Introduction to Three Dimensional Geometry Test - 17

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Introduction to Three Dimensional Geometry Test - 17

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Question 1
1 / -0

If the orthocentre, circumcentre of a triangle are $$(-3, 5, 2), (6, 2, 5)$$ respectively then the centroid of the triangle is

A
$$(3,3, 4)$$

B
$$\left (\displaystyle \dfrac{3}{2},\dfrac{7}{2},\dfrac{9}{2}\right)$$

C
$$(9,9,12)$$

D
$$\left (\displaystyle \dfrac{9}{2}\dfrac{-3}{2},\dfrac{3}{2}\right)$$

Solution

Since, the centroid divides the line joining the orthocentre and circumcentre in the ratio $$2:1$$

The coordinates of the centroid will be,

($$ \dfrac{9}{3} , \dfrac{9}{3}, \dfrac{12}{3} $$)

$$= (3,3,4)$$
Question 2
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If $$\mathrm{A}= (-1,6, 6)$$ , $$\mathrm{B}=(-4,9, 6)$$ , $$\displaystyle \mathrm{G}=\frac{1}{3}(-5,22,22)$$ and $$\mathrm{G}$$ is the centroid of the $$\Delta \mathrm{A}\mathrm{B}\mathrm{C}$$ then the name of the triangle $$\mathrm{A}\mathrm{B}\mathrm{C}$$ is

A
an isosceles triangle

B
a right angled triangle

C
an equilateral triangle

D
a right-angled isosceles triangle

Solution

The centroid of a triangle if given by
$$G=((\dfrac{x_{1}+x_{2}+x_{3}}{3}),\dfrac{y_{1}+y_{2}+y_{3}}{3}),\dfrac{z_{1}+z_{2}+z_{3}}{3}))$$
Let the coordinates of C be $$(x,y,z)$$; then
$$G=(\dfrac{-5}{3},\dfrac{22}{3},\dfrac{22}{3})=(\dfrac{x-1-4}{3},\dfrac{y+6+9}{3},\dfrac{z+6+6}{3})$$
Comparing LHS and RHS gives us
$$x-5=-5$$ or $$x=0$$; similarly $$y+15=22$$ or $$y=7$$. and $$z+12=22$$ or $$z=10$$.
Hence
$$C=(0,7,10)$$.
Thus the coordinates of the vertices of the triangle are
$$A(-1,6,6)$$ $$B=(-4,9,6)$$ and $$C(0,7,10)$$.
Now
$$AB=3\sqrt{2}$$
$$BC=6$$
$$AC=3\sqrt{2}$$
Now
$$AB^{2}+AC^{2}=(3\sqrt{2})^{2}+(3\sqrt{2})^{2}=18+18=36$$
$$=BC^{2}$$.
Hence it is a right angled isosceles triangle right angled at A
Question 3
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The extremities of a diagonal of a rectangular parallelopiped whose faces are parallel to the reference planes are $$(-2, 4, 6)$$ and $$(3, 16, 6)$$. The length of the base diagonal is

A
$$7$$

B
$$10$$

C
$$11$$

D
$$13$$

Solution

$$AC=$$ base diagonal

where,

$$A(-2,4,6),C(3,16,6)$$

$$=\sqrt{(3+2)^2+(16-4)^2+(6-6)^2}$$

$$=\sqrt{5^2+12^2}$$

$$=13$$
Question 4
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In the tetrahedron $$ABCD,\ A= (1, 2, -3)$$ and $$G(-3,4, 5)$$ is the centroid of the tetrahedron. If $$P$$ is the centroid of the $$\Delta BCD$$, then $$AP=$$

A
$$\displaystyle \dfrac{8\sqrt{21}}{3}$$

B
$$\displaystyle \dfrac{4\sqrt{21}}{3}$$

C
$$4\sqrt{21}$$

D
$$\displaystyle \dfrac{\sqrt{21}}{3}$$

Solution

Given, $$A=(1,2,-3), G(-3,4,5)$$

Therefore, $$AG=\sqrt { { (-3-1) }^{ 2 }+{ (4-2) }^{ 2 }+{ (5-(-3)) }^{ 2 } } $$

and $$ AG=\sqrt { 84 } =2\sqrt { 21 } $$

$$P$$ is the centroid of $$\triangle BCD$$

So, $$G$$ divides $$AP$$ in $$3:1$$.

Let $$AG=3x$$, then $$GP=x$$

$$3x=2\sqrt { 21 } \\ x=\dfrac { 2\sqrt { 21 } }{ 3 } $$

Now $$AP=AG+GP$$

$$\Rightarrow AP=3x+x$$

$$ \Rightarrow AP=4x$$

$$ \Rightarrow AP=4\left( \dfrac { 2\sqrt { 21 } }{ 3 } \right) =\dfrac { 8\sqrt { 21 } }{ 3 } $$

So, option A is correct.
Question 5
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In the $$\Delta $$ ABC , A $$=$$ (1, 3, -2) and G (-1, 4, 2) is the centroid of the triangle. If D is the mid point of BC then AD $$=$$

A
$$\displaystyle \dfrac{\sqrt{21}}{2}$$

B
$$\displaystyle \dfrac{3\sqrt{21}}{2}$$

C
$$\sqrt{21}$$

D
$$\dfrac{63}{2}$$

Solution

First, we calculate the distance AG,
It is $$ {(4+1+16)}^{0.5} = {21}^{0.5} $$
From the property of the centroid that it divides the line joining AD in the ratio 2:1,
The distance$$ AD = \frac{3}{2} $$AG
Hence,$$ AD = \frac{3}{2} {21}^{0.5} $$
Question 6
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The harmonic conjugate of $$(2, 3, 4)$$ with respect to the points $$(3, -2, 2)$$ and $$(6, -17, -4)$$ is

A
$$(\dfrac{18}{5}, -5,\dfrac{4}{5})$$

B
$$(11, -16,2)$$

C
$$\left (\displaystyle \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}\right)$$

D
$$(0, 0, 0)$$

Solution

Let us assume that point $$(2,3,4)$$ divides $$(3,-2,2)$$ and $$(6,-17,-4)$$ in the ratio $$\lambda:1$$
We have $$\frac{6\lambda+3}{1+\lambda}=2$$
$$\Rightarrow \lambda=-\frac{1}{4}$$
Therefore point $$(2,3,4)$$ divides given two points in the ratio $$1:4$$ externally
Let the point which divides the given two points in the ratio $$1:4$$ internally be $$(x,y,z)$$
We have $$x=\frac{6+12}{5}=\frac{18}{5}$$ , $$y=\frac{-17-8}{5}=-5$$ and $$z=\frac{-4+8}{5}=\frac{4}{5}$$
Therefore option $$A$$ is correct
Question 7
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$$A = (2, 3, 0)$$ and $$B = (2,1, 2)$$ are two points. If the points $$P, Q$$ are on the line $$AB$$ such that $$AP= PQ = QB$$, then $$PQ=$$

A
$$2\sqrt{2}$$

B
$$6\sqrt{2}$$

C
$$\sqrt{\dfrac{8}{9}}$$

D
$$\sqrt{2}$$

Solution

$$AB =\sqrt{(2-2)^2+(3-1)^2+(0-2)^2}=\sqrt 8 $$
Since $$AP=PQ=QB$$
$$\therefore PQ =\dfrac{AB}{3}=\dfrac{\sqrt 8}{3}=\sqrt{\dfrac{8}{9}}$$
Question 8
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If the points $$A,B,C,D$$ are collinear and $$C,D$$ divide $$AB$$ in the ratios $$2:3, -2:3$$ respectively, then the ratio in which $$A$$ divides $$CD$$ is

A
$$5:1$$

B
$$2:3$$

C
$$3:2$$

D
$$1:5$$

Solution

$$C$$ divide $$AB$$ in the ratio of $$2:3$$

$$\Rightarrow AC=2x,CB=3x$$

$$D$$ divide $$AB$$ in the ratio of $$-2:3$$

$$\Rightarrow DA=2y,DB=3y$$

$$\Rightarrow AB=DB-DA=y$$

$$\Rightarrow 5x=y\Rightarrow x=\dfrac { y }{ 5 } $$

$$\Rightarrow CA:AD=2x:2y=\dfrac { x }{ y } = \cfrac{\dfrac y5}{y} = \cfrac15$$

$$\Rightarrow$$ $$A$$ divides $$CD$$ in the ratio $$1:5$$

Question 9

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Then the correct matching is

List - I	List - II
A: The coordinates of the mid point of the line joining $$(-1,-1,1)$$ and $$(-1,1,-1)$$	1) $$(-2,1,1)$$
B: The coordinates of the point which divides the line segment joining $$($$2,3,1 $$)$$ and $$(5, 0,4)$$ in the ratio1:2	2) $$(-1,0,0)$$
$$\mathrm{C}$$: The points and $$P(2,1, -3)$$ are three vertices of a parallelogram $$\mathrm{P}\mathrm{Q}\mathrm{R}\mathrm{S}$$, the fourth vertex	3) $$(\displaystyle \frac{13}{3},\frac{-11}{3},6)$$
$$\mathrm{D}$$: The vertices of a triangle are $$(7,-4,7)$$ , $$(1,-6,10)$$ and $$(5, -1,1)$$ . centroid of the triangle	4) $$($$3, 2, 2 $$)$$

$$A-2,\ B-4,\ C-1,\ D-3$$

$$A-1,\ B-2,\ C-3,\ D-4$$

$$A-2,\ B-3,\ C-1,\ D-4$$

$$A-1,\ B-4,\ C-3,\ D-2$$

Solution

A: The co-ordinate of the mid point of the line joining $$(-1,-1,1) $$ and $$(-1,1,-1)$$

from mid point formula $$\left( \dfrac {-1-1}{2},\dfrac{-1+1}{2},\dfrac{1-1}{2}\right)$$

$$=(-1,0,0)$$

$$A\rightarrow 2$$

B: co-ordinates of the point which divides the line segment Joining

$$(2,3,1)$$ and $$(5,0,4) $$ in ratio $$1:2$$

so from section formula $$\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2},\dfrac{m_1z_2+m_2z_1}{m_1+m_2}\right)$$

Here $$m_1=1,m_2=2$$

so $$=\left(\dfrac {1\times +2\times 2}{3},\dfrac{1(0)+2(3)}{3},\dfrac{1(4)+2(1)}{3}\right)$$

$$=(3,2,2)$$

$$B\rightarrow 4$$

C: The point $$(2,3,1),(5,0,4)$$ and $$(2,1,-3)$$ and $$A(x,y,z)$$ vertices of a parallelogram $$PQRS$$ then fourth vertex

we know that the diagonals of parallelogram bisect each other have they mid point of $$PR=$$ mid point of $$QS$$ .... [Ref. image]

$$\Rightarrow \left(\dfrac{2+5}{2},\dfrac{1}{2},\dfrac{1}{2}\right)=\left(\dfrac{x+2}{2},\dfrac{y+3}{2},\dfrac{z+1}{2}\right)$$

$$\Rightarrow \dfrac{x+2}{2}=\dfrac{7}{2} \Rightarrow x=5$$

$$\Rightarrow \dfrac{y+3}{2}=\dfrac{-1}{2} \Rightarrow y=-2$$

$$ \Rightarrow \dfrac{z+1}{2}= \dfrac{1}{2} \Rightarrow z=0$$

D: The vertixes of a triangle $$ax$$ $$(7,-4,7),(1,-6,10)$$ and $$(5,-1,1)$$ Centroid of triangle

Then centroid $$=\left(\dfrac{7+1+5}{3},\dfrac{-4-6-1}{3},\dfrac{7+10+1}{3}\right)$$

$$=\left(\dfrac{13}{3},\dfrac{-11}{3},6\right)$$

$$D \rightarrow 3$$

Hence correct option is A

Question 10
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$$P (1,1,1 )$$ and $$Q(\lambda, \lambda, \lambda)$$ are two points in the space such that $$PQ=\sqrt{27}$$, then the value(s) of $$\lambda $$ can be

A
$$-4$$

B
$$-2,4$$

C
$$2$$

D
$$4,3$$

Solution

The square of the distance between the $$2$$ points is $$ {3\times{(\lambda - 1)}^{2}} $$
This has to be equal to $$27$$.
Hence, $$ \lambda - 1 $$ $$= 3,-3$$
$$ \lambda = -2,4 $$

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Introduction to Three Dimensional Geometry Test - 17

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Introduction to Three Dimensional Geometry Test - 17

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Question 1

$$(3,3, 4)$$

$$\left (\displaystyle \dfrac{3}{2},\dfrac{7}{2},\dfrac{9}{2}\right)$$

$$(9,9,12)$$

$$\left (\displaystyle \dfrac{9}{2}\dfrac{-3}{2},\dfrac{3}{2}\right)$$

Solution

Question 2

an isosceles triangle

a right angled triangle

an equilateral triangle

a right-angled isosceles triangle

Solution

Question 3

$$7$$

$$10$$

$$11$$

$$13$$

Solution

Question 4

$$\displaystyle \dfrac{8\sqrt{21}}{3}$$

$$\displaystyle \dfrac{4\sqrt{21}}{3}$$

$$4\sqrt{21}$$

$$\displaystyle \dfrac{\sqrt{21}}{3}$$

Solution

Question 5

$$\displaystyle \dfrac{\sqrt{21}}{2}$$

$$\displaystyle \dfrac{3\sqrt{21}}{2}$$

$$\sqrt{21}$$

$$\dfrac{63}{2}$$

Solution

Question 6

$$(\dfrac{18}{5}, -5,\dfrac{4}{5})$$

$$(11, -16,2)$$

$$\left (\displaystyle \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}\right)$$

$$(0, 0, 0)$$

Solution

Question 7

$$2\sqrt{2}$$

$$6\sqrt{2}$$

$$\sqrt{\dfrac{8}{9}}$$

$$\sqrt{2}$$

Solution

Question 8

$$5:1$$

$$2:3$$

$$3:2$$

$$1:5$$

Solution

Question 9

$$A-2,\ B-4,\ C-1,\ D-3$$

$$A-1,\ B-2,\ C-3,\ D-4$$

$$A-2,\ B-3,\ C-1,\ D-4$$

$$A-1,\ B-4,\ C-3,\ D-2$$

Solution

Question 10

$$-4$$

$$-2,4$$

$$2$$

$$4,3$$

Solution

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