Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 34

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Question 1
1 / -0

If R divides the line segment joining P(2, 3, 4) and Q (4, 5, 6) in the ratio -3 : 2, then the value of the parameter which represents R is

A
$$=(8,9,10)$$

B
$$=(10,9,8)$$

C
$$=(10,8,9)$$

D
$$=(9,10,8)$$

Solution

Given,

$$P(2,3,4),Q(4,5,6)$$

Given Ratio

$$=-\dfrac{3}{2}=\dfrac{m}{n}$$

Let

$$R$$ be $$(x_3,y_3,z_3)$$

$$R=\left ( \dfrac{mx_2+nx}{m+n},\dfrac{my_2+ny}{m+n}+\dfrac{mz_2+nz}{m+n}\right )$$

$$=\left ( \dfrac{-3(4)+2(2)}{-3+2},\dfrac{-3(5)+2(3)}{-3+2}+\dfrac{-3(6)+2(4)}{-3+2}\right )$$

$$=\left ( -\dfrac{8}{-1},-\dfrac{9}{-1},-\dfrac{10}{-1} \right )$$

$$=(8,9,10)$$
Question 2
1 / -0

If (2, 3, -1) is the midpoint of AB where A=(-1,5,3) then B =

A
(5, 1, -5)

B
(5, -1, 5)

C
(-5, 1, 5)

D
NONE OF THESE

Solution

Given, $$(2,3,-1)$$ is mid point
then it will divide $$AB$$ in $$1:1$$
let, $$B$$ be $$(x,y,z)$$
so, according to section formula
$$(\dfrac{1x+1(-1)}{1+1},\dfrac{1y+1(5)}{1+1},\dfrac{1z+1(3)}{1+1})=(2,3,-1)$$
$$(\dfrac{x-1}{2},\dfrac{y+5}{2},\dfrac{z+3}{2})=(2,3,-1)$$
$$\implies{\dfrac{x-1}{2}=2}$$
$$\implies{x-1=4}$$
$$\implies{x=5}$$
Also,$$\dfrac{y+5}{2}=3$$
$$\implies{y+5=6}$$
$$\implies{y=6-5=1}$$
Also,$$\dfrac{z+3}{2}=-1$$
$$\implies{z+3=-2}$$
$$\implies{z=-2-3=-5}$$
hence,$$B=(5,1,-5)$$
Question 3
1 / -0

The coordinates of the point where the line through $$(3, -4, -5)$$ and $$(2, -3, 1)$$ crosses the plane passing through three points $$(2, 2, 1),(3, 0, 1)$$ and $$(4, -1, 0)$$ is

A
$$(1, 2, 7)$$

B
$$(-1, 2, -7)$$

C
$$(1, -2, 7)$$

D
None of these

Solution
Question 4
1 / -0

Locate the point $$(3, 2, -1)$$ in $$3\text{D}$$ octant.

A
$$1st$$ Octant

B
$$2nd$$ Octant

C
$$7th$$ Octant

D
$$8th$$ Octant
Question 5
1 / -0

If the distance of the point P(4, 3, 5) from the Y-axis is $$\lambda $$, then the value of $${ 7\lambda }^{ 2 }$$ is

A
$$287$$

B
$$7\sqrt { 41 } $$

C
$$63$$

D
$$21$$

Solution

$$\Rightarrow$$ $$A=(x_1,y_1,z_1)=(0,3,0)$$ and $$P=(x_2,y_2,z_2)=(4,3,5)$$
Here, $$PA=\lambda$$
$$\Rightarrow$$ $$AP(\lambda)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

$$=\sqrt{(4-0)^2+(3-3)^2+(5-0)^2}$$

$$=\sqrt{16+0+25}$$

$$=\sqrt{41}$$

$$\Rightarrow$$ $$7\lambda^2=7(\sqrt{41})^2$$
$$=7\times 41$$
$$=287$$
Question 6
1 / -0

$$30$$ consider at three dimensional figure represented by $$xyz^2=2$$, then its minimum distance from origin is?

A
$$2$$

B
$$4$$

C
Both A & B

D
None of the above

Solution
Question 7
1 / -0

The distance of P(2,-3) from the x-axis is......

A
2

B
-3

C
3

D
$$\sqrt{13}$$
Question 8
1 / -0

If the distance between the points (2,-3) and (5,b) is 5 , then b=.........

A
1

B
2

C
7

D
5

Solution
Question 9
1 / -0

Find the centroid of a triangle, mid-points of whose sides are $$(!,2,-3),(3,0,1)$$ and $$(-1,1,-4)$$

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

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

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

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

Solution
Question 10
1 / -0

The point in the $$xy -$$ plane which is equidistant from $$(2, 0, 3), (0, 3, 2)$$ and $$(0,0, 1)$$ is

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

B
$$(-3, 2, 0)$$

C
$$(3, -2. 0)$$

D
$$(3, 2, 0)$$

E
$$(3, 2, 1)$$

Solution

Let the points are $$A(2,0,3),B(0,3,2)$$ and $$D(0,0,1)$$
We know that Z-coordinate of every point an xy-plane is zero so let $$p(x,y,0) $$ be a point on xy-plane such that $$PA=PB=PC$$
Now, $$PA=PB$$
$$\Rightarrow PA^2=PB^2$$
$$\Rightarrow (x-2)^2+(y-0)^2+(0-3)^2=(x-0)^2+(y-3)^2+(0-2)$$
$$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0$$......(i)
and , $$PB=PC$$
$$\Rightarrow PB^2=PC^2$$
$$\Rightarrow (x-0)^2+(y-3)^2(0-2)^2=(x-0)^2+(y-0)^2+(0-1)^2$$
$$\Rightarrow -6y+12=0$$
$$\Rightarrow y=2$$......(ii)
Putting $$y=2$$ in equation (i), we get $$x=3$$
Hence, the required point is $$(3,2,0)$$