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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Question 1

(13, 13, 13)\left ( \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )(31​,31​,31​)

(43, 23, 0)\left ( \displaystyle \frac{4}{3},\, \displaystyle \frac{2}{3},\, 0 \right )(34​,32​,0)

(−13, 13, 13)\left (- \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )(−31​,31​,31​)

(23, 13, 13)\left ( \displaystyle \frac{2}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )(32​,31​,31​)

Solution

Question 2

1:21:21:2

3:23:23:2

2:32:32:3

1:31:31:3

Solution

Question 3

Collinear

Forms a triangle

In different plane

None of the above

Solution

Question 4

−(ax1+by1+cz1+d)(ax2+by2+cz2+d)\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) } (ax2​+by2​+cz2​+d)−(ax1​+by1​+cz1​+d)​

(ax1+by1+cz1+d)(ax2+by2+cz2+d)\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) } (ax2​+by2​+cz2​+d)(ax1​+by1​+cz1​+d)​

ax1x2+by1y2+cz1z2d2\displaystyle \frac { a{ x }_{ 1 }{ x }_{ 2 }+b{ y }_{ 1 }{ y }_{ 2 }+c{ z }_{ 1 }{ z }_{ 2 } }{ { d }^{ 2 } } d2ax1​x2​+by1​y2​+cz1​z2​​

None of these

Solution

Question 5

3(p(q2+1)2q)1/3\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{2q} \right )^{1/3}3​(2qp(q2+1)​)1/3

3(p(q2+1)q)1/3\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{q} \right )^{1/3}3​(qp(q2+1)​)1/3

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

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

1:21:21:2

3:23:23:2

2:32:32:3

1:31:31:3

Solution

Question 8

Square

Rectangle

Parallelogram

Rhombus

Solution

Question 9

2

4

3

1

Solution

Question 10

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

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

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

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

Solution

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

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

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

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

$1:2$

$3:2$

$2:3$

$1:3$

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

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

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

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

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

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

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

$1:2$

$3:2$

$2:3$

$1:3$

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

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

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

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