Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 37

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Question 1
1 / -0

$$D(2, 1, 0), E(2, 0, 0), F(0, 1, 0)$$ are mid point of the sides $$BC, CA, AB$$ of $$\Delta$$ $$ABC$$ respectively, The the centroid of $$\Delta$$ABC is

A
$$\left ( \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$

B
$$\left ( \displaystyle \frac{4}{3},\, \displaystyle \frac{2}{3},\, 0 \right )$$

C
$$\left (- \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$

D
$$\left ( \displaystyle \frac{2}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$

Solution

Centroid of triangle coincide with the centroid of triangle formed by joing the mid-point of sides of triangle
So, centroid of $$\triangle ABC =$$ centroid of $$\triangle DEF$$
$$=\left(\dfrac{2+2+0}{3},\dfrac{1+0+1}{3},\dfrac{0+0+0}{3}\right)$$

$$=\left(\dfrac{4}{3},\dfrac{2}{3},0\right)$$
Question 2
1 / -0

If $$P(x,y,x)$$ is a point on the line segment joining $$Q(2,2,4)$$ and $$R(3,5,6)$$ such that the projection of $$\overline { OP } $$ on the axis are $$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 } ,\frac { 26 }{ 5 } ,$$ respectively, then $$P$$ divides $$QR$$ in the ratio

A
$$1:2$$

B
$$3:2$$

C
$$2:3$$

D
$$1:3$$

Solution

Since $$OP$$ has projection $$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 } $$ and $$\displaystyle \frac { 26 }{ 5 } $$ on the coordinate axis,

therefore $$\displaystyle OP=\frac { 13 }{ 5 }\vec i+\frac { 19 }{ 5 } \vec j+\frac { 26 }{ 5 } \vec k$$.
Suppose $$P$$ divides the join of $$Q(2,2,4)$$ and $$R(3,5,6)$$ in the ratio $$\lambda :1$$.
Then, the position vector of $$P$$ is
$$\displaystyle \left( \frac { 3\lambda +2 }{ \lambda +1 } \right)\vec i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \vec j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \vec k$$

$$\displaystyle \therefore \frac { 13 }{ 5 } \vec i+\frac { 19 }{ 5 }\vec j+\frac { 26 }{ 5 }\vec k=\left( \frac { 3\lambda +2 }{ \lambda +1 } \right) \vec i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \vec j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \vec k$$

$$\displaystyle \Rightarrow \frac { 3\lambda +2 }{ \lambda +1 } =\frac { 13 }{ 5 } ;\frac { 5\lambda +2 }{ \lambda +1 } =\frac { 19 }{ 5 } ;\frac { 6\lambda +4 }{ \lambda +1 } =\frac { 26 }{ 5 } $$

$$\displaystyle \Rightarrow 2\lambda =3$$
$$\Rightarrow \lambda =\dfrac { 3 }{ 2 } $$
Question 3
1 / -0

There are three points with position vectors $$ -2a+3b+5c, a+2b+3c $$ and$$ 7a-c$$. What is the relation between the three points?

A
Collinear

B
Forms a triangle

C
In different plane

D
None of the above

Solution

The relation between the three points are collinear
Thus option A is correct answer
Question 4
1 / -0

The plane $$ax+by+cz+d=0$$ divides the line joining the points $$\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 } \right) $$ in the ratio

A
$$\displaystyle \frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) } $$

B
$$\displaystyle \frac { \left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) } $$

C
$$\displaystyle \frac { a{ x }_{ 1 }{ x }_{ 2 }+b{ y }_{ 1 }{ y }_{ 2 }+c{ z }_{ 1 }{ z }_{ 2 } }{ { d }^{ 2 } } $$

D
None of these

Solution

Let the given plane meet the line joining the given points in $$\left( { x }_{ 3 },{ y }_{ 3 },{ z }_{ 3 } \right) $$.
Then $$a{ x }_{ 3 }+b{ y }_{ 3 }+c{ z }_{ 3 }+d=0$$ ...(1)
Also let the point $$\left( { x }_{ 3 },{ y }_{ 3 },{ z }_{ 3 } \right) $$ divide the line joining the given points in the ratio $$m:n$$
Then
$$\displaystyle { x }_{ 3 }=\frac { m{ x }_{ 1 }+n{ x }_{ 2 } }{ m+n } ,{ y }_{ 3 }=\frac { m{ y }_{ 1 }+n{ y }_{ 2 } }{ m+n } $$ and $$\displaystyle { z }_{ 3 }=\frac { m{ z }_{ 1 }+n{ z }_{ 2 } }{ m+n } $$

Substituting these values in (1), we get
$$\displaystyle a\left( \frac { m{ x }_{ 1 }+n{ x }_{ 2 } }{ m+n } \right) +b\left( \frac { m{ y }_{ 1 }+n{ y }_{ 2 } }{ m+n } \right) +c\left( \frac { m{ z }_{ 1 }+n{ z }_{ 2 } }{ m+n } \right) +d=0$$

$$\Rightarrow a\left( m{ x }_{ 1 }+n{ x }_{ 2 } \right) +b\left( m{ y }_{ 1 }+n{ y }_{ 2 } \right) +c\left( m{ z }_{ 1 }+n{ z }_{ 2 } \right) +d\left( m+n \right) =0$$
$$\Rightarrow m\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) +n\left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) =0$$

$$\displaystyle\Rightarrow \frac { n }{ m } =\frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) }$$

Hence, option A.
Question 5
1 / -0

Let $$\displaystyle a,b,c \epsilon R$$ such that $$abc = p$$ and $$qa-b = 0$$, where $$ p$$ and $$q$$ are fixed positive number, then minimum distance of the point $$(a, b, c)$$ from origin in the three dimensional coordinate system is:

A
$$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{2q} \right )^{1/3}$$

B
$$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{q} \right )^{1/3}$$

C
$$\displaystyle \sqrt{3}(p)^{1/3}$$

D
$$\displaystyle \sqrt{2}\left ( \frac{p}{q} \right )^{1/2}$$

Solution

We know that the distance between point $$(a,b,c)$$ and origin $$(0,0,0)$$ is:
$$d= \sqrt({a^2+b^2+c^2)}$$.
Using the inequality: $$AM\geq GM$$, for terms $$a^2, b^2$$ and $$c^2$$:

$$\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge \sqrt [ 3 ]{ { (abc) }^{ 2 } } $$.
$$abc = p$$ (Given)
$$\implies \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge { (p) }^{ \frac{2}{3} } $$.

Taking square root on both sides of the inequality:

$$ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } \ge \sqrt { 3 } { p }^{ 1/3 }$$.

Hence, Option C is correct.
Question 6
1 / -0

The set of points in space 4 inches from a given line and 4 inches from a given point on this line is ______ , if given point lies on the given line.

A
a set consisting of two points

B
a set consisting of four points

C
a set consisting of two circles

D
the empty set
Question 7
1 / -0

If $$P\left( x,y,z \right) $$ is a point on the line segment joining $$Q\left( 2,2,4 \right) $$ and $$R\left( 3,5,6 \right) $$ such that the projections of $$OP$$ on the axis are $$\cfrac { 13 }{ 5 } ,\cfrac { 19 }{ 5 } ,\cfrac { 26 }{ 5 } $$ respectively, then $$P$$ divides $$QR$$ in the ratio

A
$$1:2$$

B
$$3:2$$

C
$$2:3$$

D
$$1:3$$

Solution

Since, $$\overline { OP } $$ has projections $$\cfrac { 13 }{ 5 } ,\cfrac { 19 }{ 5 } ,\cfrac { 26 }{ 5 } $$ on the coordinate axes
$$\therefore \overline { OP } =\cfrac { 13 }{ 5 } \hat { i } +\cfrac { 19 }{ 5 } \hat { j } +\cfrac { 26 }{ 5 } \hat { k } $$
Suppose $$P$$ divides the line segment joining of $$Q(2,2,4)$$ and $$R(3,5,6)$$ in the ratio $$\lambda:1$$
then the position vector of $$P$$ is
$$\left( \cfrac { 3\lambda +2 }{ \lambda +1 } \right) \hat { i } +\left( \cfrac { 5\lambda +2 }{ \lambda +1 } \right) \hat { j } +\left( \cfrac { 6\lambda +4 }{ \lambda +1 } \right) \hat { k } $$
$$\therefore \cfrac { 13 }{ 5 } \hat { i } +\cfrac { 19 }{ 5 } \hat { j } +\cfrac { 26 }{ 5 } \hat { k } =\left( \cfrac { 3\lambda +2 }{ \lambda +1 } \right) \hat { i } +\left( \cfrac { 5\lambda +2 }{ \lambda +1 } \right) \hat { j } +\left( \cfrac { 6\lambda +4 }{ \lambda +1 } \right) \hat { k } $$
$$\Rightarrow \lambda =3/2$$
$$\therefore$$ Required ratio in which $$P$$ divides $$QR$$ is $$3:2$$
Question 8
1 / -0

The points $$A(1,2,3); B-(-1,-2,-1); C(2,3,2)$$ and $$D(4,7,6)$$ form

A
Square

B
Rectangle

C
Parallelogram

D
Rhombus

Solution
Question 9
1 / -0

30 consider at three dimensional figure represented by $$xy{z^2} = 2$$, then its minimum distance from origin is

A
2

B
4

C
3

D
1

Solution
Question 10
1 / -0

The distances of the point $$P(1,2,3)$$ from the coordinates axes are:

A
$$\sqrt {13} ,\sqrt {10} ,\sqrt 5 $$

B
$$\sqrt {11} ,\sqrt {10} ,\sqrt 5 $$

C
$$\sqrt {13} ,\sqrt {20} ,\sqrt {15} $$

D
$$\sqrt {23} ,\sqrt {10} ,\sqrt 5 $$

Solution

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Introduction to Three Dimensional Geometry Test - 37

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Introduction to Three Dimensional Geometry Test - 37

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Question 1

$$\left ( \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$

$$\left ( \displaystyle \frac{4}{3},\, \displaystyle \frac{2}{3},\, 0 \right )$$

$$\left (- \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$

$$\left ( \displaystyle \frac{2}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$

Solution

Question 2

$$1:2$$

$$3:2$$

$$2:3$$

$$1:3$$

Solution

Question 3

Collinear

Forms a triangle

In different plane

None of the above

Solution

Question 4

$$\displaystyle \frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) } $$

$$\displaystyle \frac { \left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) } $$

$$\displaystyle \frac { a{ x }_{ 1 }{ x }_{ 2 }+b{ y }_{ 1 }{ y }_{ 2 }+c{ z }_{ 1 }{ z }_{ 2 } }{ { d }^{ 2 } } $$

None of these

Solution

Question 5

$$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{2q} \right )^{1/3}$$

$$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{q} \right )^{1/3}$$

$$\displaystyle \sqrt{3}(p)^{1/3}$$

$$\displaystyle \sqrt{2}\left ( \frac{p}{q} \right )^{1/2}$$

Solution

Question 6

a set consisting of two points

a set consisting of four points

a set consisting of two circles

the empty set

Question 7

$$1:2$$

$$3:2$$

$$2:3$$

$$1:3$$

Solution

Question 8

Square

Rectangle

Parallelogram

Rhombus

Solution

Question 9

2

4

3

1

Solution

Question 10

$$\sqrt {13} ,\sqrt {10} ,\sqrt 5 $$

$$\sqrt {11} ,\sqrt {10} ,\sqrt 5 $$

$$\sqrt {13} ,\sqrt {20} ,\sqrt {15} $$

$$\sqrt {23} ,\sqrt {10} ,\sqrt 5 $$

Solution

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