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#### Question 9:

Show that the points (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of an isosceles right-angled triangle.

Let A(0, 7, 10), B($-$1, 6, 6), C($-$4, 9, 6) be the coordinates of △ABC
Then, we have:

Hence △ABC is a right-angled isosceles triangle.

#### Question 10:

Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.

Let A(3,3,3) , B(0,6,3) , C( 1,7,7) and D (4,4,7) are the vertices of quadrilateral $\square ABCD$

We have :

AB = $\sqrt{{\left(0-3\right)}^{2}+{\left(6-3\right)}^{2}+{\left(3-3\right)}^{2}}$
$=\sqrt{{\left(-3\right)}^{2}+{\left(3\right)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+9+0}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

BC = $\sqrt{{\left(1-0\right)}^{2}+{\left(7-6\right)}^{2}+{\left(7-3\right)}^{2}}$
$=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+16}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

CD = $\sqrt{{\left(4-1\right)}^{2}+{\left(4-7\right)}^{2}+{\left(7-7\right)}^{2}}$
$=\sqrt{{\left(3\right)}^{2}+{\left(-3\right)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+9+0}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

DA = $\sqrt{{\left(4-3\right)}^{2}+{\left(4-3\right)}^{2}+{\left(7-3\right)}^{2}}$
$=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+16}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}\phantom{\rule{0ex}{0ex}}$

AB = BC = CD = DA

AC = $\sqrt{{\left(1-3\right)}^{2}+{\left(7-3\right)}^{2}+{\left(7-3\right)}^{2}}$
$=\sqrt{{\left(-2\right)}^{2}+{\left(4\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+16+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$

BD = $\sqrt{{\left(4-0\right)}^{2}+{\left(4-6\right)}^{2}+{\left(7-3\right)}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{\left(4\right)}^{2}+{\left(-2\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+4+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$
$\therefore$AC = BD
Since, all sides and diagonals of quadrilateral $\square ABCD$ are equal
Therefore, the points are the vertices of a square.

#### Question 11:

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.

Let A(1,3,0),B($-$5,5,2), C($-$9,$-$1,2) and D($-$3,$-$3,0) be the coordinates of quadrilateral$\square ABCD$

Since, opposite pair of sides are equal .

Therefore, $\square ABCD$ is a parallelogram .

∴ ABCD is not a rectangle.

#### Question 12:

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.

Let A(1,3,4) , B($-$1,6,10) , C($-$7,4,7) and D ($-$5,1,1) be the vertices of quadrilateral

Hence, ABCD is a rhombus.

#### Question 13:

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.

The faces of a regular tetrahedron are equilateral triangles.

Let us consider $△$OAB

Hence, this face is an equilateral triangle.
Similarly, $△$OBC, $△$OAC, $△$ABC all are equilateral triangles.
Hence, the tetrahedron is a regular one.

#### Question 14:

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.

Let the points be A (3, 2, 2), B ($-$1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere
whose centre be P (1, 3, 4).
Since, AP, BP, CP  and DP are radii.
Hence, AP = BP = CP = DP

Here, we see that AP = BP = CP = DP
Hence,  A (3, 2, 2), B ($-$1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere whose radius is 3 .

#### Question 15:

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).

Let P (x, y, z) be the required point which is equidistant from the points O(0,0,0), A(2,0,0)
B(0,3,0) and C(0,0,8)

Then,
OP = AP
$⇒O{P}^{2}=A{P}^{2}$

Similarly, we have:
OP = BP
$⇒O{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}$

Similarly, we also have:

OP = CP
$⇒O{P}^{2}=C{P}^{2}$

Thus, the required point is P .

#### Question 16:

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 the 3PA = 2PB.

Let coordinates of point P be (x, y, z).
Given:
3PA = 2PB

#### Question 17:

Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).

Let P (x, y, z) be the point if $P{A}^{2}+P{B}^{2}=2{k}^{2}$

$⇒{\left(\sqrt{{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z-5\right)}^{2}}\right)}^{2}+{\left(\sqrt{{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z+7\right)}^{2}}\right)}^{2}=2{k}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-6x+9+{y}^{2}-8y+16+{z}^{2}-10z+25+{x}^{2}+2x+1+{y}^{2}-6y+9+{z}^{2}+14z+49=2{k}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}+2{z}^{2}-4x-14y+4z+109-2{k}^{2}=0$

Hence, $2{x}^{2}+2{y}^{2}+2{z}^{2}-4x-14y+4z+109-2{k}^{2}=0$ is the locus of point P.

#### Question 18:

Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.

Let A(a,b,c) , B(b,c,a) and C(c,a,b) be the vertices of $△ABC$. Then,

AB = $\sqrt{{\left(b-a\right)}^{2}+{\left(c-b\right)}^{2}+{\left(a-c\right)}^{2}}$
$=\sqrt{{b}^{2}-2ab+{a}^{2}+{c}^{2}-2bc+{b}^{2}+{a}^{2}-2ca+{c}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2{a}^{2}+2{b}^{2}+2{c}^{2}-2ab-2bc-2ca}\phantom{\rule{0ex}{0ex}}=\sqrt{2\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)}\phantom{\rule{0ex}{0ex}}$

BC = $\sqrt{{\left(c-b\right)}^{2}+{\left(a-c\right)}^{2}+{\left(b-a\right)}^{2}}$
$=\sqrt{{c}^{2}-2bc+{b}^{2}+{a}^{2}-2ca+{c}^{2}+{b}^{2}-2ab+{a}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2{a}^{2}+2{b}^{2}+2{c}^{2}-2ab-2bc-2ca}\phantom{\rule{0ex}{0ex}}=\sqrt{2\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)}\phantom{\rule{0ex}{0ex}}$

CA = $\sqrt{{\left(a-c\right)}^{2}+{\left(b-a\right)}^{2}+{\left(c-b\right)}^{2}}$
$=\sqrt{{a}^{2}-2ca+{c}^{2}+{b}^{2}-2ab+{a}^{2}+{c}^{2}-2bc+{b}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2{a}^{2}+2{b}^{2}+2{c}^{2}-2ab-2bc-2ca}\phantom{\rule{0ex}{0ex}}=\sqrt{2\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)}\phantom{\rule{0ex}{0ex}}$
$\therefore$ AB = BC = CA
Therefore,$△ABC$ is an equilateral triangle.

#### Question 19:

Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?

Let A(3,6,9), B(10,20,30) and C( 25,$-$41,5) are vertices of $△ABC$

AB = $\sqrt{{\left(10-3\right)}^{2}+{\left(20-6\right)}^{2}+{\left(30-9\right)}^{2}}$
$=\sqrt{{\left(7\right)}^{2}+{\left(14\right)}^{2}+{\left(21\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{49+196+441}\phantom{\rule{0ex}{0ex}}=\sqrt{686}\phantom{\rule{0ex}{0ex}}=7\sqrt{14}$

BC = $\sqrt{{\left(25-10\right)}^{2}+{\left(-41-20\right)}^{2}+{\left(5-30\right)}^{2}}$
$=\sqrt{{\left(15\right)}^{2}+{\left(-61\right)}^{2}+{\left(-25\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{225+3721+625}\phantom{\rule{0ex}{0ex}}=\sqrt{4571}\phantom{\rule{0ex}{0ex}}$

CA= $\sqrt{{\left(3-25\right)}^{2}+{\left(6+41\right)}^{2}+{\left(9-5\right)}^{2}}$
$=\sqrt{{\left(-22\right)}^{2}+{\left(47\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{484+2209+16}\phantom{\rule{0ex}{0ex}}=\sqrt{2709}\phantom{\rule{0ex}{0ex}}=3\sqrt[]{301}$

A triangle $△ABC$ is right-angled at B if $C{A}^{2}=A{B}^{2}+B{C}^{2}$.
But, $C{A}^{2}$$A{B}^{2}+B{C}^{2}$
Hence, the points are not vertices of a right-angled triangle.

#### Question 20:

Verify the following:
(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
(iv) (5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.

(i) Let A(0, 7, $-$10) , B(1, 6, $-$6) , C(4, 9, $-$6) be the vertices of $△ABC$.Then,
AB = $\sqrt{{\left(1-0\right)}^{2}+{\left(6-7\right)}^{2}+{\left(-6+10\right)}^{2}}$
$=\sqrt{{1}^{2}+{\left(-1\right)}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+16}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

BC = $\sqrt{{\left(4-1\right)}^{2}+{\left(9-6\right)}^{2}+{\left(-6+6\right)}^{2}}$
$=\sqrt{{3}^{2}+{3}^{2}+0}\phantom{\rule{0ex}{0ex}}=\sqrt{9+9}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

CA$\sqrt{{\left(0-4\right)}^{2}+{\left(7-9\right)}^{2}+{\left(-10+6\right)}^{2}}$
$=\sqrt{16+4+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$
Clearly, AB BC
Thus, the given points are the vertices of an isosceles triangle.

(ii) Let A(0,7,10) , B$-$1,6,6) and C$-$4,9,6) be the vertices of $△ABC$. Then ,

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

$=\sqrt{{\left(-1\right)}^{2}+{\left(-1\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+16}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

BC $\sqrt{{\left(-4+1\right)}^{2}+{\left(9-6\right)}^{2}+{\left(6-6\right)}^{2}}$
$=\sqrt{{\left(-3\right)}^{2}+{\left(3\right)}^{2}+0}\phantom{\rule{0ex}{0ex}}=\sqrt{9+9}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$

AC = $\sqrt{{\left(-4-0\right)}^{2}+{\left(9-7\right)}^{2}+{\left(6-10\right)}^{2}}$
$=\sqrt{{\left(-4\right)}^{2}+{\left(2\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+4+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$
$A{C}^{2}$$=A{B}^{2}+B{C}^{2}$
Thus, the given points are the vertices of a right-angled triangle.

(iii) Let A($-$1, 2, 1) , B(1, $-$2, 5) , C(4, $-$7, 8), D(2, $-$3, 4) be the vertices of quadrilateral $\square ABCD$

Since, each pair of opposite sides are equal.
Thus, quadrilateral $\square ABCD$ is a parallelogram.

(iv) Let A(5, $-$1, 1) , B(7, $-$4, 7) , C(1, $-$6, 10), D($-$1, $-$3, 4) be the vertices of quadrilateral $\square ABCD$

Since, all the sides are equal.
Thus, quadrilateral ABCS is a rhombus.

#### Question 21:

Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Let P (x, y, z) be any point that is equidistant from the points A (1, 2, 3) and B (3, 2, −1).
Then, we have:
PA = PB

Hence, the locus is x $-$ 2z = 0

#### Question 22:

Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(–4, 0, 0) is equal to 10.

Let P (x, y, z) be any point , the sum of whose distance from the points A(4,0,0) and B($-$4,0,0)
is equal to 10. Then,
PA + PB = 10

$⇒\sqrt{{\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}+\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}=10\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}=10-\sqrt{{\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{x}^{2}+8x+16+{y}^{2}+{z}^{2}}=10-\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒{x}^{2}+8x+16+{y}^{2}+{z}^{2}=100+{x}^{2}-8x+16+{y}^{2}+{z}^{2}-20\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}⇒16x-100=-20\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}⇒4x-25=-5\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-200x+625=25\left({x}^{2}-8x+16+{y}^{2}+{z}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-200x+625=25{x}^{2}-200x+400+25{y}^{2}+25{z}^{2}\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}+25{y}^{2}+25{z}^{2}-225=0$

#### Question 23:

Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

To show that ABCD is a parallelogram, we need to show that its two opposite sides are equal.

Thus, ABCD is not a rectangle.

#### Question 24:

Find the equation of the set of the points P such that its distances from the points A(3, 4, –5) and B(–2, 1, 4) are equal.

Let P (x, y, z) be any point which is equidistant from A (3,4,5) and B ($-$2,1,4) .Then,
PA = PB
$⇒\sqrt{{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z+5\right)}^{2}}=\sqrt{{\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}+{\left(z-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{x}^{2}-6x+9+{y}^{2}-8y+16+{z}^{2}+10z+25}=\sqrt{{x}^{2}+4x+4+{y}^{2}-2y+1+{z}^{2}-8z+16}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-6x+9+{y}^{2}-8y+16+{z}^{2}+10z+25={x}^{2}+4x+4+{y}^{2}-2y+1+{z}^{2}-8z+16\phantom{\rule{0ex}{0ex}}⇒-10x-6y+18z+29=0\phantom{\rule{0ex}{0ex}}\therefore 10x+6y-18z-29=0$

Hence, 10x + 6y $-$18z $-$29 = 0 is the required equation.

#### Question 1:

The vertices of the triangle are A(5, 4, 6), B(1, –1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

AB = $\sqrt{{4}^{2}+{5}^{2}+{3}^{2}}=\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}$
AC = $\sqrt{{1}^{2}+{1}^{2}+{4}^{2}}=\sqrt{18}=3\sqrt{2}$
AD is the internal bisector of $\angle BAC$.
$\therefore \frac{BD}{DC}=\frac{AB}{AC}=\frac{5}{3}$
Thus, D divides BC internally in the ratio 5:3.

#### Question 2:

A point C with z-coordinate 8 lies on the line segment joining the points A(2, –3, 4) and B(8, 0, 10). Find its coordinates.

Suppose C divides AB in the ratio $\lambda :1$.
Then, the coordinates of C are .
The z-coordinate of C is 8.

Hence, the coordinates of C are as follows:

Hence, the coordinates of C are (6, −1, 8).

#### Question 3:

Show that the three points A(2, 3, 4), B(–1, 2 – 3) and C(–4, 1, –10) are collinear and find the ratio in which C divides AB.

Suppose C divides AB in the ratio $\lambda :1$.
Then, the coordinates of C are .
But, the coordinates of C are ($-$4, 1, $-$10).

From these three equations, we have:
$\lambda =-2$
So, C divides AB in the ratio 2:1 (externally).

#### Question 4:

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Let A$\equiv$(2, 4, 5) and B$\equiv$(3, 5, 4)
Let the line joining A and B be divided by the yz-plane at point P in the ratio $\lambda :1$.
Then, we have:
P
Since P lies on the yz-plane, the x-coordinate of P will be zero.

Hence, the yz-plane divides AB in the ratio 2:3 (externally).

#### Question 5:

Find the ratio in which the line segment joining the points (2, –1, 3) and (–1, 2, 1) is divided by the plane x + y + z = 5.

Suppose the plane x + y + z = 5 divides the line joining the points A (2, −1, 3) and B ($-$1, 2, 1) at a point C in the ratio $\lambda :1$.
Then, coordinates of C are as follows:

Now, the point C lies on the plane x + y + z = 5.
Therefore, the coordinates of C must satisfy the equation of the plane.
$\frac{-\lambda +2}{\lambda +1}+\frac{2\lambda -1}{\lambda +1}+\frac{\lambda +3}{\lambda +1}=5\phantom{\rule{0ex}{0ex}}⇒-\lambda +2+2\lambda -1+\lambda +3=5\lambda +5\phantom{\rule{0ex}{0ex}}⇒2\lambda +4=5\lambda +5\phantom{\rule{0ex}{0ex}}⇒-3\lambda =1\phantom{\rule{0ex}{0ex}}\therefore \lambda =\frac{-1}{3}$
So, the required ratio is 1:3 (externally).

#### Question 6:

If the points A(3, 2, –4), B(9, 8, –10) and C(5, 4, –6) are collinear, find the ratio in which C divides AB.

Suppose C divides AB in the ratio $\lambda :1$.
Then, coordinates of C are as follows:

But, the coordinates of C are (5, 4, $-$6).

These three equations give $\lambda =\frac{1}{2}$.
So, C divides AB in the ratio 1:2.

#### Question 7:

The mid-points of the sides of a triangle ABC are given by (–2, 3, 5), (4, –1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Let be the vertices of the given triangle.
And, let be the mid-points of the sides EF, FA and DE, respectively.
Now, A is the mid-point of EF.

B is the mid-point of FD.

C is the mid-point of DE.

Adding the first three equations in (i), (ii) and (iii):
$2\left({x}_{1}+{x}_{2}+{x}_{3}\right)=-4+8+12\phantom{\rule{0ex}{0ex}}⇒{x}_{1}+{x}_{2}+{x}_{3}=8$
Solving the first three equations in (i), (ii) and (iii) with ${x}_{1}+{x}_{2}+{x}_{3}=8$:

Adding the next three equations in (i), (ii) and (iii):
$2\left({y}_{1}+{y}_{2}+{y}_{3}\right)=6-2+10\phantom{\rule{0ex}{0ex}}⇒{y}_{1}+{y}_{2}+{y}_{3}=7$
Solving the next three equations in (i), (ii) and (iii) with ${y}_{1}+{y}_{2}+{y}_{3}=7$:

Adding the last three equations in (i), (ii) and (iii):
$2\left({z}_{1}+{z}_{2}+{z}_{3}\right)=10+14+6\phantom{\rule{0ex}{0ex}}⇒{z}_{1}+{z}_{2}+{z}_{3}=15$
Solving the last three equations in (i), (ii) and (iii) with ${z}_{1}+{z}_{2}+{z}_{3}=15$:

Thus, the vertices of the triangle are (12, 1, 5), (0, 9, 1), (−4, −3, 9).

#### Question 8:

A(1, 2, 3), B(0, 4, 1), C(–1, –1, –3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

AB = $\sqrt{{\left(-1\right)}^{2}+{2}^{2}+{\left(-2\right)}^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$
AC = $\sqrt{{\left(-2\right)}^{2}+{\left(-3\right)}^{2}+{\left(-6\right)}^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$

AD is the internal bisector of $\angle BAC$.
$\therefore \frac{BD}{DC}=\frac{AB}{AC}=\frac{3}{7}$
Thus, D divides BC internally in the ratio 3:7.

#### Question 9:

Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).

Let the sphere  meet the line joining the given points at the point (x1, y1, z1).

Then, we have:

Suppose the point (x1, y1, z1) divides the line joining the points  (12, – 4, 8) and (27, – 9, 18) in the ratio λ:1.

Substitute these values in equation (1):

Thus, the sphere divides the line joining the given points in the ratio 2:3 and 2:– 3.

Hence, the given sphere divides the line joining the points (12, – 4, 8) and (27, – 9, 18) internally in the ratio 2:3 and externally in the ratio −2:3.

#### Question 10:

Show that the plane ax + by + cz + d = 0 divides the line joining the points (x1, y1, z1) and (x2, y2, z2) in the ratio $-\frac{a{x}_{1}+b{y}_{1}+c{z}_{1}+d}{a{x}_{2}+b{y}_{2}+c{z}_{2}+d}$.

Let:
A = (x1, y1, z1
B = (x2, y2, z2)

Now, let the line joining A and B be divided by the plane ax + by + cz + d = 0 at point P in the ratio $\lambda :1$.

P =
Since P lies on the given plane,

ax + by + cz + d = 0

Thus,

#### Question 11:

Find the centroid of a triangle, mid-points of whose sides are (1, 2, –3), (3, 0, 1) and (–1, 1, –4).

#### Question 12:

The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, –6) respectively, find the coordinates of the point C.

Let G be the centroid of $∆$ABC.
Given: G
Let C

#### Question 13:

Find the coordinates of the points which tisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6).

Let P$\equiv$(4, 2, $-$6) and Q$\equiv$(10, $-$16, 6)
Let A and B be the point of trisection.
Then, we have:
PA = AB = BQ
∴ PA:AQ = 1:2
Thus, A divides PQ internally in the ratio 1:2.
∴ A$\equiv$

$⇒$ A$\equiv$(6, $-$4, $-$2)
AB = BQ

Therefore, B is the mid-point of AQ.
∴ B$\equiv$

$⇒$ B$\equiv$(8, $-$10, 2)

#### Question 14:

Using section formula, show that he points A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) are collinear.

Let, C divides AB in the ratio $\lambda$:1.
Then, coordinates of C are $\left(\frac{-\lambda +2}{\lambda +1},\frac{2\lambda -3}{\lambda +1},\frac{\lambda +4}{\lambda +1}\right)$.
But, the coordinates of C are .
$\frac{-\lambda +2}{\lambda +1}$=0 , $\frac{2\lambda -3}{\lambda +1}$=$\frac{1}{3}$, $\frac{\lambda +4}{\lambda +1}$ = 2
From each of these equations, we get $\lambda =2$.
Since each of these equations give the same value of $\lambda$.
Therefore, the given points are collinear.

#### Question 15:

Given that  P(3, 2, –4), Q(5, 4, –6) and R(9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Let Q divide PR in the ratio $\lambda :1$
Thus, the coordinates of Q are as follows:

But, the coordinates of Q are (5, 4, −6) .

These three equation gives $\lambda =\frac{1}{2}$.
So, Q divides PR in the ratio $\frac{1}{2}:1$ or 1:2

#### Question 16:

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, –8) is divided by the yz-plane.

Let A = (4, 8, 10) , B = (6, 10, $-$8)
Let the line joining A and B be divided by the YZ-plane at point P in the ratio $\lambda :1$.
P=
Since P lies on the YZ-plane, the x-coordinate of P will be zero.

Hence, the YZ-plane divides AB in the ratio 2:3 (externally)

#### Question 1:

Write the distance of the point P (2, 3,5) from the xy-plane.

The distance of the point P (2, 3, 5) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 5.
Hence, the distance of the point P (2, 3,5) from the xy-plane is 5.

#### Question 2:

Write the distance of the point P(3, 4, 5) from z-axis.

The distance of the point P(3, 4, 5) from z-axis is given by
$\sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5$

#### Question 3:

If the distance between the points P(a, 2, 1) and Q (1, −1, 1) is 5 units, find the value of a.

PQ = 5

#### Question 4:

The coordinates of the mid-points of sides AB, BC and CA of  △ABC are D(1, 2, −3), E(3, 0,1) and F(−1, 1, −4) respectively. Write the coordinates of its centroid.

Let the coordintes of the triangles be A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3).
Now,
Mid point of AB is D(1, 2, −3)

Mid point of BC is E(3, 0,1)

Mid point of AC is F(−1, 1, −4)

Adding (1), (2) and (3),we get

Hence, the centroid of the traingle ABC is (1, 1, −2).

#### Question 5:

Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.

We know that the x  and z coordinates on y - axis are 0
The coordinates of the foot of the perpendicular from a point (1, 2, 3) on y - axis are (0, 2, 0)

#### Question 6:

Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.

The distance of the point P(3, 5, 12) from x-axis is given by
$\sqrt{{5}^{2}+{\left(12\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{169}\phantom{\rule{0ex}{0ex}}=13$

#### Question 7:

Write the coordinates of third vertex of a triangle having centroid at the origin and two vertices at (3, −5, 7) and (3, 0, 1).

Let the coordinates of third vertex  be (x1, y1, z1)
Now,

Hence, the coordinates of third vertex of a triangle  is (−6, 5, −8).

#### Question 8:

What is the locus of a point for which y = 0, z = 0?

We know that on x - axis both y = 0, z = 0.
Hence, the locus of a point for which y = 0, z = 0 is x - axis.

#### Question 9:

Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.

Let the yz - plane divide the line sgement joining the points (2, 4,5) and (3, −5, 4) in m : 1.
Now, we know that on yz-plane the coordinate of x is 0.
$\therefore \frac{m×3+1×2}{m+1}=0\phantom{\rule{0ex}{0ex}}⇒3m+2=0\phantom{\rule{0ex}{0ex}}⇒m=-\frac{2}{3}$
Hence, yz - plane divide the line sgement joining the points (2, 4,5) and (3, −5, 4) in 2 : 3 externally.

#### Question 10:

Find the point on y-axis which is at a distance of $\sqrt{10}$ units from the point (1, 2, 3).

We know that the x and z coordinates of the point on the y-axis are 0.
So, let the required point be (0, y, 0)
Now,

Hence, the required point is (0, 2, 0)

#### Question 11:

Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).

We know that the y and z coordinates of the point on the x-axis are 0.
So, let the required point be C (x, y, z)
Now, CA = CB

$\sqrt{{\left(3-x\right)}^{2}+{\left(2-0\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{{\left(5-x\right)}^{2}+{\left(5-0\right)}^{2}+{\left(4-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒9-6x+{x}^{2}+4+4=25-10x+{x}^{2}+25+16\phantom{\rule{0ex}{0ex}}⇒17-6x+{x}^{2}=66-10x+{x}^{2}\phantom{\rule{0ex}{0ex}}⇒4x=49\phantom{\rule{0ex}{0ex}}⇒x=\frac{49}{4}$
Hence, the required point is

#### Question 12:

Find the coordinates of a point equidistant from the origin and points A (a, 0, 0), B (0, b, 0) andC(0, 0, c).

Let the point be P(x, y, z).
Now, PO = PA
$\sqrt{{\left(0-x\right)}^{2}+{\left(0-y\right)}^{2}+{\left(0-z\right)}^{2}}=\sqrt{{\left(a-x\right)}^{2}+{\left(0-y\right)}^{2}+\left(0-z\right)}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+{z}^{2}={a}^{2}-2ax+{x}^{2}+{y}^{2}+{z}^{2}\phantom{\rule{0ex}{0ex}}⇒0={a}^{2}-2ax\phantom{\rule{0ex}{0ex}}⇒x=\frac{a}{2}$
Also, PO = PB
$\sqrt{{\left(0-x\right)}^{2}+{\left(0-y\right)}^{2}+{\left(0-z\right)}^{2}}=\sqrt{{\left(0-x\right)}^{2}+{\left(b-y\right)}^{2}+\left(0-z\right)}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+{z}^{2}={x}^{2}+{b}^{2}-2by+{y}^{2}+{z}^{2}\phantom{\rule{0ex}{0ex}}⇒0={b}^{2}-2by\phantom{\rule{0ex}{0ex}}⇒y=\frac{b}{2}$
Again, PO = PC
$\sqrt{{\left(0-x\right)}^{2}+{\left(0-y\right)}^{2}+{\left(0-z\right)}^{2}}=\sqrt{{\left(0-x\right)}^{2}+{\left(0-y\right)}^{2}+\left(c-z\right)}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+{z}^{2}={x}^{2}+{y}^{2}+{c}^{2}-2cz+{z}^{2}\phantom{\rule{0ex}{0ex}}⇒0={c}^{2}-2cz\phantom{\rule{0ex}{0ex}}⇒z=\frac{c}{2}$

Hence, the coordinates of the point is

#### Question 13:

Write the coordinates of the point P which is five-sixth of the way from A(−2, 0, 6) to B(10, −6, −12).

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

Hence, the coordinates of the point is

#### Question 14:

If a parallelopiped is formed by the planes drawn through the points (2,3,5) and (5, 9, 7) parallel to the coordinate planes, then write the lengths of edges of the parallelopiped and length of the diagonal.

Lengths of edges of the parallelopiped are given by
5 − 2, 9 − 3, 7 − 5
= 3, 6, 2
Length of the diagonal is given by

#### Question 15:

Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3,2) and C(0, 0,1).

The coordiante of x point on yz-plane is 0
Let the point be P(0, y, z).
Now, PA = PB

Also, PA = PC

Solving (1) and (2),we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).

#### Question 16:

If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(−2, b −5) and C (4, 7, c), find the values of a, b, c.

We have A(a, 1, 3), B(−2, b −5) and C (4, 7, c)
Now,

#### Question 1:

The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
(a) 2 : 3
(b) 3 : 2
(c) –2 : 3
(d) 4 : –3

(c) $-$2 : 3

Let A$\equiv$(2, 4, 5) and B$\equiv$(3, 5, $-$9)
Let the line joining A and B be divided by the yz-plane at point P in the ratio $\lambda :1$.
Then, we have,

P

Since P lies on the yz-plane, the x-coordinate of P will be zero.

$\therefore \frac{3\lambda +2}{\lambda +1}=0\phantom{\rule{0ex}{0ex}}⇒3\lambda +2=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{-2}{3}$
Hence, the yz-plane divides AB in the ratio $-$2 : 3

#### Question 2:

The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
(a) a : b
(b) b : c
(c) c : a
(d) c : b

(d) c : b

Let A$\equiv$(a, b, c) and B$\equiv$($-$a, $-$c, $-$b)
Let the line joining A and B be divided by the xy-plane at point P in the ratio $\lambda :1$.
Then, we have,

P

Since P lies on the xy-plane, the z-coordinate of P will be zero.

$\therefore \frac{-b\lambda +c}{\lambda +1}=0\phantom{\rule{0ex}{0ex}}⇒-b\lambda +c=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{c}{b}$
Hence, the xz-plane divides AB in the ratio c : b

#### Question 3:

If P(0, 1, 2), Q(4, –2, 1) and O(0, 0, 0) are three points, then POQ =

(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{2}$

(d) $\frac{\mathrm{\pi }}{2}$

#### Question 4:

If the extremities of the diagonal of a square are (1, –2, 3 and (2, –3, 5), then the length of the side is

(a) $\sqrt{6}$

(b) $\sqrt{3}$

(c) $\sqrt{5}$

(d) $\sqrt{7}$

(b) $\sqrt{3}$

Length of the diagonal = $\sqrt{{\left(2-1\right)}^{2}+{\left(-3+2\right)}^{2}+{\left(5-3\right)}^{2}}=\sqrt{1+1+4}=\sqrt{6}$

∴ Length of the side =

#### Question 5:

The points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) are the vertices of
(a) a rectangle
(b) a square
(c) a parallelogram
(d) none of these

(d) None of these

Suppose:
A(5, –4, 2)
B(4, –3, 1)
C(7, 6, 4)
D(8, –7, 5)

We see that none of the sides are equal.

#### Question 6:

In a three dimensional space the equation x2 – 5x + 6 = 0 represents
(a) points
(b) planes
(c) curves
(d) pair of straight lines

(c) curves

Since, there is only one variable in the given equation. Also, it is quadratic equation.
Hence, It represents curves in yz plane.

#### Question 7:

Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to
(a) 2
(b) 3
(c) 6
(d) 7

(b) 3

Suppose d is the diameter of the sphere. Then

Hence, radius of the sphere is 3 units.

#### Question 8:

XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
(a) 3 : 7
(b) 2 : 7
(c) –3 : 7
(d) –2 : 7

(c) −3:7

Let A$\equiv$(2, 3, 1) and B$\equiv$(6, 7, 1)
Let the line joining A and B be divided by the xz-plane at point P in the ratio $\lambda :1$.
Then, we have,

P

Since P lies on the xz-plane, the y-coordinate of P will be zero.

$\therefore \frac{7\lambda +3}{\lambda +1}=0\phantom{\rule{0ex}{0ex}}⇒7\lambda +3=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{-3}{7}$
Hence, the xz-plane divides AB in the ratio $-$3 : 7

#### Question 9:

What is the locus of a point for which y = 0, z = 0?
(a) x - axis
(b) y - axis
(c) z - axis
(d) yz - plane

We know that on x - axis both y = 0, z = 0.
Hence, the correct answer is option (a)

#### Question 10:

The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz - plane are
(a) (3, 4, 0)
(b) (0, 4, 5)
(c) (3, 0, 5)
(d) (3, 0, 0)

We know that the x - coordinate on yz - plane is 0.
The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz - plane are (0, 4, 5).
Hence, the correct answer is option (b).

#### Question 11:

The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are

(a) (6, 0, 0)
(b) (0, 7, 0)
(c) (0, 0, 8)
(d) (0, 7, 8)

We know that the y  and z coordinates on x - axis are 0
The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are (6, 0, 0)
Hence, the correct answer is option (a).

#### Question 12:

The perpendicular distance of the point P (6, 7, 8) from xy - plane is

(a) 8
(b) 7
(c) 6
(d) 10

The distance of the point P (6, 7, 8) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 8.
Hence, the correct answer is option (a).

#### Question 13:

The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is
(a) 10
(b) $\sqrt{34}$
(c) $\sqrt{113}$
(d) 512

The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is given by

$\sqrt{{3}^{2}+{5}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{34}$
Hence, the correct answer is option (b)

#### Question 14:

The perpendicular distance of the point P(3, 3,4) from the x-axis is

(a) $3\sqrt{2}$
(b) 5
(c) 3
(d) 4

The perpendicular distance of the point P(3, 3,4) from the x-axis is given by

$\sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5$
Hence, the correct answer is option (b)

#### Question 15:

The length of the perpendicular drawn from the point P(a, b, c) from z-axis is

(a) $\sqrt{{a}^{2}+{b}^{2}}$

(b) $\sqrt{{b}^{2}+{c}^{2}}$

(c) $\sqrt{{a}^{2}+{c}^{2}}$

(d) $\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}$

The length of the perpendicular drawn from the point P(x, y, z) from z-axis is given by $\sqrt{{y}^{2}+{x}^{2}}$
Thus, the length of the perpendicular drawn from the point P(a, b, c) from z-axis is $\sqrt{{a}^{2}+{b}^{2}}$
Hence, the correct answer is option (a)

#### Question 1:

Name the octants in which the following points lie:
(i) (5, 2, 3)
(ii) (–5, 4, 3)
(iii) (4, –3, 5)
(iv) 7, 4, –3)
(v) (–5, –4, 7)
(vi) (–5, –3, –2)
(vii) (2, –5, –7)
(viii) (–7, 2 – 5)

(i)  The x-coordinate, the y-coordinate and the z-coordinate of the point (5, 2, 3) are all positive.
Therefore, this point lies in XOYZ octant.

(ii) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, 4, 3) are negative, positive and positive, respectively.
Therefore, this point lies in X'OYZ octant.

(iii) The x-coordinate, the y-coordinate and the z-coordinate of the point (4, −3, 5) are positive, negative and positive, respectively.
Therefore, this point lies in XOY'Z octant.

(iv) The x-coordinate, the y-coordinate and the z-coordinate of the point (7, 4, −3) are positive, positive and negative, respectively.
Therefore, this point lies in XOYZ' octant.

(v) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, −4, 7) are negative, negative and positive, respectively.
Therefore, this point lies in X'OY'Z octant .

(vi) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, −3, −2) are all negative.
Therefore, this point lies in X'OY'Z' octant

(vii) The x-coordinate, the y-coordinate and the z-coordinate of the point (2, −5, −7) are positive, negative and negative, respectively.
Therefore, this point lies in XOY'Z' octant.

(viii) The x-coordinate, the y-coordinate and the z-coordinate of the point(−7, 2, −5) are negative, positive and negative, respectively.
Therefore, this point lies in X'OYZ' octant.

#### Question 2:

Find the image  of:
(i) (–2, 3, 4) in the yz-plane.
(ii) (–5, 4, –3) in the xz-plane.
(iii) (5, 2, –7) in the xy-plane.
(iv) (–5, 0, 3) in the xz-plane.
(v) (–4, 0, 0) in the xy-plane.

(i)
(ii) $\left(-5,-4,-3\right)$
(iii)
(iv)
(v)

#### Question 3:

A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.

Let P$\equiv$(1, 0, −1)
The length of each side of the cube is 5.
The three edges from vertex of the cube are drawn from P towards the negative x and y axes and the positive z-axis.
Therefore, the coordinates of the vertex of the cube will be as follows:
x-coordinate = 1, 1$-$5 = $-$4, i.e. 1, $-$4
y-coordinate = 0, 0$-$5 = $-$5, i.e. 0, $-$5
z-coordinate = $-$1, $-$1 + 5 = 4, i.e. $-$1, 4
Hence, the remaining seven vertices of the cube are as follows:

#### Question 4:

Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.

Let P$\equiv$(3, 0, −1),  Q$\equiv$(−2, 5, 4)
PE = Distance between the parallel planes ABCP and FQDE
= $\left|4+1\right|=5$
(These planes are perpendicular to the z-axis)
PA = Distance between the parallel planes ABQF and PCDE
= $\left|-2-3\right|=5$
(These planes are perpendicular to the x-axis)
Similarly, PC = $\left|5-0\right|=5$
Thus, the length of the edges of the parallelepiped are 5, 5 and 5

#### Question 5:

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 parallelepiped so formed.

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

#### Question 6:

Find the distances of the point P(–4, 3, 5) from the coordinate axes.

Let PQ be the perpendicular to the xy-plane and QA be perpendicular from Q to the y-axis.
PA will be perpendicular to the x-axis.
Also, QA = $\left|3\right|$ and PQ = $\left|5\right|$
Now, distance of P from the x-axis:
PB = $\sqrt{B{Q}^{2}+Q{P}^{2}}$
$=\sqrt{{3}^{2}+{5}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+25}=\sqrt{34}$
Similarly,
From the right-angled $∆PAQ$:
distance of P from the y-axis:
PA = $\sqrt{A{Q}^{2}+Q{P}^{2}}$
$=\sqrt{{\left(-4\right)}^{2}+{\left(5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+25}=\sqrt{41}$
Similarly, the length of the perpendicular from P to the z-axis =$\sqrt{{\left(-4\right)}^{2}+{\left(3\right)}^{2}}$
$=\sqrt{16+9}\phantom{\rule{0ex}{0ex}}=\sqrt{25}=5$

#### Question 7:

The coordinates of a point are (3, –2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

The seven coordinates are as follows:

#### Question 1:

Find the distance between the following pairs of points:
(i) P(1, –1, 0) and Q(2, 1, 2)
(ii) A(3, 2, –1) and B(–1, –1, –1).

(i) PQ = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}$
= $\sqrt{{\left(2-1\right)}^{2}+{\left(1+1\right)}^{2}+{\left(2-0\right)}^{2}}$
= $\sqrt{{1}^{2}+{2}^{2}+{2}^{2}}$
= $\sqrt{1+4+4}$
= $\sqrt{9}$
= 3

(ii) AB = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}$
$=\sqrt{{\left(-1-3\right)}^{2}+{\left(-1-2\right)}^{2}+{\left(-1+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-4\right)}^{2}+{\left(-3\right)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+9+0}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5$

#### Question 2:

Find the distance between the points P and Q having coordinates (–2, 3, 1) and (2, 1, 2).

PQ = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}$

#### Question 3:

Using distance formula prove that the following points are collinear:
(i) A(4, –3, –1), B(5, –7, 6) and C(3, 1, –8)
(ii) P(0, 7, –7), Q(1, 4, –5) and R(–1, 10, –9)
(iii) A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6)

(i)  AB = $\sqrt{{\left(5-4\right)}^{2}+{\left(-7+3\right)}^{2}+{\left(6+1\right)}^{2}}$
$=\sqrt{{\left(1\right)}^{2}+{\left(-4\right)}^{2}+{\left(7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+16+49}\phantom{\rule{0ex}{0ex}}=\sqrt{66}$

BC = $\sqrt{{\left(3-5\right)}^{2}+{\left(1+7\right)}^{2}+{\left(-8-6\right)}^{2}}$
= $\sqrt{{\left(-2\right)}^{2}+{\left(8\right)}^{2}+{\left(-14\right)}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{4+64+196}\phantom{\rule{0ex}{0ex}}=\sqrt{264}\phantom{\rule{0ex}{0ex}}=2\sqrt{66}$

AC = $\sqrt{{\left(3-4\right)}^{2}+{\left(1+3\right)}^{2}+{\left(-8+1\right)}^{2}}$
$=\sqrt{{\left(-1\right)}^{2}+{\left(4\right)}^{2}+{\left(-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+16+49}\phantom{\rule{0ex}{0ex}}=\sqrt{66}$

Hence, the points are collinear.

(ii) PQ =
$=\sqrt{{\left(1\right)}^{2}+{\left(-3\right)}^{2}+{\left(2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+9+4}\phantom{\rule{0ex}{0ex}}=\sqrt{14}$

QR = $\sqrt{{\left(-1-1\right)}^{2}+{\left(10-4\right)}^{2}+{\left(-9+5\right)}^{2}}$
$=\sqrt{{\left(-2\right)}^{2}+{\left(6\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+36+16}\phantom{\rule{0ex}{0ex}}=\sqrt{56}\phantom{\rule{0ex}{0ex}}=2\sqrt{14}$

PR$=\sqrt{{\left(-1-0\right)}^{2}+{\left(10-7\right)}^{2}+{\left(-9+7\right)}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{\left(-1\right)}^{2}+{\left(3\right)}^{2}+{\left(-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+9+4}\phantom{\rule{0ex}{0ex}}=\sqrt{14}$

Hence, the points are collinear.

(iii) AB = $\sqrt{{\left(-1-3\right)}^{2}+{\left(0+5\right)}^{2}+{\left(8-1\right)}^{2}}$
$=\sqrt{{\left(-4\right)}^{2}+{\left(5\right)}^{2}+{\left(7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+25+49}\phantom{\rule{0ex}{0ex}}=\sqrt{90}\phantom{\rule{0ex}{0ex}}=3\sqrt{10}$

BC = $\sqrt{{\left(7+1\right)}^{2}+{\left(-10-0\right)}^{2}+{\left(-6-8\right)}^{2}}$
$=\sqrt{{\left(8\right)}^{2}+{\left(-10\right)}^{2}+{\left(-14\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{64+100+196}\phantom{\rule{0ex}{0ex}}=\sqrt{360}\phantom{\rule{0ex}{0ex}}=6\sqrt{10}$

AC = $\sqrt{{\left(7-3\right)}^{2}+{\left(-10+5\right)}^{2}+{\left(-6-1\right)}^{2}}$
$=\sqrt{{\left(4\right)}^{2}+{\left(-5\right)}^{2}+{\left(-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+25+49}\phantom{\rule{0ex}{0ex}}=\sqrt{90}\phantom{\rule{0ex}{0ex}}=3\sqrt{10}$

Hence, the points are collinear.

#### Question 4:

Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

(i) We know that the z-coordinate of every point on the xy-plane is zero.
So, let P (x, y, 0) be a point on the xy-plane such that PA = PB = PC
Now, PA = PB

(ii)  We know that the x-coordinate of every point on the yz-plane is zero.
So, let P (0, y, z) be a point on the yz-plane such that PA = PB = PC
Now, PA = PB
$⇒{\left(0-1\right)}^{2}+{\left(y+1\right)}^{2}+{\left(z-0\right)}^{2}={\left(0-2\right)}^{2}+{\left(y-1\right)}^{2}+{\left(z-2\right)}^{2}$

Hence, the required point is $\left(0,\frac{11}{8},\frac{-3}{8}\right)$.

(iii)  We know that the y-coordinate of every point on the zx-plane is zero.
So, let P (x, 0, z) be a point on the zx-plane such that PA = PB = PC
Now, PA = PB
$⇒{\left(x-1\right)}^{2}+{\left(0+1\right)}^{2}+{\left(z-0\right)}^{2}={\left(x-2\right)}^{2}+{\left(0-1\right)}^{2}+{\left(x-2\right)}^{2}$

Hence, the required point is .

#### Question 5:

Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).

Let M be the point on the z-axis.
Then, the coordinates of M will be .

Let M be equidistant from the points A and B

AM = $\sqrt{{\left(0-1\right)}^{2}+{\left(0-5\right)}^{2}+{\left(z-7\right)}^{2}}$
$=\sqrt{{\left(-1\right)}^{2}+{\left(-5\right)}^{2}+{\left(z-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+25+{z}^{2}-14z+49}\phantom{\rule{0ex}{0ex}}=\sqrt{{z}^{2}-14z+75}$

BM = $\sqrt{{\left(0-5\right)}^{2}+{\left(0-1\right)}^{2}+{\left(z+4\right)}^{2}}$
$=\sqrt{{\left(-5\right)}^{2}+{\left(-1\right)}^{2}+{z}^{2}+8z+16}\phantom{\rule{0ex}{0ex}}=\sqrt{25+1+{z}^{2}+8z+16}\phantom{\rule{0ex}{0ex}}=\sqrt{{z}^{2}+8z+42}$

Then, AM = BM

Thus, the coordinates of M are .

#### Question 6:

Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).

Let the point on the y-axis be Y which is equidistant from the points P and Q .
Then ,PY=QY
Now,  $\sqrt{{\left(0-3\right)}^{2}+{\left(y-1\right)}^{2}+{\left(0-2\right)}^{2}}$=$\sqrt{{\left(0-5\right)}^{2}+{\left(y-5\right)}^{2}+{\left(0-2\right)}^{2}}$

$⇒\sqrt{{\left(-3\right)}^{2}+{y}^{2}+2y+1+{\left(-2\right)}^{2}}=\sqrt{{\left(-5\right)}^{2}+{\left(y-5\right)}^{2}+{\left(-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{9+{y}^{2}+2y+1+4}=\sqrt{25+{y}^{2}+10y+25+4}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{y}^{2}+2y+14}=\sqrt{{y}^{2}+10y+54}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒{y}^{2}+2y+14={y}^{2}+10y+54$

$2y-10y=54-14\phantom{\rule{0ex}{0ex}}-8y=40\phantom{\rule{0ex}{0ex}}-y=\frac{40}{8}\phantom{\rule{0ex}{0ex}}y=-5$

Thus, the required point on the y-axis is (0, −5, 0).

#### Question 7:

Find the points on z-axis which are at a distance $\sqrt{21}$ from the point (1, 2, 3).

Let the point be A (0, 0, z).Then,

AP = $\sqrt{21}$

Hence, the coordinates of the required point are (0, 0, 7)  and (0, 0, −1).

#### Question 8:

Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.

Let A (1, 2, 3) , B (2, 3, 1) and C (3, 1, 2) are the coordinates of the triangle $△ABC$
AB = $\sqrt{{\left(2-1\right)}^{2}+{\left(3-2\right)}^{2}+{\left(1-3\right)}^{2}}$
$=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+4}\phantom{\rule{0ex}{0ex}}=\sqrt{6}$

BC = $\sqrt{{\left(3-2\right)}^{2}+{\left(1-3\right)}^{2}+{\left(2-1\right)}^{2}}$
$=\sqrt{{\left(1\right)}^{2}+{\left(-2\right)}^{2}+{\left(1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+4+1}\phantom{\rule{0ex}{0ex}}=\sqrt{6}$

AC = $\sqrt{{\left(3-1\right)}^{2}+{\left(1-2\right)}^{2}+{\left(2-3\right)}^{2}}$
$=\sqrt{{\left(2\right)}^{2}+{\left(-1\right)}^{2}+{\left(-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+1+1}\phantom{\rule{0ex}{0ex}}=\sqrt{6}$
Now, AB = BC = AC

Therefore, it is an equilateral triangle.

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