Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 12 - Introduction To Three Dimensional Geometry

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Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 12 Introduction To Three Dimensional Geometry are provided here with simple step-by-step explanations. These solutions for Introduction To Three Dimensional Geometry are extremely popular among class 11 Science students for Maths Introduction To Three Dimensional Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 11 Science Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 220:

Question 1:

Locate the following points:
(i) (1, –1, 3)
(ii) (–1, 2, 4)
(iii) (–2, –4, –7)
(iv) (–4, 2, –5)

Answer:

The location of the points A(1, –1, 3), B(–1, 2, 4), C(–2, –4, –7) and D(–4, 2, –5) in the plane is as follows:
Mark the axes in capital X, X', Y, Y', Z, Z'

Page No 220:

Question 2:

Name the octant in which each of the following points lies.
(i) (1, 2, 3)
(ii) (4, –2, 3)
(iii) (4, –2, –5)
(iv) (4, 2, –5)
(v) (–4, 2, 5)
(vi) (–3, –1, 6)
(vii) (2, –4, –7)
(viii) (–4, 2, –5)

Answer:

(i) 1st Octant
(ii) 4th Octant
(iii) 8th Octant
(iv) 5th Octant
(v) 2nd Octant
(vi) 3rd Octant
(vii) 8th Octant
(viii) 6th Octant

Page No 220:

Question 3:

Let A, B, C be the feet of perpendiculars from a point P on the x, y, z-axis respectively. Find the coordinates of A, B and C in each of the following where the point P is :
(i) (3, 4, 2)
(ii) (–5, 3, 7)
(iii) (4, –3, –5)

Answer:

(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)

Page No 221:

Question 4:

Let A, B, C be the feet of perpendiculars from a point P on the xy, yz and zx-planes respectively. Find the coordinates of A, B, C in each of the following where the point P is
(i) (3, 4, 5)
(ii) (–5, 3, 7)
(iii) (4, –3, –5)

Answer:

(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)

Page No 221:

Question 5:

How far apart are the points (2, 0, 0) and (–3, 0, 0)?

Answer:

Given that, the two points are A(2, 0, 0) and B(–3, 0, 0).
Distance between two points $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
$= \sqrt{{(2 + 3)}^{2} + {(0 - 0)}^{2} + {(0 - 0)}^{2}} = \sqrt{25}$
= 5 (âˆµ distance cannot be negative)
Hence, the distance between the two points is 5 units.

Page No 221:

Question 6:

Find the distance from the origin to (6, 6, 7).

Answer:

The two given points are O(0, 0, 0) and P(6, 6, 7).
Distance between two points = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
$= \sqrt{{(6 - 0)}^{2} + {(6 - 0)}^{2} + {(7 - 0)}^{2}} = \sqrt{36 + 36 + 49} = \sqrt{121}$
= 11 units (âˆµ distance cannot be negative)
Hence, the distance between the two points is 11 units.

Page No 221:

Question 7:

Show that if x² + y² = 1, then the point $(x, y, \sqrt{1 - x^{2} - y^{2}})$ is at a distance 1 unit from the origin.

Answer:

Given that, x²+ y²= 1
Consider two points O(0, 0, 0) and $A (x, y, \sqrt{1 - x^{2} - y^{2}}) .$
Distance between two points $= \sqrt{{(x - 0)}^{2} + {(y - 0)}^{2} + {(\sqrt{1 - x^{2} - y^{2}} - 0)}^{2}}$
$= \sqrt{x^{2} + y^{2} + 1 - x^{2} - y^{2}} = \sqrt{1}$
= 1 unit
Hence, the distance between them is 1 unit.

Page No 221:

Question 8:

Show that the point A(1, –1, 3), B(2, –4, 5) and (5, –13, 11) are collinear.

Answer:

Given: A(1, –1, 3)
B(2, –4, 5)
C(5, –13, 11)
Now, Distance between two points $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
AB = $\sqrt{{(2 - 1)}^{2} + {(- 4 + 1)}^{2} + {(5 - 3)}^{2}}$

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

= $\sqrt{14}$ units
BC $= \sqrt{{(5 - 2)}^{2} + {(- 13 + 4)}^{2} + {(11 - 5)}^{2}}$
$= \sqrt{9 + 81 + 36} = \sqrt{126} = 3 \sqrt{14} units$
$AC = \sqrt{{(5 - 1)}^{2} + {(- 13 + 1)}^{2} + {(11 - 3)}^{2}}$

$= \sqrt{16 + 144 + 64} = \sqrt{24} = 4 \sqrt{14} units$
Thus,
AB + BC = AC
Hence, the three points are collinear.

Page No 221:

Question 9:

Three consecutive vertices of a parallelogram ABCD are A(6, –2, 4), B(2, 4, –8), C(–2, 2, 4). Find the coordinates of the fourth vertex.

Answer:

Given that, ABCD is a parallelogram with vertices as A(6, –2, 4), B(2, 4, –8) and C(–2, 2, 4).
Let the fourth vertex be D(x, y, z).
Now,
Midpoint of AC $= (\frac{6 - 2}{2}, \frac{- 2 + 2}{2}, \frac{4 + 4}{2})$
= (2, 0, 4)
Midpoint of BD = $(\frac{x + 2}{2}, \frac{y + 4}{2}, \frac{z - 8}{2})$
Since the diagonals of a parallelogram bisect each other,
(2, 0, 4) = $(\frac{x + 2}{2}, \frac{y + 4}{2}, \frac{z - 8}{2})$
$\Rightarrow \frac{x + 2}{2} = 2 \Rightarrow x = 2 \frac{y + 4}{2} = 0 \Rightarrow y = - 4 \frac{z - 8}{2} = 4 \Rightarrow z = 16$

Thus, the coordinates of the fourth vertex are: D(2, –4, 16).

Page No 221:

Question 10:

Show that the triangle ABC with vertices A(0, 4, 1), B(2, 3, –1) and C(4, 5, 0) is right angled.

Answer:

Given that, the three vertices of a triangle are A(0, 4, 1), B(2, 3, –1) and C(4, 5, 0).
Distance between two points $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
AB = $\sqrt{{(2 - 0)}^{2} + {(3 - 4)}^{2} + {(- 1 - 1)}^{2}}$
$= \sqrt{4 + 1 + 4}$
= 3 units
AC = $\sqrt{{(4 - 0)}^{2} + {(5 - 4)}^{2} + {(0 - 1)}^{2}}$
$= \sqrt{16 + 1 + 1} = 3 \sqrt{2} units$
Now,
AC²= 9 × 2 = 18 square units
AB²+ BC²= AC²
Hence, Δ ABC is a right-angled triangle.

Page No 221:

Question 11:

Find the third vertex of triangle whose centroid is origin and two vertices are (2, 4, 6) and (0, –2, –5).

Answer:

Given that, the centroid of a triangle with vertices A(2, 4, 6) and B(0, –2, –5) is the origin O(0, 0, 0).
Thus, (0, 0, 0) = $(\frac{2 + 0 + x}{3}, \frac{4 - 2 + y}{3}, \frac{6 - 5 + z}{3})$
$= (\frac{2 + x}{3}, \frac{2 + y}{3}, \frac{1 + z}{3})$
$\Rightarrow \frac{2 + x}{3} = 0 \Rightarrow x = - 2 \frac{2 + y}{3} = 0 \Rightarrow y = - 2 \frac{1 + z}{3} = 0 \Rightarrow z = - 1$
Hence, the third vertex of the triangle has coordinates (–2, –2, –1).

Page No 221:

Question 12:

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

Answer:

Given that, the midpoints of three sides of a triangle are D(1, 2, –3), E(3, 0, 1) and F(–1, 1, –4).
Now, $\frac{x_{1} + x_{2}}{2} = 1$
⇒ x₁+ x₂= 2 .....(1)
y₁+ y₂= 4 .....(2)
z₁+ z₂= –6 .....(3)

Similarly,
x₂ + x₃= –2 .....(4)
y₂ + y₃= 2 .....(5)
z₂ + z₃= –8 .....(6)
x₁ + x₃= 6 .....(7)
y₁ + y₃= 0 .....(8)
z₁ + z₃= 2 .....(9)
Using (1), (4) and (7),
2(x₁+ x₂+ x₃) = 2 – 2 + 6
⇒ x₁+ x₂+x₃= 3
Similarly, y₁+ y₂+ y₃= $\frac{6}{2} = 3$
z₁ + z₂+z₃= –6
Now, the centroid of a triangle is given as: $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$
s $= (\frac{3}{3}, \frac{3}{3}, \frac{- 6}{3})$
= (1, 1, –2)
Hence, the coordinates of the centroid are (1, 1, –2).

Page No 221:

Question 13:

The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, –1). Find its vertices.

Answer:

Given that, the midpoints of the sides of a triangle are P(5, 7, 11), Q(0, 8, 5) and R(2, 3, –1).
Let the vertices of the coordinates be A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃).

Using midpoint formula,
x₁+ x₂= 10 .....(1)
y₁+ y₂= 14 .....(2)
z₁ + z₂= 22 .....(3)
x₂ + x₃= 4 .....(4)
y₂ + y₃= 6 .....(5)
z₂ + z₃= –2 .....(6)
x₁ + x₃= 0 .....(7)
y₁ + y₃= 16 .....(8)
z₁ + z₃= 10 .....(9)
Solving (1), (4) and (7),
x₂= 3 ⇒ x₁= 7 and x₃= –7
Solving (2), (5) and (8)
y₂ = 12
⇒ y₁ = 2 and y₃ = 4
Solving (3), (6) and (9),
z₂= 17
⇒ z₁ = 5 and z₃ = –7
Hence, the required vertices are A(7, 2, 5), B(3, 12, 17) and C(–3, 4, –7).

Page No 221:

Question 14:

Three vertices of a Parallelogram ABCD are A(1, 2, 3), B(–1, –2, –1) and C(2, 3, 2). Find the fourth vertex D.

Answer:

Given that, vertices of a parallelogram are A(1, 2, 3), B(–1, –2, –1) and C(2, 3, 2).
Let the fourth vertex be D(x, y, z) .
Now,
Midpoint of AC = $(\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2})$

$= (\frac{3}{2}, \frac{5}{2}, \frac{5}{2})$

Midpoint of BD = $(\frac{x - 1}{2}, \frac{y - 2}{2}, \frac{z - 1}{2})$
The diagonals of a parallelogram bisect each other.
Therefore,
$\frac{3}{2} = \frac{x - 1}{2}$
⇒ x – 1 = 3
⇒ x = 4
$\frac{y - 2}{2} = \frac{5}{2}$
⇒ y – 2 = 5
⇒ y = 7
and $\frac{z - 1}{2} = \frac{5}{2}$
⇒ z = 6
Hence, the fourth vertex of parallelogram is D(4, 7, 6).

Page No 221:

Question 15:

Find the coordinate of the points which trisect the line segment joining the points A(2, 1, –3) and B(5, –8, 3).

Answer:

Given that, the line segment AB is trisected with A(2, 1, –3) and B(5, –8, 3).
Let point O(x₁, y₁,z₁) divide AB in the ratio 1 : 2.

Thus,
$x_{1} = \frac{1 \times 5 + 2 \times 2}{1 + 2} = \frac{5 + 4}{3} = 3 y_{1} = \frac{1 \times (- 8) + 2 \times 1}{1 + 2} = \frac{- 6}{3} = - 2 z_{1} = \frac{1 \times 3 + 2 \times (- 3)}{1 + 2} = \frac{- 3}{3} = - 1$
Thus, O(3, –2, –1) is one of the points.
Similarly, let P(x₂, y₂, z₂) be the point that divides AB in 2 : 1.
Thus,

$x_{2} = \frac{2 \times 5 + 1 \times 2}{2 + 1} = 4 y_{2} = \frac{2 \times (- 8) + 1 \times 1}{2 + 1} = - 5 z_{2} = \frac{2 \times 3 + 1 \times (- 3)}{2 + 1} = 1$
Thus, P(4, –5, 1).
Hence, the two points that trisect AB are O(3, –2, –1) and P(4, –5, 1).

Page No 221:

Question 16:

If the origin is the centriod 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.

Answer:

Given that, the triangle with vertices A(a, 1, 3), B(–2, b, –5) and C(4, 7, c) has centroid as the origin.
The coordinates of the centroid are given as:
$(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3}) \Rightarrow (0, 0, 0) = (\frac{a - 2 + 4}{3}, \frac{1 + b + 7}{3}, \frac{3 - 5 + c}{3}) = (\frac{a + 2}{3}, \frac{b + 8}{3}, \frac{c - 2}{3})$

$\Rightarrow \frac{a + 2}{3} = 0 \Rightarrow a = - 2 \frac{b + 8}{3} = 0 \Rightarrow b = - 8 \frac{c - 2}{3} = 0 \Rightarrow c = 2$
Hence, a = –2, b = –8, c = 2.

Page No 221:

Question 17:

Let A(2, 2, –3), B(5, 6, 9) and C(2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D.

Answer:

Given that, the vertices of the triangle are A(2, 2, –3), B(5, 6, 9) and C(2, 7, 0).
Using the distance formula,
AB = $\sqrt{{(5 - 2)}^{2} + {(6 - 2)}^{2} + {(9 + 3)}^{2}}$
$= \sqrt{9 + 16 + 144}$
= 13 units
AC = $\sqrt{{(2 - 2)}^{2} + {(7 - 2)}^{2} + {(9 + 3)}^{2}}$
$= \sqrt{25 + 144}$
= 13 units
Thus, ΔABC is an isosceles triangle.
Therefore, the bisector of ∠A meets BC at its midpoint.
$\Rightarrow D (\frac{5 + 2}{2}, \frac{6 + 7}{2}, \frac{9 + 9}{2}) = D (\frac{7}{2}, \frac{15}{2}, 9)$
Hence, the coordinates of D are $(\frac{7}{2}, \frac{13}{2}, 9) .$

Page No 221:

Question 18:

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.

Answer:

Given that, A(2, 3, 4), B(–1, 2, –3) and C(–4, 1, –10).
Using the distance formula,

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

$= \sqrt{9 + 1 + 49} = \sqrt{59} units$

$BC = \sqrt{{(- 4 + 1)}^{2} + {(1 - 2)}^{2} + {(- 10 + 3)}^{2}}$

$= \sqrt{9 + 1 + 49} = \sqrt{59} units$

$\begin{array}{rcl} A C & = & \sqrt{{(2 + 4)}^{2} + {(3 - 1)}^{2} + {(4 + 10)}^{2}} \\ = & \sqrt{36 + 4 + 196} \\ = & \sqrt{236} \\ = & 2 \sqrt{59} units \end{array}$
⇒AB + BC = AC
Thus, the three given points are collinear. Now, it is given that C divides AB. Let the ratio in which C divides AB be m : n.
Fig.
Using the section formula,
$- 4 = \frac{- m + 2 n}{m + n}$
⇒ – 4m – 4n = – m + 2n
⇒ – 6n = 3m
⇒ m = – 2n i.e. divides externally
⇒ m : n = 2 : 1
Hence, C divides AB externally in the ratio 2 : 1.

Page No 222:

Question 19:

The mid-point of the sides of a triangle are (1, 5, –1), (0, 4, –2) and (2, 3, 4). Find its vertices. Also find the centriod of the triangle.

Answer:

Given that, the midpoints of the sides of a triangle are P(1, 5, –1), Q(0, 4, –2) and R(2, 3, 4).
Let the vertices of the triangle be
A(x, y_1,z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃).

Using the midpoint theorem,
x₁+ x₂= 2
x₂+ x₃= 4
x₃+ x₁= 0
⇒ x₂= –1
x₁= 3
x₃= – 3
Again,
y₁+ y₂= 10
y₂+ y₃= 6
y₃+ y₁= 8
⇒ y₂= 6
y₁= 4 and y₃ = 4
Also,
z₁ + z₂= –2
z₂ + z₃= 8
z₃ + z₁= –4
⇒ z₂= –7
z₁= 5 and z₃= 3
Therefore, the vertices of the given triangle are A(–1, 6, –7), B(3, 4, 5) and C(1, 2, 3). Now the centroid of the triangle is given as:
$(\frac{- 1 + 3 + 1}{3}, \frac{6 + 4 + 2}{3}, \frac{- 7 + 5 + 3}{3}) = (1, 4, \frac{1}{3})$

Page No 222:

Question 20:

Prove that the points (0, –1, –7), (2, 1, –9) and (6, 5, –13) are collinear. Find the ratio in which the first point divides the join of the other two.

Answer:

Let the three points be A(0, –1, –7), B(2, 1, –9) and C(6, 5, –13).
Using the distance formula,

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

      $= \sqrt{4 + 4 + 4} = 2 \sqrt{3} units$

BC = $\sqrt{{(6 - 2)}^{2} + {(5 - 1)}^{2} + {(- 13 + 9)}^{2}}$

       $= \sqrt{16 + 16 + 16} = 4 \sqrt{3} units$

AC = $\sqrt{{(6 - 0)}^{2} + {(5 + 1)}^{2} + {(- 13 + \ 72)}^{2}}$

       $= \sqrt{36 + 36 + 36} = 6 \sqrt{3} units$

⇒ AB + BC = AC
Hence, the three points are collinear.

Now, AB : AC = $2 \sqrt{3} : 6 \sqrt{3}$
                        = 1 : 3
Hence, the ratio in which the first point divides the join of the other two is 1 : 3 externally.

Page No 222:

Question 21:

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

Answer:

Given that, the cube has edge of 2 units.
Thus, the coordinates of its vertices are:
(0, 0, 0), (2, 0, 0), (2, 2, 0), (0, 2, 0), (0, 2, 2), (2, 2, 2), (2, 0, 2) and (0, 0, 2).

Page No 222:

Question 22:

Choose the correct answer from the given four options:
The distance of point P(3, 4, 5) from the yz-plane is
(A) 3 units
(B) 4 units
(C) 5 units
(D) 550

Answer:

In the yz-plane, a point is of the type (0, y, z. Therefore, the distance is determined by the x-coordinate of the point. Thus, the point P(3, 4, 5) is at a distance of 3 units from the yz-plane.
Hence, the correct answer is option A.

Page No 222:

Question 23:

Choose the correct answer from the given four options:
What is the length of foot of perpendicular drawn from the point P(3, 4, 5) on y-axis
(A) $\sqrt{41}$
(B) $\sqrt{34}$
(C) 5
(D) none of these

Answer:

Given that, a perpendicular is drawn from P(3, 4, 5) on 4-axis.

The length of the perpendicular = $\sqrt{{(3 - 0)}^{2} + {(4 - 4)}^{2} + {(5 - 0)}^{2}}$

$= \sqrt{9 + 25} = \sqrt{34} units$

Hence, the correct answer is option B.

Page No 222:

Question 24:

Choose the correct answer from the given four options:
Distance of the point (3, 4, 5) from the origin (0, 0, 0) is
(A) $\sqrt{50}$
(B) 3
(C) 4
(D) 5

Answer:

Let P(3, 4, 5) and O(0, 0, 0).
Using the distance formula,
$\begin{array}{rcl} O P & = & \sqrt{{(3 - 0)}^{2} + {(4 - 0)}^{2} + {(5 - 0)}^{2}} \\ = & \sqrt{9 + 16 + 25} \\ = & \sqrt{50} units \end{array}$

Hence, the correct answer is option (A).

Page No 222:

Question 25:

Choose the correct answer from the given four options:
If the distance between the points (a, 0, 1) and (0, 1, 2) is $\sqrt{27}$ , then the value of a is
(A) 5
(B) ± 5
(C) – 5
(D) none of these

Answer:

Let A(a, 0, 1) ans B(0, 1, 2).
Given that, the distance between them is $\sqrt{27}$ units.
Using the distance formula,
$\sqrt{27} = \sqrt{{(a - 0)}^{2} + {(0 - 1)}^{2} + {(1 - 2)}^{2}} \Rightarrow 27 = a^{2} + 1 + 1 \Rightarrow a^{2} = 25 \Rightarrow a = \pm 5$

Hence, the correct answer is option (B).

Page No 222:

Question 26:

Choose the correct answer from the given four options:
x-axis is the intersection of two planes
(A) xy and xz
(B) yz and zx
(C) xy and yz
(D) none of these

Answer:

x-axis is the intersection of two planes that X can form, i.e., xy and xz plane.

Hence, the correct answer is option (A).

Page No 222:

Question 27:

Choose the correct answer from the given four options:
Equation of y-axis is considered as
(A) x = 0, y = 0
(B) y = 0, z = 0
(C) z = 0, x = 0
(D) none of these

Answer:

On the y-axis, x = 0 and z = 0.

Hence, the correct answer is option (C).

Disclaimer: The answer in NCERT Exemplar is incorrect. The correct answer is as per the given solution.

Page No 222:

Question 28:

Choose the correct answer from the given four options:
The point (–2, –3, –4) lies in the
(A) First octant
(B) Seventh octant
(C) Second octant
(D) Eighth octant

Answer:

The point (–2, –3, –4) lies in the seventh octant.

Hence, the correct answer is option (B).

Page No 222:

Question 29:

Choose the correct answer from the given four options:
A plane is parallel to yz-plane so it is perpendicular to :
(A) x-axis
(B) y-axis
(C) z-axis
(D) none of these

Answer:

The yz-plane is perpendicular to the x-axis. Thus, a line parallel to the yz-plane is perpendicular to the x-axis.

Hence, the correct answer is option (A).

Page No 222:

Question 30:

Choose the correct answer from the given four options:
The locus of a point for which y = 0, z = 0 is
(A) equation of x-axis
(B) equation of y-axis
(C) equation at z-axis
(D) none of these

Answer:

The locus of a point for which y = 0 and z = 0 is the x-axis.

Hence, the correct answer is option (A).

Page No 223:

Question 31:

Choose the correct answer from the given four options:
The locus of a point for which x = 0 is
(A) xy-plane
(B) yz-plane
(C) zx-plane
(D) none of these

Answer:

The locus of a point for which x = 0 is the yz-plane.

Hence, the correct answer is option (B).

Page No 223:

Question 32:

Choose the correct answer from the given four options:
If a parallelopiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelopiped is
(A) $2 \sqrt{3}$
(B) $3 \sqrt{2}$
(C) $\sqrt{2}$
(D) $\sqrt{3}$

Answer:

The length of the diagonal of the parallelopiped is given by the distance between two points, A(5, 8, 10) and B(3, 6, 8).
Using the distance formula,
$\begin{array}{rcl} A B & = & \sqrt{{(5 - 3)}^{2} + {(8 - 6)}^{2} + {(10 - 8)}^{2}} \\ = & \sqrt{4 + 4 + 4} \\ = & 2 \sqrt{3} units \end{array}$

Hence, the correct answer is option (A).

Page No 223:

Question 33:

Choose the correct answer from the given four options:
L is the foot of the perpendicular drawn from a point P (3, 4, 5) on the xy-plane. The coordinates of point L are
(A) (3, 0, 0)
(B) (0, 4, 5)
(C) (3, 0, 5)
(D) none of these

Answer:

Given that, point P(3, 4, 5) has the foot of its perpendicular as L in the xy-plane.
Thus, L(3, 4, 0)

Hence, the correct answer is option (D).

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Question 34:

Choose the correct answer from the given four options:
L is the foot of the perpendicular drawn from a point (3, 4, 5) on x-axis. The coordinates of L are
(A) (3, 0, 0)
(B) (0, 4, 0)
(C) (0, 0, 5)
(D) none of these

Answer:

On the x-axis, the coordinates of the foot of the perpendicular L are (3, 0, 0).

Hence, the correct answer is option (A).

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Question 35:

Fill in the blank.
The three axes OX, OY, OZ determine ________ .

Answer:

three coordinate planes.

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Question 36:

Fill in the blank.
The three planes determine a rectangular parallelopiped which has ________ of rectangular faces.

Answer:

three pairs

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Question 37:

Fill in the blank.
The coordinates of a point are the perpendicular distance from the ________ on the respectives axes.

Answer:

given point

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Question 38:

Fill in the blank.
The three coordinate planes divide the space into ________ parts.

Answer:

eight

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Question 39:

Fill in the blank.
If a point P lies in yz-plane, then the coordinates of a point on yz-plane is of the form ________.

Answer:

(0, y, z)

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Question 40:

Fill in the blank.
The equation of yz-plane is ________.

Answer:

x = 0

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Question 41:

Fill in the blank.
If the point P lies on z-axis, then coordinates of P are of the form ________.

Answer:

(0, 0, z)

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Question 42:

Fill in the blank.
The equation of z-axis, are ________.

Answer:

x = 0 and y = 0

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Question 43:

Fill in the blank.
A line is parallel to xy-plane if all the points on the line have equal ________.

Answer:

z-coordinates

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Question 44:

Fill in the blank.
A line is parallel to x-axis if all the points on the line have equal ________.

Answer:

y and z-coordinates

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Question 45:

Fill in the blank.
x = a represent a plane parallel to ________.

Answer:

yz-plane

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Question 46:

Fill in the blank.
The plane parallel to yz - plane is perpendicular to ________.

Answer:

x-axis

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Question 47:

Fill in the blank.
The length of the longest piece of a string that can be stretched straight in a rectangular room whose dimensions are 10, 13 and 8 units are ______.

Answer:

Let l = 10, b = 13 and h = 8 of the rectangular room.
Considering the right angled triangle with sides 10 and 13 units,
$H_{1}^{2} = 100 + 169 = 269 \Rightarrow H_{1} = \sqrt{269} units$
Now, considering the right angled triangle with sides $\sqrt{269}$ and 8 units,
${H_{2}}^{2} = 269 + 64 = 333 \Rightarrow H_{2} = \sqrt{333} units$

Hence, the length of the largest string that can be stretched straight in a rectangular room whose dimensions are 10, 13 and 8 units is $\sqrt{333}$
units.

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Question 48:

Fill in the blank.
If the distance between the points (a, 2, 1) and (1, –1, 1) is 5, then a _______.

Answer:

Let A(a, 2, 1) and B(a, 2, 1) and and AB = 5 units.
Using the distance formula,
$\begin{array}{rcl} A B & = & \sqrt{{(1 - a)}^{2} + {(- 1 - 2)}^{2} + {(1 - 1)}^{2}} \\ = & \sqrt{1 + a^{2} - 2 a + 9 + 0} \end{array}$
$\Rightarrow 25 = a^{2} - 2 a + 10 \Rightarrow a^{2} - 2 a - 15 = 0 \Rightarrow a^{2} - 5 a + 3 a - 15 = 0 \Rightarrow a (a - 5) + 3 (a - 5) = 0 \Rightarrow a = - 3 and a = 5$

Hence, if the distance between two points (a, 2, 1) and (1, –1, 1) is 5, then a = –3 or 5.

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Question 49:

Fill in the blank.
If the mid-points of the sides of a triangle AB; BC; CA are D (1, 2, – 3), E (3, 0, 1) and F (–1, 1, – 4), then the centriod of the triangle ABC is ________.

Answer:

The centroid of $∆ A B C$ is the same as that of $∆ D E F .$
Therefore, the centroid O of $∆ D E F$ is given as.
$(\frac{1 + 3 - 1}{3}, \frac{2 + 0 + 1}{3}, \frac{- 3 + 1 - 4}{3}) = (1, 1, - 2)$

Hence, if the midpoints of the sides of a triangle AB, BC and CA are D(1, 2, –3), E(3, 0, 1) and F(–1, 1, –4), then the centroid of the $∆$ ABC is (1, 1, –2).

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Question 50:

Match each item given under the column C₁ to its correct answer given under column C₂.

	Column C₁		Column C₂
(a)	In xy-plane	(i)	I^st octant
(b)	Point (2, 3,4) lies in the	(ii)	yz-plane
(c)	Locus of the points having x coordinate 0 is	(iii)	z-coordinate is zero
(d)	A line is parallel to x-axis if and only	(iv)	z-axis
(e)	If x = 0, y = 0 taken together will represent the	(v)	plane parallel to xy-plane
(f)	z = c represent the plane	(vi)	if all the points on the line have equal y and z-coordinates.
(g)	Planes x = a, y = b represent the line	(vii)	from the point on the respective
(h)	Coordinates of a point are the distances from the origin to the feet of perpendiculars	(viii)	parallel to z - axis.
(i)	A ball is the solid region in the space enclosed by a	(ix)	disc
(j)	Region in the plane enclosed by a circle is known as a	(x)	sphere

Answer:

(a) In xy-plane, z-coordinate is zero.
(b) Point (2, 3, 4) lies in the 1st octant.
(c) Locus of the points having x-coordinates O is yz-plane.
(d) A line is parallel to x-axis if and only if all the points on the line have equal y and z-coordinates.
(e) If x = 0, y = 0 taken together will represent the z-axis.
(f) z = c represents the plane parallel to the xy-plane.
(g) Planes x = a, y = b represent the line parallel to z-axis.
(h) Coordinates of a point are the distances from the origin to the feet of perpendiculars from the point on the respective axes.
(i) A ball is the solid region in the space enclosed by a sphere.
(j) Region in the plane enclosed by a circle is known as a disc.
Thus,
(a) - (iii)
(b) - (i)
(c) - (ii)
(d) - (vi)
(e) - (iv)
(f) - (v)
(g) - (vii)
(h) - (vii)
(i) - (x)
(j) - (ix)

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