)}^2}\\ & =\sqrt{16+1+1}=\sqrt{18}\end{array}$

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

Hence, vertices $\mathrm{△}ABC$ is a right angled triangle.

Solution

Let third vertex of $\mathrm{△}ABC$ i.e., is $A(x,y,z)$.

$(2,4,6)$

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

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

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 $\mathrm{△}ABC$ 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 $BC$ are $(1,2,-3)$.

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

Similarly for the sides $AB$ and $AC$,

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

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

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

$x_1+2x_2+x_3=0$

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

$y_1+2y_2+y_3=6$

From Eqs. (vii) and (x),

$z_1+2z_2+z_3=-14$

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

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

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

From Eqs. (xi) and (viii),

$2y_2=6\Rightarrow y_2=3$

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

From Eqs. (xii) and (ix),

$\begin{array}{c}2z_2=-16\Rightarrow z_2=-8\\ z_2=-8,\text{ then }{z}_1=0\\ z_1=0,\text{ then }{z}_3=2\\ z_1=0,z_2=-8,z_3=2\end{array}$

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 $\mathrm{△}ABC$ 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 $BC(5,7,11)$.

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

Similarly for the sides $AB$ and $AC$,

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

From Eqs. (i) and (iv),

$x_1+2x_2+x_3=14$

From Eqs. (ii) and (v),

From Eqs. (iii) and (vi),

$y_1+2y_2+y_3=20$

From Eqs. (vii) and (x),

From Eqs. (viii) and (xi),

$z_1+2z_2+z_3=20$

$\begin{array}{rl}2x_2& =14\Rightarrow x_2=7\\ x_2& =7,\text{ then }{x}_3=10-7=3\\ x_3& =3,\text{ then }{x}_1=-3\\ x_1& =-3,x_2=7,x_3=3\end{array}$

$\begin{array}{rl}2y_2& =4\Rightarrow y_2=2\\ y_2& =2,\text{ then }{y}_1=4\\ y_1& =4,\text{ then }{y}_3=12\\ y_1& =4,y_2=2,y_3=12\end{array}$

From Eqs. (ix) and (xii),

$\begin{array}{rl}2z_2& =10\Rightarrow z_2=5\\ z_2& =5,\text{ then }{z}_1=-7\\ z_1& =-7,\text{ then }{z}_3=17\\ z_1& =-7,z_2=5,z_3=17\end{array}$

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 $ABCD$ is $D(x,y,z)$. Then, the mid-point of $AC$ 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 $BD$,

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

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 $AB$.

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

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

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

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

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 $\mathrm{△}ABC$ i.e., $G(0,0,0)$.

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

Solution

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

$\begin{array}{rl}AB& =\sqrt{9+16+144}=\sqrt{169}=13\\ AC& =\sqrt{0+25+144}=\sqrt{169}=\sqrt{13}\\ \frac{AB}{AC}& =\frac{13}{13}\Rightarrow AB=AC\\ \frac{BD}{DC}& =\frac{1}{1}\Rightarrow BD=DC\end{array}$

$\Rightarrow \frac{AB}{AC}=\frac{13}{13}\Rightarrow AB=AC$

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

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

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 $C(-4,1,-10)$.

$\begin{array}{rl}& \therefore AB=\sqrt{(2+1{)}^2+(3-2{)}^2+(4+3{)}^2}\\ & =\sqrt{9+1+49}=\sqrt{59}\\ & BC=\sqrt{(-1+4{)}^2+(2-1{)}^2+(-3+10{)}^2}\\ & =\sqrt{9+1+49}=\sqrt{59}\\ & AC=\sqrt{(2+4{)}^2+(3-1{)}^2+(4+10{)}^2}\\ & =\sqrt{36+4+196}\\ & =\sqrt{236}=2\sqrt{59}\\ & AB+BC=\sqrt{59}+\sqrt{59}=2\sqrt{59}\end{array}$

Solution

Let the vertices of $\mathrm{△}ABC$ 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 $BC$ is $D(1,5,-1)$.

Then,

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

Similarly, the mid-points of $AB$ and $AC$ are $F(2,3,4)$ and $E(0,4,-2)$,

and

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

Now,

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

From Eqs. (i) and (iv),

From Eqs. (ii) and (v),

$x_1+2x_2+x_3=6$

From Eqs. (iii) and (vi),

$y_1+2y_2+y_3=16$

From Eqs. (vii) and (x),

Then,

$z_1+2z_2+z_3=6$

From Eqs. (viii) and (xi),

Then,

$\begin{array}{rl}2x_2& =6\Rightarrow x_2=3\\ x_2& =3,\text{ then }{x}_3=-1\\ x_3& =-1\\ x_1& =1\Rightarrow x_1=1,x_2=3,x_2=-1\end{array}$

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

$\Rightarrow$

From Eqs. (ix) and (xii),

Then,

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

$z_1=3$

$z_1=3,$

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 $C(6,5,-13)$

$\begin{array}{rl}& AB=\sqrt{(0-2{)}^2+(-1-1{)}^2+(-7+9{)}^2}=\sqrt{4+4+4}=2\sqrt{3}\\ & BC=\sqrt{(2-6{)}^2+(1-5{)}^2+(-9+13{)}^2}=\sqrt{16+16+16}=4\sqrt{3}\\ & AC=\sqrt{(0-6{)}^2+(-1-5{)}^2+(-7+13{)}^2}=\sqrt{36+36+36}=6\sqrt{3}\end{array}$

$\begin{array}{cc}\because & AB+BC=2\sqrt{3}+4\sqrt{3}=6\sqrt{3}\\ \text{ So, }& AB+BC=AC\\ \text{ Hence, the points }A,B\text{ and }C\text{ are collinear. }\\ & ABB\\ & AB:AC=2\sqrt{3}:6\sqrt{3}=1:3\end{array}$

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, $[\because YZ$-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 $A(0,4,0)$,

$\begin{array}{rl}PA& =\sqrt{(0-3{)}^2+(4-4{)}^2+(0-5{)}^2}\\ & =\sqrt{9+0+25}=\sqrt{34}\end{array}$

Solution

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

$\begin{array}{rl}PO& =\sqrt{(0-3{)}^2+(0-4{)}^2+(0-5{)}^2}\\ & =\sqrt{9+16+25}=\sqrt{50}\end{array}$

Solution

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

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

Solution

(a) We know that, on the $XY$ and $XZ$-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 $YZ$-plane.

Solution

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

$\begin{array}{rl}\therefore AB& =\sqrt{(5-3{)}^2+(6-8{)}^2+(10-8{)}^2}\\ & =\sqrt{4+4+4}=2\sqrt{3}\end{array}$

Solution

(d) We know that, on the $XY$-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 $OX,OY$ and $OZ$ 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 $YZ$-plane.

Solution

The plane parallel to $YZ$-plane is perpendicular to $X$-axis.

Solution

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

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

Solution

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

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

Solution

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