Introduction to Three Dimensional Geometry Test

Result

Introduction to Three Dimensional Geometry Test - 16

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 distance between the circumcentre and the ortho centre of the triangle formed by the points $$(2, 1, 5), (3, 2, 3)$$ and $$(4, 0, 4)$$ is

A
$$\sqrt{6}$$

B
$$\displaystyle \dfrac{\sqrt{6}}{2}$$

C
$$2\sqrt{6}$$

D
$$0$$

Solution

let the points $$A(2,1,5) B(3,2,3)$$ and $$C(4,0,4)$$
Now using distance formula in $$3D$$, we have
$$AB=\sqrt{6}$$, $$BC=\sqrt{6}$$ and $$AC=\sqrt{6}$$
Therefore, it is a equilateral triangle and as we know that the circumcentre and orthocentre in equilateral triangle coincide each other.
Therefore distance between them is $$0$$.
Question 2
1 / -0

The ratio in which $$yz$$-plane divides the line segment joining $$(-3, 4, 2), (2, 1, 3)$$ is

A
$$-4 : 1$$

B
$$3 : 2$$

C
$$-2 : 3$$

D
$$1: 4$$

Solution

Let the plane divide the line in the ratio $$p:1$$.
A point that divides the line joining these $$2$$ points in the ratio $$p : 1$$ is given by $$\left ( \dfrac{2p-3}{p+1} , \dfrac{p+4}{p+1}, \dfrac{3p+2}{p+1} \right)$$
Since, this point has to lie on the $$zy$$ plane, so
$$2p -3 =0$$
$$p = \dfrac{3}{2} $$
Question 3
1 / -0

If the $$zx$$-plane divides the line segment joining $$(1, -1, 5)$$ and $$(2, 3, 4)$$ in the ratio $$p : 1$$, then $$p + 1=$$

A
$$\displaystyle \dfrac{1}{3}$$

B
$$1 $$

C
$$\displaystyle \dfrac{3}{4}$$

D
$$\displaystyle \dfrac{4}{3}$$

Solution

Let the points be given by $$(1,-1,5)$$ and $$(2,3,4)$$.
A point that divides the line joining these $$2$$ points in the ratio $$p : 1$$ is given by $$\left ( \dfrac{2p+1}{p+1} , \dfrac{3p-1}{p+1}, \dfrac{4p+5}{p+1} \right)$$.
Since, this point has to lie on the $$zx$$ plane, so
$$3p - 1 =0$$
$$\Rightarrow p = \dfrac{1}{3} $$
$$\Rightarrow 1+p = \dfrac{4}{3} $$
Question 4
1 / -0

The circum centre of the triangle formed by the points $$(2, 5, 1), (1, 4, -3)$$ and $$(-2, 7, -3)$$ is

A
$$(6,0,1)$$

B
$$(0,6,-1)$$

C
$$(-1,6,2)$$

D
$$(6,1,-2)$$

Solution

Let the points $$A(2,5,1) B(1,4,-3)$$ and $$C(-2,7,-3)$$
Now using distance formula in $$3D$$, we have
$$AB {=}$$$$\sqrt{18}$$, $$BC {=}$$$$\sqrt{18}$$, and $$AC {=}$$$$\sqrt{36}$$
Since, $${AB}^{2}+{BC}^{2}$$$${=}$$$${AC}^{2}$$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e $$AC$$.
Therefore by using section formula (1:1), circumcentre $${=}(0,6,-1)$$
Question 5
1 / -0

The distance from the origin to the centroid of the tetrahedron formed by the points $$(0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c)$$ is:

A
$$\displaystyle \dfrac{\sqrt{a+{b}+{c}}}{4}$$

B
$$\displaystyle \dfrac{\sqrt{{a}+{b}+{c}}}{3}$$

C
$$\displaystyle \dfrac{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}{16}$$

D
$$\displaystyle \dfrac{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}{4}$$

Solution

Let centroid of the tetrahedron having vertices $$(0,0,0),(a,0,0),(0,b,0)$$ and $$(0,0,c)$$ be $$G$$.

So, coordinate of $$G$$ is given by,

$$G =\left(\dfrac{0+a+0+0}{4},\dfrac{0+0+b+0}{4}, \dfrac{0+0+0+c}{4}\right)=\left(\dfrac{a}{4},\dfrac{b}{4},\dfrac{c}{4} \right)$$

Hence, distance of $$G$$ from origin is $$=OG=\sqrt{\left(\dfrac{a}{4}-0\right)^2+\left(\dfrac{b}{4}-0\right)^2+\left(\dfrac{c}{4}-0\right)^2}=\dfrac{\sqrt{a^2+b^2+c^2}}{4}$$
Question 6
1 / -0

The circum radius of the triangle formed by the points $$(1, 2, -3), (2, -3, 1)$$ and $$(-3, 1, 2)$$ is:

A
$$\sqrt{14}$$

B
$$14$$

C
$$\sqrt{13}$$

D
$$0$$

Solution

Let $$A(1,2,-3) , B(2,-3,1) , C(-3,1,2)$$
$$ |AB | = \sqrt{1^2+ 5^2+4^2} = \sqrt{42} $$
$$|BC| = \sqrt {5^2+4^2+1^2} = \sqrt{42}$$
$$|AC| = \sqrt { 4^2+1^2+5^2} = \sqrt{42}$$
It is an equilateral triangle. Hence, the value of $$R$$ will be $$\displaystyle \dfrac{a}{\sqrt{3}}$$ $$ = \sqrt {14}$$
Question 7
1 / -0

If $$(4, 2, p)$$ is the centroid of the tetrahedron formed by the points $$(k, 2, -1), (4, 1, 1), (6,2, 5)$$ and
$$(3, 3, 3)$$ then $$k + p=$$

A
$$\displaystyle \frac{17}{3}$$

B
$$1$$

C
$$\displaystyle \frac{5}{3}$$

D
$$5$$

Solution

Centroid of the tetrahedron formed by the points $$(k, 2, -1), (4, 1, 1), (6,2, 5)$$ and $$(3, 3, 3)$$ is,

$$=\left(\dfrac{k+4+6+3}{4},\dfrac{2+1+2+3}{4},\dfrac{-1+1+5+3}{4} \right)=\left( \dfrac{k+ 13}{4},2,2\right)=(4,2,p)$$ (given)
$$\therefore \dfrac{k+13}{4}=4\Rightarrow k=3$$
and $$p=2$$
Hence $$k+p=3+2=5$$
Question 8
1 / -0

$$G(1, 1, -2)$$ is the centroid of the triangle $$ABC$$ and $$D$$ is the mid point of $$BC$$. If $$A = (-1, 1, -4)$$, then $$D =$$

A
$$\left (\displaystyle \dfrac{1}{2},1, \dfrac{-5}{2}\right)$$

B
$$(5,1,2)$$

C
$$(-5,-1,-2)$$

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

Solution

Let the coordinates of $$D$$ be $$(p,q,r)$$.
Since, the centroid divides the line joining $$AD$$ in the ratio $$2:1$$, the coordinates of centroid should be,
$$\left ( \dfrac{2p-1}{3} , \dfrac{2q+1}{3} , \dfrac{2r-4}{3} \right) $$
Comparing it with the coordinates of the centroid given, $$D(2,1,-1)$$.
Question 9
1 / -0

If the centroid of tetrahedron $$OABC$$ where $$A,B,C$$ are given by $$(a,2,3), (1,b,2)$$ and $$(2,1,c)$$ respectively is $$(1,2,-2)$$, then distance of $$P(a,b,c)$$ from origin is

A
$$\sqrt{195}$$

B
$$\sqrt{14}$$

C
$$\sqrt{\dfrac{107}{14}}$$

D
$$\sqrt{13}$$

Solution

The Centroid of tetrahedron $$OABC$$ with vertices $$O(0,0,0), A(a,2,3), B(1,b,2)$$ and $$C(2,1,c)$$ is $$G\left(\dfrac{a+3}{4},\dfrac{b+3}{4},\dfrac{5+c}{4}\right)$$.
Comparing the coordinates of the Centroid $$G$$ with the given coordinates of centroid $$(1,2,-2)$$, we get
$$a = 1, b= 5, c = -13$$
$$\therefore$$ distance of $$P(1,5,-13)$$ from origin $$O(0,0,0)$$ is
$$D: \sqrt{1^2 + 5^2 + (-13)^2} = \sqrt{195}$$
Question 10
1 / -0

The circum radius of the triangle formed by the points $$(2, -1, 1), (1, -3, -5)$$ and $$(3, -4, -4)$$ is

A
$$\displaystyle \dfrac{\sqrt{6}}{2}$$

B
$$\displaystyle \dfrac{\sqrt{35}}{2}$$

C
$$\displaystyle \dfrac{\sqrt{41}}{2}$$

D
$$\sqrt{41}$$

Solution

Let the points $$A(2,-1,1) B(1,-3,-5)$$ and $$C(3,-4,-4)$$
Now using distance formula in $$3D$$, we have
$$AB=\sqrt{41}$$, $$BC=\sqrt{6}$$ and $$AC=\sqrt{35}$$
Since, $${BC}^{2}+{AC}^{2}$$$${=}$$$${AB}^{2}$$
Hence, it is right angle triangle and as we know that the circum radius of right angled triangle is equal to half of hypotenuse.
Thus, radius $$=\dfrac {\sqrt {41}}{2}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Introduction to Three Dimensional Geometry Test - 16

Result

Introduction to Three Dimensional Geometry Test - 16

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\sqrt{6}$$

$$\displaystyle \dfrac{\sqrt{6}}{2}$$

$$2\sqrt{6}$$

$$0$$

Solution

Question 2

$$-4 : 1$$

$$3 : 2$$

$$-2 : 3$$

$$1: 4$$

Solution

Question 3

$$\displaystyle \dfrac{1}{3}$$

$$1 $$

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

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

Solution

Question 4

$$(6,0,1)$$

$$(0,6,-1)$$

$$(-1,6,2)$$

$$(6,1,-2)$$

Solution

Question 5

$$\displaystyle \dfrac{\sqrt{a+{b}+{c}}}{4}$$

$$\displaystyle \dfrac{\sqrt{{a}+{b}+{c}}}{3}$$

$$\displaystyle \dfrac{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}{16}$$

$$\displaystyle \dfrac{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}{4}$$

Solution

Question 6

$$\sqrt{14}$$

$$14$$

$$\sqrt{13}$$

$$0$$

Solution

Question 7

$$\displaystyle \frac{17}{3}$$

$$1$$

$$\displaystyle \frac{5}{3}$$

$$5$$

Solution

Question 8

$$\left (\displaystyle \dfrac{1}{2},1, \dfrac{-5}{2}\right)$$

$$(5,1,2)$$

$$(-5,-1,-2)$$

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

Solution

Question 9

$$\sqrt{195}$$

$$\sqrt{14}$$

$$\sqrt{\dfrac{107}{14}}$$

$$\sqrt{13}$$

Solution

Question 10

$$\displaystyle \dfrac{\sqrt{6}}{2}$$

$$\displaystyle \dfrac{\sqrt{35}}{2}$$

$$\displaystyle \dfrac{\sqrt{41}}{2}$$

$$\sqrt{41}$$

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.