Introduction to Three Dimensional Geometry Test

Result

Introduction to Three Dimensional Geometry Test - 28

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

Plane $$ax + by + cz = 1$$ intersect axes in $$A, B, C$$ respectively. If $$G\left (\dfrac {1}{6}, -\dfrac {1}{3}, 1\right )$$ is a centroid of $$\triangle ABC$$ then $$a + b + 3c =$$ _________.

A
$$\dfrac {4}{3}$$

B
$$4$$

C
$$2$$

D
$$\dfrac {5}{6}$$

Solution

let $$A\left (\dfrac {1}{a}, 0, 0\right ) B\left (0, \dfrac {1}{b}, 0\right ) C\left (0, 0, \dfrac {1}{c}\right )$$
$$centroid \Rightarrow \left (\dfrac {1}{3a}, \dfrac {1}{3b}, \dfrac {1}{3c}\right ) = \left (\dfrac {1}{6}, \dfrac {-1}{3}, 1\right )$$
On comparing we get,
$$ 3a=6 \Rightarrow a=2\\3b=-3\Rightarrow b=-1\\3c=1\Rightarrow c=\dfrac{1}{3}$$
$$\therefore a = 2, b = -1, c = \dfrac{1}{3}$$
$$\therefore a + b + 3c = 2$$
Only (C) is correct.
Question 2
1 / -0

If $$xy-$$plane and $$yz-$$plane divides the line segment joining $$A(2,4,5)$$ and $$B(3,5,-4)$$ in the ratio $$a:b$$ and $$p:q$$ respectively then value of $$\left(\dfrac {a}{b}+\dfrac {p}{q}\right)$$ may be

A
$$\dfrac {23}{12}$$

B
$$\dfrac {7}{5}$$

C
$$\dfrac {7}{12}$$

D
$$\dfrac {21}{10}$$

Solution

Let the point on $$xy$$-plane which divide line AB in the ratio a:b be $$\left( {{x_1},{y_1},0} \right)$$ and the point on $$yz$$-plane which divide line AB in the ratio p:q be $$\left( {0,{y_2},{z_2}} \right)$$.
Then $$\left( {{x_1},{y_1},0} \right) = \left( {\dfrac{{3a + 2b}}{{a + b}},\dfrac{{5a + 4b}}{{a + b}},\dfrac{{ - 4a + 5b}}{{a + b}}} \right)$$
$$ \Rightarrow \dfrac{{ - 4a + 5b}}{{a + b}} = 0$$
$$ \Rightarrow 4a = 5b$$
$$ \Rightarrow \dfrac{a}{b} = \dfrac{5}{4}$$
and $$\left( {0,{y_2},{z_2}} \right) = \left( {\dfrac{{3p + 2q}}{{p + q}},\dfrac{{5p + 4q}}{{p + q}},\dfrac{{ - 4p + 5q}}{{p + q}}} \right)$$
$$ \Rightarrow \dfrac{{3p + 2q}}{{p + q}} = 0$$
$$ \Rightarrow 3p = - 2q$$
$$ \Rightarrow \dfrac{p}{q} = \dfrac{{ - 2}}{3}$$
Now, $$\dfrac{a}{b} + \dfrac{p}{q} = \dfrac{5}{4} - \dfrac{2}{3} = \dfrac{{15 - 8}}{{12}} = \dfrac{7}{{12}}$$
Question 3
1 / -0

Locus of a point $$P$$ which such that $$PA = PB$$ where $$A = (0, 3, 2)$$ and $$B = (2, 4, 1)$$ is

A
$$2x + y - z = 4$$

B
$$x - 2y + z + 1 = 0$$

C
$$9x - 2y + 4z - 5 = 0$$

D
None of these

Solution

Let the co-ordinate of $$P$$ be $$(x,y,z)$$.
Given the points $$A(0,3.2)$$ and $$B(2,4,1)$$.
Given,
$$PA=PB$$.
or, $$\sqrt{x^2+(y-3)^2+(z-2)^2}=\sqrt{(x-2)^2+(y-4)^2+(z-1)^2}$$
or, $$-6y+9-4z+4=-4x+4-8y+16-2z+1$$ [ Squaring and then eliminating the higher power of $$x,y,z$$]
or,$$4x+2y-2z=8$$
or, $$2x+y-z=4$$.
So locus of $$P$$ is $$2x+y-z=4$$.
Question 4
1 / -0

Points $$(5, 0, 2), (2, -6, 0), (4, -9, 6)$$ & $$(7, -3, 8)$$ are vertices of a

A
Square

B
Rhombus

C
Rectangle

D
Parallelogram

Solution

$$\textbf{Step -1: Find the distance between the given points.}$$
$$\text{Let }A(5,0,2),B(2,-6,0),C(4,-9,6)\text{ and }D(7,-3,8).$$
$$AB=\sqrt{(2-5)^2+(-6-0)^2+(0-2)^2}$$
$$=\sqrt{9+36+4}$$
$$=\sqrt{49}$$
$$=7$$
$$BC=\sqrt{(4-2)^2+(-9-(-6))^2+(6-0)^2}$$
$$=\sqrt{4+9+36}$$
$$=\sqrt{49}$$
$$=7$$
$$CD=\sqrt{(7-4)^2+(-3-(-9))^2+(8-6)^2}$$
$$=\sqrt{9+36+4}$$
$$=\sqrt{49}$$
$$=7$$
$$DA=\sqrt{(5-7)^2+(0-(-3))^2+(2-8)^2}$$
$$=\sqrt{4+9+36}$$
$$=\sqrt{49}$$
$$=7$$
$$AC=\sqrt{(4-5)^2+(-9-0)^2+(6-2)^2}$$
$$=\sqrt{1+81+16}$$
$$=\sqrt{98}$$
$$BD=\sqrt{(7-2)^2+(-3-(-6))^2+(8-0)^2}$$
$$=\sqrt{25+9+64}$$
$$=\sqrt{98}$$
$$\textbf{Step -2: Find the correct option.}$$
$$\because AB=BC=CD=DA$$
$$\therefore\text{All sides are equal.}$$
$$\text{and }AC=BD$$
$$\text{They must be its diagonal and they are also equal.}$$
$$\text{So, the figure formed from the given vertices is a square.}$$
$$\textbf{Hence , the correct option is A.}$$
Question 5
1 / -0

The values of a for which $$(8,-7,a),(5,2,4)$$ and $$(6,-1,2)$$ are collinear , is given by

A
$$2$$

B
$$-2$$

C
$$-1$$

D
$$1$$

Solution
Question 6
1 / -0

Given $$A(3, 2, -4), B(5, 4, -6)$$ & $$C(9, 8, -10)$$ are collinear. Ratio in which $$B$$ divides $$AC$$

A
$$1:2$$

B
$$2:1$$

C
$$-1:2$$

D
None of these

Solution

Given A $$(3,2,-4)$$
B $$(5,4,-6)$$
C $$(9,8,-10)$$

Let the ratio be $$\lambda :1$$

$$ \therefore 5= \cfrac { (3\times 1)+(9\times \lambda ) }{ \lambda +1 } \\ $$

$$\Rightarrow 5\lambda +5=3+9\lambda $$

$$\Rightarrow 4\lambda =2$$

$$\therefore \lambda =\cfrac { 1 }{ 2 } \\ \text{Hence the } Ratio=\lambda :1$$

$$ \quad \quad \quad =1:2$$
Question 7
1 / -0

A cube of side 5 has one vertex at the point (1,0,-1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.

A
(1,0,1),

B
(0,-1,0),

C
(0,0,-1),

D
(1,0,0)

Solution

Consider the problem
Below, are four complete cube face on $$XZ-plane,\,(y=0)$$

Given point
$$(1,0,-1)$$

End of the edge parallel to negative $$x-axis$$
$$(0,0-1)$$

Origin
$$(0,0,0)$$

End of the edge parallel to positive $$z-axis $$
$$(1,0,0)$$

And, below
Four point complete the opposite face of cube.

consider $$P$$, end of edge parallel to negative $$y-axis $$
$$(1,-1,-1)$$

Edge from $$P$$ parallel to positive $$z-axis $$
$$(0,-1,0)$$

Edge from $$P$$ parallel to negative $$x-axis $$
$$(0,-1,-1)$$

And
$$(0,-1,0)$$
Question 8
1 / -0

If $$O=(0,0,0),OP=5$$ and the d.rs of OP are $$1,2,2$$ then $$P_x+P_y+P_z=$$

A
$$25$$

B
$$\dfrac{25}{9}$$

C
$$\dfrac{25}{3}$$

D
$$\left(\dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3}\right)$$

Solution
Question 9
1 / -0

Find the co-ordinates of a point lying on the line $$\dfrac{x -2}{3} = \dfrac{y + 3}{4} = \dfrac{z - 1}{7}$$ which is at a distance $$10$$ units from $$(2, -3, 1)$$.

A
$$(32,37,71)$$

B
$$(-28,-43,-69)$$

C
$$(-32,-37,-71)$$

D
None of these

Solution

Given that the required point lies on the line $$\dfrac{x-2}{3}=\dfrac{y+3}{4}=\dfrac{z-1}{7}$$.

The required point is at a distance of $$10$$ units from $$(2,-3,1)$$

By option verification,

(A) Substituting the point $$(32,37,71)$$

$$\implies \dfrac{32-2}{3}=\dfrac{37+3}{4}=\dfrac{71-1}{7}$$

$$\implies 10=10=10$$

Therefore, distance is $$\sqrt{(32-2)^2+(37+3)^2+(71-1)^2} =86.02$$ units

Hence, $$(32,37,71)$$ is not the required point.

(B) Substituting the point $$(-28,-43,-69)$$

$$\implies \dfrac{-28-2}{3}=\dfrac{-43+3}{4}=\dfrac{-69-1}{7}$$

$$\implies -10=-10=-10$$

Therefore, distance is $$\sqrt{(-28-2)^2+(-43+3)^2+(-69-1)^2} =86.02$$ units

Hence, $$(-28,-43,-69)$$ is not the required point.

(C) Substituting the point $$(-32,-37,-71)$$

$$\implies \dfrac{-32-2}{3}=\dfrac{-37+3}{4}=\dfrac{-71-1}{7}$$

$$\implies 11.3=11.3=10.28$$

Therefore, distance is $$\sqrt{(-32-2)^2+(-37+3)^2+(-71-1)^2} =86.57$$ units

Hence, $$(-32,-37,-71)$$ is not the required point.
Question 10
1 / -0

The ratio in which the join of $$(1, -2, 4)$$ and $$(4, 2, -1)$$ divided by the $$XY$$ plane is

A
$$1:3$$

B
$$3:1$$

C
$$4:1$$

D
$$1:4$$

Solution

Let $$P$$ be the point where the line joining the given two points $$(1,-2,4)$$ and $$(4,2,-1)$$ intersects the $$X-Y$$ plane in $$m:n$$ ratio. We are to find $$m:n$$.
Now the co-ordinate of the point $$P$$ be $$\left(\dfrac{4m+n}{m+n},\dfrac{2m-2n}{m+n},\dfrac{-m+4n}{m+n}\right)$$.
As the point $$P$$ lies on the $$X-Y$$ plane,
$$\dfrac{-m+4n}{m+n}=0$$
or, $$-m+4n=0$$
or, $$\dfrac{m}{n}=\dfrac{4}{1}$$
or, $$m:n=4:1$$

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 - 28

Result

Introduction to Three Dimensional Geometry Test - 28

Score

-

Rank

-

SHARING IS CARING

Question 1

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

$$4$$

$$2$$

$$\dfrac {5}{6}$$

Solution

Question 2

$$\dfrac {23}{12}$$

$$\dfrac {7}{5}$$

$$\dfrac {7}{12}$$

$$\dfrac {21}{10}$$

Solution

Question 3

$$2x + y - z = 4$$

$$x - 2y + z + 1 = 0$$

$$9x - 2y + 4z - 5 = 0$$

None of these

Solution

Question 4

Square

Rhombus

Rectangle

Parallelogram

Solution

Question 5

$$2$$

$$-2$$

$$-1$$

$$1$$

Solution

Question 6

$$1:2$$

$$2:1$$

$$-1:2$$

None of these

Solution

Question 7

(1,0,1),

(0,-1,0),

(0,0,-1),

(1,0,0)

Solution

Question 8

$$25$$

$$\dfrac{25}{9}$$

$$\dfrac{25}{3}$$

$$\left(\dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3}\right)$$

Solution

Question 9

$$(32,37,71)$$

$$(-28,-43,-69)$$

$$(-32,-37,-71)$$

None of these

Solution

Question 10

$$1:3$$

$$3:1$$

$$4:1$$

$$1:4$$

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.