Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 11

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Question 1
1 / -0

$$(-1,-5,-7)$$ lies in Octant

A
I

B
VII

C
V

D
III

Solution

Here all the three $$x,y,z$$ coordinate are negative of the given point.
Therefore, it will lie in the seventh Octant.
Question 2
1 / -0

A rectangular wooden prism is made up of three pieces, each consisting of four cubes of wood glued together. Which of the following pieces has the same shape as the black piece?

A

B

C

D

Solution

Simply imagine it
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
$$\dfrac{1}{3}$$

B
$$1:3$$

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

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

Solution

Given points are $$(1,-1,5)$$ and $$(2,3,4)$$
since ZX-plane divides the line segment in the ratio $$p:1,$$y-coordinate will be $$ 0 $$
the y-coordinate of the point dividing the line segment will be $$\dfrac{3{p}-1}{p+1}=0$$
$$\ p=\dfrac{1}{3}$$
$$p+1=\dfrac{1}{3}+1=\dfrac{4}{3}$$
Question 4
1 / -0

The point $$(0 , -2 , 5)$$ lies on the

A
z axis

B
x axis

C
xy plane

D
yz plane

E
xz plane

Solution
Question 5
1 / -0

An ordered triplet corresponds to ___________ in three dimensional space.

A
three points

B
a unique point

C
a point in each octant

D
infinite number of points

Solution

It is fundamental fact that:
The ordered triplet $$(x,y,z)$$ represents an unique point in three dimensional space
Question 6
1 / -0

The ratio in which the plane $$2x+3y-2z+7=0$$ divides the line segment joining the points $$(-1, 1, 3)$$, $$(2, 3, 5)$$ is

A
$$3:5$$

B
$$7:5$$

C
$$9:11$$

D
$$1:5$$ externally

Solution

Let the plane divide the line $$AB$$ at $$C$$ in the ratio $$k:1$$
The coordinayes of $$C$$ $$ = \left( {\dfrac{{2k - 1}}{{k + 1}},\dfrac{{3k + 1}}{{k + 1}},\dfrac{{5k + 3}}{{k + 1}}} \right)$$
Since this point lies on the plane $$2x+3y-2z+7=0$$
Therefore
$$2\left( {\dfrac{{2k - 1}}{{k + 1}}} \right) + 3\left( {\dfrac{{3k + 1}}{{k + 1}}} \right) - 2\left( {\dfrac{{5k + 3}}{{k + 1}}} \right) + 7 = 0$$
$$\Rightarrow 4k-2+9k+3-10k-6+7k+7=0$$
$$\Rightarrow 10k+2=0$$
$$\Rightarrow k= \dfrac{-2}{10}= \dfrac{-1}{5}$$
If the divides the line in the ratio $$1:5$$ externally
Question 7
1 / -0

The points $$(2, 5)$$ and $$(5, 1)$$ are the two opposite vertices of a rectangle. If the other two vertices are points on the straight line $$y = 2x + k$$, then the value of k is

A
4

B
3

C
-4

D
-3

E
1

Solution

Points $$(2,5)$$ and $$(5,1)$$ form a diagonal of the rectangle.
Hence, the mid point of these points will lie on the other diagonal.
Mid Point $$ = \left( \dfrac{2+5}{2}, \dfrac{5+1}{2} \right) = \left( \dfrac{7}{2}, 3 \right)$$
Equation of the other diagonal is $$ y = 2x + k$$
Therefore, $$3 = 2 \times \dfrac{7}{2} +k \Rightarrow k = -4$$
Question 8
1 / -0

The ratio in which $$xy-$$plane divides the line joining the points $$(1, 0, -3)$$ and $$(1, -5, 7)$$ is given by

A
$$7:3$$

B
$$3:7$$

C
$$3:4$$

D
$$4:7$$

Solution

Let $$xy-plane $$ divide the line joining the given points in the ratio $$k:1$$ and the point of intersection is $$x,y,0$$

then,
$$\dfrac{7k-3}{k+1}=0$$

$$7k-3=0$$

$$k=\dfrac{3}{7}$$

Therefore,
The ratio is $$3:7$$
Question 9
1 / -0

The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is

A
$$\sqrt{17}$$

B
$$\sqrt{29}$$

C
$$\sqrt{34}$$

D
$$\sqrt{41}$$

Solution

$$P(1,2,3),$$ Plane :$$x-y+z=5$$
$$F$$ is foot of perpendicular form $$P$$ to plane and $$I$$ is image,then $$PF=FI$$
$$\therefore$$ If $$(x,y,z)=(r+1,-r+2,r+3)$$ are foot of perpendicular.
$$ \Rightarrow (r+1)-(-r+2)+r+3=5\quad \quad \Rightarrow r=1\\ \therefore F=(2,1,4)\\ \therefore I=(3,0,5)$$
$$ \therefore$$ distance of $$I$$ from origin $$=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 34 } $$
Question 10
1 / -0

The distance between the origin and the centroid of the tetrahedron whose vertices are $$(0, 0, 0)$$, $$(a, 0, 0), (0, b, 0), (0, 0, c)$$ is?

A
$$\sqrt{a^2+b^2+c^2}$$

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

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

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

Solution

Given,
Centroid of tatrahedron =$$\left( {\frac{a}{4},\,\frac{b}{4}\,,\frac{c}{4}} \right)$$
distance between origin & centroid of tetrahedron..................
$$\left( {\frac{a}{4},\,\frac{b}{4}\,,\frac{c}{4}} \right)$$ (0,0,0)
$$\begin{array}{l} =\sqrt { { { \left( { \frac { a }{ 4 } } \right) }^{ 2 } }+{ { \left( { \frac { b }{ 4 } } \right) }^{ 2 } }+{ { \left( { \frac { c }{ 4 } } \right) }^{ 2 } } } \\ \\ =\frac { { \sqrt { { a_{ 2 } }+{ b_{ 2 } }+{ c_{ 2 } } } } }{ 4 } \end{array}$$
So that the correct option is C.