Introduction to Three Dimensional Geometry Test

Result

Introduction to Three Dimensional Geometry Test - 18

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The value(s) of $$\lambda $$, for which the triangle with vertices $$(6,10,10),(1,0,-5)$$ and $$(6,-10,\lambda)$$ will be a right angled triangle is/ are

A
$$1$$

B
$$\displaystyle \dfrac{70}{3},0$$

C
$$35$$

D
$$\displaystyle 0,-\dfrac{70}{3}$$

Solution

Let the given points be $$A,B$$ and $$C$$ respectively.
Then, $${ AB }^{ 2 }=350,{ AC }^{ 2 }=500\lambda -20\lambda +{ \lambda }^{ 2 }{ BC }^{ 2 }=150+10\lambda +{ \lambda }^{ 2 }$$
Now, $$\displaystyle{ AB }^{ 2 }+{ AC }^{ 2 }={ BC }^{ 2 }\Rightarrow 350+500-20\lambda +{ \lambda }^{ 2 }=150+10\lambda +{ \lambda }^{ 2 }\Rightarrow \lambda =\dfrac { 70 }{ 3 } $$
Next, $${ AB }^{ 2 }+BC^{ 2 }={ AC }^{ 2 }\Rightarrow 250+150+10\lambda +{ \lambda }^{ 2 }=500-20\lambda +{ \lambda }^{ 2 }\Rightarrow \lambda =0$$
Further, $${ BC }^{ 2 }+AC^{ 2 }={ AB }^{ 2 }\Rightarrow 150+10\lambda +12+500-20\lambda +{ \lambda }^{ 2 }=350\Rightarrow { \lambda }^{ 2 }-5\lambda +150=0$$,
which has no real solution.
$$\therefore $$ the triangle is right angle for $$\displaystyle\lambda=0,\dfrac { 70 }{ 3 } $$
Question 2
1 / -0

$$P(0,5,6),Q(1,4,7),R(2,3,7)$$ and $$S(3,5,16)$$ are four points in the space. The point nearest to the origin $$O(0,0,0)$$ is

A
$$P$$

B
$$Q$$

C
$$R$$

D
$$S$$

Solution

The $$4$$ points are as given.
We calculate their individual distance from the origin.
$$OP =$$ $$ {({5}^{2} + {6}^{2})}^{0.5} $$ = $$ {(61)}^{0.5} $$
$$OQ =$$ $$ {({1}^{2} + {4}^{2} + {7}^{2} )}^{0.5} $$ = $$ {(66)}^{0.5} $$
$$OR =$$ $$ {({2}^{2} + {3}^{2} + {7}^{2} )}^{0.5} $$ = $$ {(62)}^{0.5} $$
$$OS =$$ $$ {({3}^{2} + {5}^{2} + {16}^{2} )}^{0.5} $$= $$ {(290)}^{0.5} $$
Hence, $$P$$ is the nearest to the origin.
Question 3
1 / -0

If $$A(\cos\alpha,\sin\alpha, 0),B(\cos\beta,\sin\beta, 0)$$, $$C(\cos\gamma,\sin\gamma,0)$$ are vertices of $$\Delta ABC$$ and let
$$\cos \alpha+\cos\beta+\cos\gamma=3{a}$$, $$\sin\alpha+\sin\beta+\sin\gamma =3b$$, then correct matching of the following is:
List : I
List : II
A. Circumcentre
$$1. (3a,3b,0)$$
B. Centroid
$$2. (0,0,0)$$
C. Ortho centre
$$3. (a,b,0)$$

A
$$4\ 3\ 2$$

B
$$2\ 3\ 1$$

C
$$1\ 2\ 3$$

D
$$2\ 3\ 4$$

Solution

We know, $${ \sin }^{ 2 }\alpha +{ \cos }^{ 2 }\alpha =1$$
Similarly it is true for other three coordinates.
These are points on a circle of unit radius.
By observation $$(0,0,0)$$ would be the circumcentre of the circle.
Centroid $$=\cfrac { (\cos{ \alpha }_{ 1 }+\cos{ \alpha }_{ 2 }+\cos{ \alpha }_{ 3 }) }{ 3 } ,\cfrac { (\sin{ \alpha }_{ 1 }+\sin{ \alpha }_{ 2 }+\sin{ \alpha }_{ 3 }) }{ 3 } ,0$$
Centroid $$=(\cfrac { 3a }{ 3 } ,\cfrac { 3b }{ 3 } ,0)$$=$$(a,b,0)$$
Centroid divides line segment joining orthocentre and circumcentre in the ratio 2:1.
Orthocentre $$=(3a,3b,0)$$
Question 4
1 / -0

Arrange the points: $$\mathrm{A}(1,2-3), \mathrm{B}(-1,2,-3), \mathrm{C}(-1,-2-3)$$ and $$\mathrm{D}(1,-2, -3)$$ in the increasing order of their octant numbers:

A
$$A,B,C,D$$

B
$$B,C,D,A$$

C
$$C,D,A,B$$

D
$$D,C,B,A$$

Solution

Octant $$I$$ $$II$$ $$III$$ $$IV$$ $$V$$ $$VI$$ $$VII$$ $$VIII$$
Signs: $$+,+,+$$ $$-,+,+$$ $$-,-,+$$ $$+,-,+$$ $$+,+,-$$ $$-,+,-$$ $$-,-,-$$ $$+,-,-$$
Based on this, increasing order is
$$ A,B ,C,D$$
Question 5
1 / -0

In the $$\Delta ABC$$, if $$AB=\sqrt{2}; AC=\sqrt{20}, B=(3,2,0)$$ and $$C=(0,1,4)$$, then the length of the median passing through $$A$$ is

A
$$\displaystyle \dfrac{3}{2}$$

B
$$\displaystyle \dfrac{9}{2}$$

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

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

Solution

Using Apollonius theorem: In any triangle $$ABC$$, if $$AD$$ is a median,
then $${ AB }^{ 2 }+{ AC }^{ 2 }=2({ BD }^{ 2 }+{ AD }^{ 2 })$$
$$D$$ is the midpoint of $$BC$$
Coordinates of $$D$$ are $$\left(\dfrac{3}{2},\dfrac{3}{2},2\right)$$ and $$(BD)^2=\dfrac{26}{4}$$
$$\Rightarrow { AB }^{ 2 }+{ AC }^{ 2 }=2({ BD }^{ 2 }+{ AD }^{ 2 })$$
$$\Rightarrow AD = \dfrac { 3 }{ \sqrt { 2 } } $$
Question 6
1 / -0

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
$$13$$

B
$$\sqrt{13}$$

C
$$2\sqrt{13}$$

D
$$169$$

Solution

By looking $$(-2, 4, 6)$$ and $$(3, 16, 6)$$ extremities of a diagonal both lie in same plane ($$xy$$- plane).

Base diagonal is diagonal joining these two extremities.

So, length of diagonal $$= \sqrt { { (3+2 })^{ 2 }+{ (16-4) }^{ 2 }+{ (6-6) }^{ 2 } }$$
$$=13$$
Question 7
1 / -0

Assertion (A): The points $$A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$$ are the vertices of a rhombus
Reason (R): $$AB = BC = CD = DA$$ and $$AC = BD$$

A
Both A and R are individually true and R is the correct explanation of A

B
Both A and R individually true but R is not the correct explanation of A

C
A is true but R is false

D
Both A and R false

Solution

Given: The points $$A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$$ are the vertices of a rhombus.
So using distance formula Reason is not true.
Thus both A and R false.
Question 8
1 / -0

If the plane a $$2x-3y+5_{Z}-2=0$$ divides the line segment joining $$(1, 2, 3)$$ and $$(2, 1, k)$$ in the ratio $$9 : 11$$, then $$k$$ is

A
$$1$$

B
$$-2$$

C
$$-10$$

D
$$\displaystyle -\dfrac{1}{2}$$

Solution

Coordinate of the point which divides the line segment joining the points $$ (1,2,3)$$ and $$(2,1,k)$$ in the ratio $$9:11$$ are $$\left(\dfrac{29}{20}, \dfrac{31}{20}, \dfrac{9k+33}{20}\right)$$
Also, this point will lie on the given plane

$$\Rightarrow 2\times\dfrac{29}{20}-3\times\dfrac{31}{20}+5\times\dfrac{9k+33}{20}-2 =0$$

$$\Rightarrow k = -2$$
Hence, option B is correct.
Question 9
1 / -0

The point equidistant from the points $$(0,0,0), (1,0,0), (0,2,0)$$ and $$(0,0,3)$$ is

A
$$(1,2,3)$$

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

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

D
$$(1,-2,3)$$

Solution

Equation of sphere passes through origin is given by

$$x^2+y^2+z^2+ax+by+cz=0 ...(1)$$

Given, it also passes through $$(1,0,0),(0,2,0),0,0,3)$$

$$1+a=0\Rightarrow a = -1$$

$$4+2b=0\Rightarrow b = -2$$

and $$ 9+3c=0\Rightarrow c = -3$$

$$\therefore$$the equation of the sphere becomes $$x^2+y^2+z^2-x-2y-3z=0$$

Comparing with $$x^2+y^2+z^2+2gx+2fy+2kz+C=0$$

Therefore, the point equidistant from all the given four point will be the centre of the sphere $$(1)$$ passing through all these points which is $$\left(\dfrac{1}{2},1, \dfrac{3}{2}\right)i.e. the \ centre$$.
Question 10
1 / -0

Assertion(A): If centroid and circumcentre of a triangle are known its orthocentre can be found.
Reason (R) : Centriod, orthocentre and circumcentre of a triangle are collinear

A
Both A and R are individually true and R is the correct explanation of A

B
Both A and R individually true but R is not the correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

Centroid, orthocentre and circumcentre are collinear and centroid divides
the line joining orthocentre and circumcentre in the ratio $$2:1$$ so if any two points are given then this one can be found.
Hence option 'A' is the correct choice

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

List : I	List : II
A. Circumcentre	$$1. (3a,3b,0)$$
B. Centroid	$$2. (0,0,0)$$
C. Ortho centre	$$3. (a,b,0)$$

Octant	$$I$$	$$II$$	$$III$$	$$IV$$	$$V$$	$$VI$$	$$VII$$	$$VIII$$
Signs:	$$+,+,+$$	$$-,+,+$$	$$-,-,+$$	$$+,-,+$$	$$+,+,-$$	$$-,+,-$$	$$-,-,-$$	$$+,-,-$$

Introduction to Three Dimensional Geometry Test - 18

Result

Introduction to Three Dimensional Geometry Test - 18

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1$$

$$\displaystyle \dfrac{70}{3},0$$

$$35$$

$$\displaystyle 0,-\dfrac{70}{3}$$

Solution

Question 2

$$P$$

$$Q$$

$$R$$

$$S$$

Solution

Question 3

$$4\ 3\ 2$$

$$2\ 3\ 1$$

$$1\ 2\ 3$$

$$2\ 3\ 4$$

Solution

Question 4

$$A,B,C,D$$

$$B,C,D,A$$

$$C,D,A,B$$

$$D,C,B,A$$

Solution

Question 5

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

$$\displaystyle \dfrac{9}{2}$$

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

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

Solution

Question 6

$$13$$

$$\sqrt{13}$$

$$2\sqrt{13}$$

$$169$$

Solution

Question 7

Both A and R are individually true and R is the correct explanation of A

Both A and R individually true but R is not the correct explanation of A

A is true but R is false

Both A and R false

Solution

Question 8

$$1$$

$$-2$$

$$-10$$

$$\displaystyle -\dfrac{1}{2}$$

Solution

Question 9

$$(1,2,3)$$

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

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

$$(1,-2,3)$$

Solution

Question 10

Both A and R are individually true and R is the correct explanation of A

Both A and R individually true but R is not the correct explanation of A

A is true but R is false

A is false but R is true

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.