Download App

Maths Solve the Following Question.(4 Marks)

Introduction To 3D Coordinate Geometry · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 18 questions.

Questions

Solve the Following Question.(4 Marks)

Take a timed test

18 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Answer
Here, A(a, b, c), B(b, c, a), C(c, a, b)
$\text{AB}=\sqrt{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2 + (\text{c}-\text{a})^2}$
$=\sqrt{\text{a}^2 +\text{b}^2 -2\text{ab} + \text{b}^2 +\text{c}^2 - 2\text{bc} +\text{c}^2 +\text{a}^2 -2\text{ac}}$
$\text{AB}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab} - 2\text{bc}- 2\text{ac}}$
$\text{BC}=\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2 + (\text{a}-\text{b})^2}$
$=\sqrt{\text{b}^2 +\text{c}^2 - 2\text{bc}+ \text{c}2 + \text{a}^2 -2\text{ca} +\text{a}^2 +\text{b}^2 -2\text{ab}}$
$\text{BC}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab}-2\text{bc} -2\text{ca}}$
$\text{CA}=\sqrt{(\text{a}-\text{c})^2+(\text{b}-\text{a})^2 + (\text{c}-\text{b})^2}$
$=\sqrt{\text{a}^2 +\text{c}^2 - 2\text{ac}+ \text{b}2 + \text{a}^2 -2\text{ab} +\text{b}^2 +\text{c}^2 -2\text{bc}}$
$\text{CA}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab}-2\text{bc} -2\text{ca}}$
Since, AB = BC = CA, so
$\triangle\text{ABC}$ is an isosceles A.
View full question & answer→
Question 24 Marks
Show that the points A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) are the vertices of a rhombus.
Answer
Here,
$\text{AB}=\sqrt{(1+1)^2+(3-6)^2+(4-10)^2}$
$=\sqrt{4 +9+ 36}$
$=7\text{ units}$
$\text{BC}=\sqrt{(-1+7)^2+(6-4)^2+(0-7)^2}$
$=\sqrt{36+4+9}$
$=7\text{ units}$
$\text{CD}=\sqrt{(-7+5)^2+(4-1)^2+(7-1)^2}$
$=\sqrt{4 +9+ 36}$
$=7\text{ units}$
$\text{DA}=\sqrt{(-5-1)^2+(1-3)^2+(1-4)^2}$
$=\sqrt{36+4+9}$
$=\sqrt{52}$
$=7\text{ units}$
Since, AB = BC = CD = DA
So, ABCD is a rhombus.
View full question & answer→
Question 34 Marks
Show that the points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.
Answer
Let A = (0, 7, 10), B = (-1, 6, 6) and C = (-4, 9, 6)

$\text{AB}=\sqrt{(0+1)^2+(7-6)^2+(10-6)^2}$

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

$=\sqrt{18}$

$=3\sqrt{2}\text{ units}$

$\text{BC}=\sqrt{(-1+4)^2+(6-9)^2+(6-6)^2}$

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

$=\sqrt{18}$

$=3\sqrt{2}\text{ units}$

$\text{AC}=\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}$

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

$=\sqrt{36}$

$=6\text{ units}$

$(\text{AB})^2+\text{(BC)}^2$

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

$=18+18$

$=36$

$=\text{(AC)}^2$

Also (AB) = (BC)

Hence (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.
View full question & answer→
Question 44 Marks
Using distance formula prove that the following points are collinear:
P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)
Answer
P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

$\text{PQ}=\sqrt{(0-1)^2+(7-4)^2+(-7+5)^2}$

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

$=\sqrt{1+9+4}$

$=\sqrt{14}\text{ units}$

$\text{QR}=\sqrt{(1+1)^2+(4-10)^2+(-5+9)^2}$

$=\sqrt{(2)^2+(-6)^2+(4)^2}$

$=\sqrt{4+36+16}$

$=2\sqrt{14}\text{ units}$

$\text{PR}=\sqrt{(0+1)^2+(7-10)^2+(-7+9)^2}$

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

$=\sqrt{1+9+4}$

$=\sqrt{14}\text{ units}$

Since, PQ + PR = QR

so, P, Q, R are collinear.
View full question & answer→
Question 54 Marks
Find the coordinates of the point which is equidistant from the four points $O(0, 0, 0), A(2, 0, 0), B(0, 3, 0)$ and $C(0, 0, 8).$
Answer
Let the required point be $P(x_1y_1z)$ Here, $0(0, 0, 0), A(2, 0, 0), B(0, 3, 0), C(0, 0, 8)$
Since, $(OP)^2 = (PA)^2$
$(x - 0)^2 + (Y - 0)^2 + (z - 0)^2 = (x - 2)^2 + (y - 0)^2 + (z- 0)^2$
$x^2 + y^2 + z^2 = x^2 - 4x + 4 + y^2 + z^2$
$4x = 4$
$X = 1$
$(OP)^2 = (PB)^2$
$(x - 0)^2 + (y - 0)^2 + (z - 0)^2 = (x - 0)^2 + (y - 3)^2 + (z - 0)^2$
$x^2 + y^2 + z^2 = x^2 + y^2- 6y + 9 + z^2$
$6y = 9$
$\text{y}=\frac{3}{2}$
$(OP)^2 = (PC)^2$
$(x - 0)^2 + (y - 0)^2 + (z - 0)^2 = (x - 0)^2 + (Y - 0)^2 + (z - 8)^2$
$x^2+ y^2+ z^2= x^2+ y^2+ z^2 - 16z + 64$
$16z = 64$
$z = 4$
The required point $=\Big(1,\ \frac{3}{2},\ 4\Big)$
View full question & answer→
Question 64 Marks
Show that the points (3, 2, 2), (-1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.
Answer
Here,
$\text{OA}=\sqrt{(1-3)^2+(3-2)^2+(4-2)^2}$
$=\sqrt{4+1+4}$
$=3\text{ units}$
$\text{OB}=\sqrt{(1+1)^2+(3-1)^2+(4-3)^2}$
$=\sqrt{4+4+1}$
$=3\text{ units}$
$\text{OC}=\sqrt{(1-0)^2+(3-5)^2+(4-6)^2}$
$=\sqrt{1+4+4}$
$=3\text{ units}$
$\text{OD}=\sqrt{(1-2)^2+(3-1)^2+(4-2)^2}$
$=\sqrt{1+4+4}$
$=3\text{ units}$
Since, OA = OC = OD = OB, points A, B, C, O lie on a sphere with centre O.
Radius = 3 units.
View full question & answer→
Question 74 Marks
Prove that the tetrahedron with vertices at the points O(0, 0, 0), A(0, 1, 1), B(1, 0, 1) and C(1, 1, 0) is a regular one.
Answer
Here,
$\text{AB}=\sqrt{(0-1)^2+(1-0)^2+(1-1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{BC}=\sqrt{(1-1)^2+(0-1)^2+(1-0)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{CA}=\sqrt{(1-0)^2+(1-1)^2+(0-1)^2}$
$=\sqrt{1+0+1}$
$=\sqrt{2}\text{ units}$
$\text{DA}=\sqrt{(0-0)^2+(0-1)^2+(0-1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{OB}=\sqrt{(0-1)^2+(0-0)^2+(0-1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{DA}=\sqrt{(0-1)^2+(0-1)^2+(0-0)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
Since, OA = OB = OC = AB = BC = CA
So, O, A, B, C represent a regular tetrahedron.
View full question & answer→
Question 84 Marks
Prove that the point A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.
Answer
Here

$\text{AB}=\sqrt{(1+5)^2+(3-5)^2+(0-2)^2}$

$=\sqrt{36+4+4}$

$=\sqrt{44}$

$=2\sqrt{11}\text{ units}$

$\text{BC}=\sqrt{(-5+9)^2+(5+1)^2+(2-2)^2}$

$=\sqrt{16+36}$

$=\sqrt{52}$

$=2\sqrt{13}\text{ units}$

$\text{CD}=\sqrt{(-9+3)^2+(-1+3)^2+(2-0)^2}$

$=\sqrt{36+4+4}$

$=2\sqrt{11}\text{ units}$

$\text{DA}=\sqrt{(-3-4)^2+(-3-3)^2+0}$

$=\sqrt{16+36}$

$=\sqrt{52}$

$=2\sqrt{13}\text{ units}$

$\text{AC}=\sqrt{(1+9)^2+(3+1)^2+(0-2)^2}$

$=\sqrt{150+16+4}$

$=\sqrt{120}$

$=4\sqrt{5}\text{ units}$

$\text{BD}=\sqrt{(-3+5)^2+(-3-5)^2+(0-2)^2}$

$=\sqrt{4+64+4} $

$=\sqrt{72}$

$=6\sqrt{2}\text{ units}$

Since,

AB = CD and BC = DA

⇒ ABCD is a parallelgram = BD

but, $\text{AC}\neq\text{BD}$

⇒ ABCD is not a rectangles.
View full question & answer→
Question 94 Marks
If $A(-2, 2, 3)$ and $B(13, -3, 13)$ are two points. Find the locus of a point $P$ which moves in such a way that $3PA = 2PB$.
Answer
Let $P$ be $(x_1\ y_1\ z),$
Here, $A(-2, 2, 3), B(13, -3, 13)$ and $3PA = 2PB$
$\Rightarrow3\sqrt{(\text{x}+2)^2(\text{y}-2)^2+(\text{z}-3)^2}$
$=2\sqrt{(\text{x}-13)^2+(\text{y}+3)^2+(\text{z}-13)^2}$
squaring both the sides,
$\Rightarrow 9[x^2+ 4x + 4 + y^2+ 4 - 4Y + z^2+ 9 - 6z]$
$= 4[x^2 + 169 - 26x + y^2 + 9 + 6y + z^2 + 169 - 26z]$
$\Rightarrow 9x^2 - 4x^2+ 36x + 104x + 36 - 676 + 9y^2 - 4y^2 + 36 - 36 -36y - 24y + 9z^2- 4z^2 + 81 - 676 - 54z + 6yz = 0$
$\Rightarrow 5x^2 + 5y^2 + 5z2 + 140x - 60y + 50z - 1235 = 0$
$\Rightarrow 5(x^2 + y^2 + z^2) + 140x - 60y + 50z - 1235 = 0$
View full question & answer→
Question 104 Marks
Using distance formula prove that the following points are collinear:
A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
Answer
A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

$\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

$=\sqrt{(4-5)^2+(-3+7)^2+(-1-6)^2}$

$=\sqrt{(-1)^2+(4)^2+(-7)^2}$

$=\sqrt{1+16+49}$

$=\sqrt{66}\text{ units}$

$\text{BC}=\sqrt{(5-3)^2+(-7-1)^2+(6+8)^2}$

$=\sqrt{(2)^2+(-8)^2+(14)^2}$

$=\sqrt{4+64+196}$

$=\sqrt{264}$

$=2\sqrt{66}\text{ units}$

$\text{AC}=\sqrt{(4-3)^2+(-3-1)^2+(-1+8)^2}$

$=\sqrt{(1)^2+(-4)^2+(7)^2}$

$=\sqrt{1+16+49}$

$=\sqrt{66}\text{ units}$

Since, AC + AB = BC

so, A, B, C are collinear.
View full question & answer→
Question 114 Marks
If $A(-2, 2, 3)$ and $B(13, -3, 13) $ are two points. Find the locus of a point $P$ which moves in such a way that $3PA = 2PB.$
Answer
Let $P$ be $(x_1\ y_1\ z),$
Here, $A(-2, 2, 3), B(13, -3, 13)$ and $3PA = 2PB$
$\Rightarrow3\sqrt{(\text{x}+2)^2(\text{y}-2)^2+(\text{z}-3)^2}$
$=2\sqrt{(\text{x}-13)^2+(\text{y}+3)^2+(\text{z}-13)^2}$
squaring both the sides,
$\Rightarrow 9[x^2+ 4x + 4 + y^2+ 4 - 4Y + z^2+ 9 - 6z]$
$= 4[x^2 + 169 - 26x + y^2 + 9 + 6y + z^2 + 169 - 26z]$
$\Rightarrow 9x^2 - 4x^2+ 36x + 104x + 36 - 676 + 9y^2 - 4y^2 + 36 - 36 -36y - 24y + 9z^2- 4z^2 + 81 - 676 - 54z + 6yz = 0$
$\Rightarrow 5x^2 + 5y^2 + 5z2 + 140x - 60y + 50z - 1235 = 0$
$\Rightarrow 5(x^2 + y^2 + z^2) + 140x - 60y + 50z - 1235 = 0$
View full question & answer→
Question 124 Marks
Find the locus of $P$ if $PA^2 + PB^2 = 2k^2,$ where A and B are the points $(3, 4, 5)$ and $(-1, 3, -7).$
Answer
Let $P(x_1, y_1, z),$
Here, $A(3, 4, 5), B(-1, 3, -7)$
$PA^2 + PB^2 = 2k^2$
$\Rightarrow (x - 3)^2 + (y - 4)^2 + (z - 5)^2 + (x + 1)^2 + (y - 3)^2 + (2 + 7)^2 = 2k^2$
$\Rightarrow x^2 + 9 - 6x + y^2 + 16 - 8y + z^2 + 25 - 10z + x^2 + 1 + 2x + y^2 + 9 - 6y + z^2+ 49 + 14z = 2k^2$
$\Rightarrow 2x^2 + 2y^2 + 2z^2- 4x - 14y + 4z + 109 = 2k^2$
$\Rightarrow 2(x^2 + y^2 + z^2) - 4x -14y + 4z + 109 - 2k^2 = 0$
View full question & answer→
Question 134 Marks
Show that the plane $ax+ by+ cz + d = 0$ divides the line joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+d}{\text{ax}_2+\text{by}_2+\text{cz}_2+d}.$
Answer
Assume ratio is $\lambda:1$
Plane is $ax + by + cz + d = 0$
points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$
Assume point of intersection of line and plane is $D$
$\text{D}=\Big(\frac{\lambda\text{x}_2+\text{x}_1}{\lambda+1},\frac{\lambda\text{y}_2+\text{y}_1}{\lambda+1},\frac{\lambda\text{z}_2+\text{z}_1}{\lambda+1}\Big)$
As D lies on plane, substitute D in plane equation, we get
$\lambda(\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d})+\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}=0$
$\Rightarrow\lambda=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$
 
View full question & answer→
Question 144 Marks
Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelopiped so formed.
Answer
IMAGE

Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.

$\therefore$ PA = Distance between planes PBEC and QDAF

= 5 - 3

= 2

PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and -2, respectively, from the zx-plane.

$\therefore$ PB = 0 - (-2)

= 2

PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.

$\therefore$ PC = 2 - 5

= -3
View full question & answer→
Question 154 Marks
Determine the points $yz$-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
Answer
Let $Q(0, y, z)$ be the required point.
So
$(AQ)^2 = (BQ)^2 \Rightarrow (0 - 1)^2 + (y + 1)^2 - (z - 0)^2 = (0 - 2) + (y + 1)^2 + (z - 2)^2$
$\Rightarrow 1 + y^2 + 1 + 2y + z^2 = 4 + y^2 + 1 - 2y + z^2 + 4 - 42$
$\Rightarrow 4y + 4z = 7 ... (i)$
$(BQ)^2 = (CQ)^2 \Rightarrow (0 - z)^2+ (y - 1)^2 + (z - 2)^2 = (0 - 3)^2 + (y - 2)^2 (2 + 1)^2$
$\Rightarrow 4 + y^2+ 1 - 2y + z^2+ 4 - 4z - 9 + y^2+ 4 - 4y + z^2+ 1 + 2z$
$\Rightarrow 2y - 6z = 5 ... (ii)$
$(AQ)^2= (CQ)^2 \Rightarrow (0 - 1)^2 + (y + 1)^2+ (z - 0)^2= (0 - 3)^2 + (y - 2)^2 (z +^1)^2$
$\Rightarrow 1 + y^2 + 2y + 1 + z^2 = 9 + y^2 - 4y + 4 + z^2+ 1 +2z$
$\Rightarrow 6y - 2z = 12 ... (iii)$
Solving equation $(i)$ and $(ii)$ we get
$\text{z}=\frac{-3}{16}$ and $\text{y} = \frac{31}{16}$
Put the value of $y$ and $z$ in equation $(iii)$
$6y - 2z = 12 = 12$
$6\Big(\frac{31}{16}\Big)-2\Big(\frac{-3}{16}\Big) = 12$
$\frac{186}{16}+\frac{6}{16}=12$
$\frac{192}{16}=12$
$12=12$
LHS = RHS.
so,
$\text{y}=\frac{31}{16},\ \text{z}=\frac{13}{16}$
Required point $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$
 
View full question & answer→
Question 164 Marks
Determine the points $xy-$plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
Answer
Let the point on $xy-$plane be $P(x, y, 0).$
Now $P$ is equidistance from $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
So, $AP - BP - CP$
Now,
$(AP)^2= (x - 1)^2 + (y + 1)^2 + (0 - 0)^2$
$(BP)^2 = (x - 2)^2 + (y - 1)^2 + (0 - 2)^2$
$(CP)^2 = (x - 3)^2 + (y - 2)^2 + (0 + 1)^2$
$(AP)^2 = (BP)^2 \Rightarrow (x - 1)^2 + (y + 1)^2 - (x - 2)^2 + (y - 1)^2 + 4$
$\Rightarrow x^2 + 1 - 2 + y^2 + 1 + 2y + z^2 - x^2 + 4 - 4x + y^2 + 1 - 2y + 4$
$\Rightarrow 2x + 4y = 7 ... (i)$
$(BP)^2 = (CP)^2 \Rightarrow (x - 2)^2+ (y - 1)^2 + 4 = (x - 3)^2 + (y - 2)^2 + 1$
$\Rightarrow x^2+ 4 - 4x + y^2 + 1 - 2y + z^2 + 4 = x^2 + 9 - 6x + y^2 + 4 - 4y +1$
$\Rightarrow 2x + 2y = 5 ... (ii)$
$(AP)^2= (CP)^2 \Rightarrow (x - 1)^2 + (y + 1)^2+ (x - 3)^2+ (y - 2)^2+^1$
$\Rightarrow x^2+ 1 - 2x + y^2 + 1 + 2y = x^2 + 9 - 6x + y^2 + 4 - 4y + 1$
$\Rightarrow 4x + 5y = 12 ... (iii)$
Solving equation $(i)$ and $(ii)$ we get
$y = 1,$
$\text{x} = \frac{3}{2}$
Put $x$ and $y$ in equation $(iii)$
$4\Big(\frac{3}{2}\Big) + 6(1) = 12$
$12-12$
So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$
 
View full question & answer→
Question 174 Marks
Determine the points $zx-$plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
Answer
Let $R(x, 0, z)$ be the required point.
So
$(AR)^2 = (BR)^2 \Rightarrow (1 - x)^2 + (-1 - 0)^2 + (0 - z)^2 = (2 - x) + (1 - 0)^2 + (2 - z)^2$
$\Rightarrow 1 + x^2 - 2x + 1 + z^2 = 4 + x^2 - 4x + 1 + z^2 + 4z$
$\Rightarrow 2x + 4z = 7 ... (i)$
$(BR)^2 = (CR)^2 \Rightarrow (z - z)^2+ (1 - 0)^2 + (2 - z)^2 = (3 - x)^2 + (2 - 0)^2 (-1 - z)^2$
$\Rightarrow 4 + x^2- 4x + 4 + z^2- 4z = 9 + x^2- 6x + 4 + 1^+ z^2 + 2z$
$\Rightarrow 2x - 6z = 5 ... (ii)$
$(AR)^2= (CR)^2 \Rightarrow (1 - x)^2 + (1 - 0)^2+ (0 - z)^2= (3 - x)^2 + (2 - 0)^2 + (-1 - z)^2$
$\Rightarrow 1 + x^2 - 2x + 1 + z^2 = 9 + 6x + 4 + 1 + z^2$
$\Rightarrow 4x - 2z = 12 ... (iii)$
Solving equation (i) and (ii) we get
$\text{z}=\frac{1}{5},\ \text{x}=\frac{31}{10}$
Put the value of x and z in equation $(iii)$
$4x - 2z = 12$
$4\Big(\frac{31}{10}\Big)-2\Big(\frac{1}{5}\Big) = 12$
$\frac{124}{10}+\frac{2}{10}=12$
$\frac{120}{10}=12$
$12=12$
LHS = RHS.
so,
$\text{x}=\frac{31}{10},\ \text{z}=\frac{1}{5}$
Required point $=\Big(\frac{31}{10},\ 0,\ \frac{1}{5}\Big)$
 
View full question & answer→
Question 184 Marks
Using distance formula prove that the following points are collinear:
A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
Answer
A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

$\text{AB}=\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}$

$=\sqrt{(4)^2+(-5)^2+(-7)^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}\text{ units}$

$\text{BC}=\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}$

$=\sqrt{(-8)^2+(10)^2+(14)^2}$

$=\sqrt{64+100+196}$

$=\sqrt{360}$

$=6\sqrt{10}\text{ units}$

$\text{CA}=\sqrt{(3-7)^2+(-5+10)^2+(1+6)^2}$

$=\sqrt{(-4)^2+(5)^2+(7)^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}\text{ units}$

Since, AB + AC = BC

so, A, B and C are collinear.
View full question & answer→
Generate Paper FreeAll Introduction To 3D Coordinate Geometry topics