Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 9

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Question 1
1 / -0

Let $$A(3, 0, -1), B(2, 10, 6)$$ and $$C(1, 2, 1)$$ be the vertices of a triangle and $$M$$ be the midpoint of $$AC$$. If $$G$$ divides $$BM$$ in the ratio, $$2 : 1$$, then $$\cos (\angle GOA)$$ ($$O$$ being the origin) is equal to:

A
$$\dfrac{1}{\sqrt{30}}$$

B
$$\dfrac{1}{6\sqrt{10}}$$

C
$$\dfrac{1}{\sqrt{15}}$$

D
$$\dfrac{1}{2\sqrt{15}}$$

Solution

$$G$$ is the centroid of $$\Delta ABC$$
$$G \equiv (2, 4, 2)$$
$$\overrightarrow{OG} = 2\hat{i} + 4\hat{i} + 2\hat{k}$$
$$\overrightarrow{OA} = 3\hat{i} - \hat{k}$$

$$\cos (\angle GOA) = \dfrac{\overrightarrow{OG} \cdot \overrightarrow{OA}}{|\overrightarrow{OG}| |\overrightarrow{OA}|} = \dfrac{1}{\sqrt{15}}$$
Question 2
1 / -0

In the given figure, co-ordinates of the midpoint of $$AB$$ are

A
$$(0,2)$$

B
$$(0,3)$$

C
$$(1,2)$$

D
$$(3,1)$$

Solution

Here we are given two points A(0, 0) and B(6, 2)
Since the midpoint falls midway between A and
B, it has x-co-ordinates of A and B
x-co-ordinates of midpoint of $$\displaystyle AB=\frac{0+6}{2}=3$$
and y-co-ordinates of Midpoint of $$\displaystyle AB=\frac{0+2}{2}=1$$
Thus the midpoint of AB has co-ordinates (3, 1)
Question 3
1 / -0

Name three undefined terms.

A
Point

B
Line

C
Plane

D
All of the above

Solution

The basic undefined term is point.
Line is formed from points and plane is formed from many lines.
Undefined terms are point, line and plane.
Question 4
1 / -0

Distance between $$A(4,5,6)$$ from origin $$O$$ is

A
$$25\sqrt3$$

B
$$\sqrt {77}$$

C
$$3\sqrt5$$

D
Data Insufficient

Solution

Origin is $$O(0,0,0)$$ and given point is $$A(4,5,6)$$
So, distance $$=$$ $$\sqrt {(4-0)^2+(5-0)^2+(6-0)^2}$$
$$=\sqrt {4^2 + 5^2 + 6^2} = \sqrt {77}$$
Question 5
1 / -0

Distance between the points $$(12,4,7)$$ and $$(10,5,3)$$ is

A
$$\sqrt{21}$$

B
$$\sqrt{5}$$

C
$$\sqrt{17}$$

D
none of these

Solution

Consider the problem,
Let the given points
$$A(12,4,7)$$ and $$B(10,5,3)$$
So, distance between $$A$$ and $$B$$ by distance formula.
$$AB=\sqrt{(10-12)^2+(5-4)^2+(3-7)^2}=\sqrt{(-2)^2+1^2+(-4)^2}$$
$$=\sqrt{4+1+16}=\sqrt{21}$$
So, distance between the points $$(12,4,7)$$ and $$(10,5,3)$$ is $$\sqrt{21}$$ sq. units.
Question 6
1 / -0

The three planes divides the space into

A
four parts

B
six parts

C
eight parts

D
sixteen parts

Solution

Three planes divides the space into eight regions.
Question 7
1 / -0

A point on XOZ-plane divides the join of $$(5, -3, -2)$$ and $$(1, 2, -2)$$ at

A
$$\left(\displaystyle\frac{13}{5}, 0, -2\right)$$

B
$$\left(\displaystyle\frac{13}{5}, 0, 2\right)$$

C
$$(5, 0, 2)$$

D
$$(5, 0, -2)$$

Solution

$$\begin{array}{l} The\, \, line\, \, through\, \, the\, \, po{ { int } }\, \, \left( { 5,-3,-2 } \right) \\ and\left( { 1,2,-2 } \right) \\ \vec { r } =\hat { i } +2\hat { j } -2\hat { k } +\lambda \left( { 4\hat { i } -5\hat { j } } \right) \\ \vec { r } =\left( { 1+4\lambda } \right) \hat { i } +\left( { 2-5\lambda } \right) \hat { j } -2\vec { k } \\ equation\, \, of\, \, XOZ\, \, plane \\ y=0 \\ \therefore 2-5\lambda =0 \\ \therefore \lambda =\dfrac { 2 }{ 5 } \\ \therefore \vec { r } =\dfrac { { 13 } }{ 5 } \hat { i } -2\hat { k } \\ \therefore \left( { \dfrac { { 13 } }{ 5 } ,0,-2 } \right) \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$$
Question 8
1 / -0

In three dimensions, the coordinate axes of a rectangular cartesian coordinate system are

A
three mutually parallel lines

B
three mutually perpendicular lines

C
two mutually perpendicular lines and any two parallel

D
None of these

Solution

In three dimensions, the coordinate axes, i.e. x, y and z axes of a rectangular cartesian coordinate system are three mutually perpendicular lines.
The word rectangular is used to indicate perpendicularity among the axes.
Question 9
1 / -0

If G is centroid of $$\triangle ABC$$, then

A
$$\overrightarrow {G}=\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow{c}$$

B
$$\displaystyle \overrightarrow {G} = \frac {\overrightarrow{a} + \overrightarrow {b} +\overrightarrow{c}}{2}$$

C
$$3\overrightarrow {G}=\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow {c}$$

D
$$\displaystyle 3\overrightarrow {G}=\frac{\overrightarrow {a}+\overrightarrow{b}+\overrightarrow{c}}{2}$$

Solution

We have,

In a $$\Delta ABC$$

Let

$$ \overrightarrow{A}=\overrightarrow{a} $$

$$ \overrightarrow{B}=\overrightarrow{b} $$

$$ \overrightarrow{C}=\overrightarrow{c} $$

Then,

We know that,

$$ G\,=\dfrac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3} $$

$$ 3\overrightarrow{G}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} $$

Hence, this is the answer.
Question 10
1 / -0

Find the distance between $$(12,3,4)$$ and $$(4,5,2)$$

A
$$\sqrt {72}$$

B
$$\sqrt {62}$$

C
$$\sqrt {64}$$

D
None of these

Solution

Consider the problem,
Let the given points
$$A(12,3,4)$$ and $$B(4,5,2)$$
So, distance between $$A$$ and $$B$$ by distance formula.
$$AB=\sqrt{(4-12)^2+(5-3)^2+(2-4)^2}=\sqrt{(-8)^2+2^2+(-2)^2}$$
$$=\sqrt{64+4+4}=\sqrt{72}$$
So, distance between the points $$(12,3,4)$$ and $$(4,5,2)$$ is $$\sqrt{72}$$ sq. units.

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Introduction to Three Dimensional Geometry Test - 9

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Introduction to Three Dimensional Geometry Test - 9

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Question 1

$$\dfrac{1}{\sqrt{30}}$$

$$\dfrac{1}{6\sqrt{10}}$$

$$\dfrac{1}{\sqrt{15}}$$

$$\dfrac{1}{2\sqrt{15}}$$

Solution

Question 2

$$(0,2)$$

$$(0,3)$$

$$(1,2)$$

$$(3,1)$$

Solution

Question 3

Point

Line

Plane

All of the above

Solution

Question 4

$$25\sqrt3$$

$$\sqrt {77}$$

$$3\sqrt5$$

Data Insufficient

Solution

Question 5

$$\sqrt{21}$$

$$\sqrt{5}$$

$$\sqrt{17}$$

none of these

Solution

Question 6

four parts

six parts

eight parts

sixteen parts

Solution

Question 7

$$\left(\displaystyle\frac{13}{5}, 0, -2\right)$$

$$\left(\displaystyle\frac{13}{5}, 0, 2\right)$$

$$(5, 0, 2)$$

$$(5, 0, -2)$$

Solution

Question 8

three mutually parallel lines

three mutually perpendicular lines

two mutually perpendicular lines and any two parallel

None of these

Solution

Question 9

$$\overrightarrow {G}=\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow{c}$$

$$\displaystyle \overrightarrow {G} = \frac {\overrightarrow{a} + \overrightarrow {b} +\overrightarrow{c}}{2}$$

$$3\overrightarrow {G}=\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow {c}$$

$$\displaystyle 3\overrightarrow {G}=\frac{\overrightarrow {a}+\overrightarrow{b}+\overrightarrow{c}}{2}$$

Solution

Question 10

$$\sqrt {72}$$

$$\sqrt {62}$$

$$\sqrt {64}$$

None of these

Solution

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