Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 24

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Question 1
1 / -0

If $$P(x,y,z)$$ is a point on the line segment joining $$A (2,2,4)$$ and $$B(3,5,6)$$ such that projection of $$\vec{OP}$$ on axes are $$\displaystyle \frac{13}{5},\frac{19}{5},\frac{26}{5}$$ respectively, then P divide AB in the ratio

A
$$3:2$$

B
$$2:3$$

C
$$1:2$$

D
$$1:3$$

Solution

$$\vec{OP} = \dfrac{13}{5}\hat{i} + \dfrac{19}{5}\hat{j} + \dfrac{26}{5}\hat{k} = \vec{P} - \vec{O}$$
where $$ \vec{P} =$$ origin vector
$$= 0\hat{i} + 0\hat{j} + 0\hat{k}$$
$$OP = (\dfrac{13}{5}, \dfrac{19}{5}. \dfrac{26}{5})$$ so, $$\vec{P} = \dfrac{13}{5}\hat{i} + \dfrac{19}{5}\hat{j} + \dfrac{26}{5}\hat{k}$$
If $$(x, y, z)$$ divides $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2)$$ in the ratio $$\alpha : 1$$
$$\therefore x = \dfrac{\alpha x_2 + x_1}{\alpha + 1}, y = \dfrac{\alpha y_2 + y_1}{\alpha + 1}, z = \dfrac{\alpha z_2 + z_1}{\alpha + 1}$$
Then,
$$x = \dfrac{3\alpha + 2}{\alpha + 1}, y = \dfrac{5\alpha + 2}{\alpha + 1}, z = \dfrac{6\alpha + 4}{\alpha + 1}$$
$$(x, y, z)$$ given $$= (\dfrac{13}{5}, \dfrac{19}{5}, \dfrac{26}{5})$$
$$\therefore \dfrac{3\alpha + 2}{\alpha + 1} = \dfrac{13}{5} \Rightarrow 13\alpha + 13 = 15\alpha + 10$$
$$\Rightarrow 3 = 2\alpha \Rightarrow \alpha = \dfrac{3}{2}$$
For $$z$$ co-ordinates,
$$\dfrac{19}{5} = \dfrac{5\alpha + 2}{\alpha + 1} \Rightarrow 19\alpha + 19 = 25\alpha + 10$$
$$\Rightarrow 6\alpha = 9 \Rightarrow \alpha = \dfrac{3}{2}$$
$$\alpha : 1 = 3 : 2$$
Question 2
1 / -0

Find the distance between the points whose position vectors are given as follows
$$-2\hat i+3\hat j+5\hat k, 7\hat i-\hat k$$

A
$$\displaystyle 3\sqrt{14}$$

B
$$\displaystyle \sqrt{54}$$

C
$$\displaystyle 3\sqrt{19}$$

D
$$\displaystyle \sqrt{57}$$

Solution

Distance Between two position vectors $$a\hat i+b\hat j+c\hat k$$ and $$x\hat i+y\hat j+z\hat k$$ is given by
$$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $$
Given $$-2\hat i+3\hat j+5\hat k$$, $$7\hat i-\hat k$$
Distance $$=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 3-0 \right) ^{ 2 }+\left( 5+1 \right) ^{ 2 } } $$
$$=\sqrt { 126 } $$
$$=3\sqrt { 14 } $$
Question 3
1 / -0

The points $$A(5, 1, 1), B(7, 4, 7), C(1, 6, 10)$$ and $$D(-1, 3, 4)$$ are the vertices of a

A
Parallelogram

B
Rectangle

C
Rhombus

D
Square

Solution

Given points are $$ A(5, 1, 1), B(7, 4, 7), C(1, 6, 10)$$ and $$D(-1, 3, 4)$$

$$AB=\sqrt{4+9+36}=7$$

$$BC=\sqrt{36+4+9}=7$$

$$CD=\sqrt{4+9+36}=7$$

$$AD=\sqrt{36+4+9}=7$$

$$AC= \sqrt{16+25+81}=\sqrt{122}$$

$$BD=\sqrt{64+1+9}=\sqrt{74}$$

All sides are equal and diagonals are not equal.

Hence, $$ABCD$$ is a Rhombus.

Hence, option C.
Question 4
1 / -0

If $$P\left( x,y,z \right) $$ is a point on the line segment joining $$Q\left( 2,2,4 \right) $$ and $$R\left( 3,5,6 \right) $$ such that the projection of $$\overline { OP } $$ on the axes are $$\displaystyle\frac { 13 }{ 5 } ,\frac { 19 }{ 5 } ,\frac { 26 }{ 5 } ,$$ respectively, then $$P$$ divides $$QR$$ in the ratio

A
$$1:2$$

B
$$3:2$$

C
$$2:3$$

D
$$1:3$$

Solution

Since $$OP$$ has projections $$\displaystyle\frac { 13 }{ 5 } ,\frac { 19 }{ 5 } $$ and $$\displaystyle\frac { 26 }{ 5 } $$ on the coordinate axes,
therefore $$\displaystyle OP=\frac { 13 }{ 5 } \hat i+\frac { 19 }{ 5 } \hat j+\frac { 26 }{ 5 } \hat k$$
Suppose $$P$$ divides the join of $$Q\left( 2,2,4 \right) $$ and $$R\left( 3,5,6 \right) $$ in the ratio $$\lambda:1$$
Then, the position vector of $$P$$ is
$$\displaystyle \left( \frac { 3\lambda +2 }{ \lambda +1 } \right)\hat i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \hat j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \hat k$$

$$\displaystyle\therefore \frac { 13 }{ 5 }\hat i+\frac { 19 }{ 5 }\hat j+\frac { 26 }{ 5 } \hat k=\left( \frac { 3\lambda +2 }{ \lambda +1 } \right) \hat i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \hat j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \hat k$$

$$\displaystyle\Rightarrow \frac { 3\lambda +2 }{ \lambda +1 } =\frac { 13 }{ 5 } ,\frac { 5\lambda +2 }{ \lambda +1 } =\frac { 19 }{ 5 } ,\frac { 6\lambda +4 }{ \lambda +1 } =\frac { 26 }{ 5 } $$

$$\displaystyle\Rightarrow 2\lambda =3$$
$$\Rightarrow \lambda =\dfrac { 3 }{ 2 } $$
Question 5
1 / -0

Find the distance between the points whose position vectors are given as follows
$$(-1,1,3), (0,5,6)$$

A
$$\displaystyle \sqrt{118}$$

B
$$8$$

C
$$\displaystyle \sqrt{26}$$

D
none of these

Solution

Distance Between two points $$(a,b,c)$$ and $$(x,y,z)$$ is given by
$$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $$
Given $$(-1,1,3)$$, $$(0,5,6)$$
Distance $$=\sqrt { \left( -1-0 \right) ^{ 2 }+\left( 1-6 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } $$
$$=\sqrt { 26 } $$
Question 6
1 / -0

The plane $$XOZ$$ divides the join of $$(1, - 1, 5)$$ and $$(2, 3, 4)$$ in the ratio $$\lambda : 1$$, then $$\lambda$$ is -

A
$$- 3$$

B
$$-1 / 3$$

C
$$3$$

D
$$1 / 3$$

Solution

The plane XOZ divides the join of $$(1, - 1, 5)$$ and $$(2, 3, 4)$$ in the ratio $$λ : 1$$
$$i.e.,y = 0 $$ divide the join of $$(1, -1, 5)$$ and $$(2, 3, 4)$$ in the ratio $$λ : 1. $$
$$\therefore \dfrac {3λ−1}{λ+1} = 0 ⇒λ=\dfrac{1}{3}$$.
Question 7
1 / -0

Find the distance between the pairs of points whose cartesian coordinates are $$(2,3,-1), (2,6,2).$$

A
$$\displaystyle 3\sqrt{2}.$$

B
$$\displaystyle 2\sqrt{3}.$$

C
$$\displaystyle 5\sqrt{2}.$$

D
$$\displaystyle 2\sqrt{5}.$$

Solution

Distance Between two points $$(a,b,c)$$ and $$(x,y,z)$$ is given by
$$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $$
Given $$(2,3,-1)$$, $$(2,6,2)$$
Distance $$=\sqrt { \left( 2-2 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } } $$
$$=\sqrt { 18 } $$
$$=3\sqrt { 2 }$$
Question 8
1 / -0

Find the distance between the points whose position vectors are given as follows
$$4\hat i+3\hat j-6\hat k, -2\hat i+\hat j-\hat k$$

A
$$\displaystyle \sqrt{65}$$

B
$$\displaystyle \sqrt{69}$$

C
$$13$$

D
none of these

Solution

Distance Between two position vectors $$a\hat i+b\hat j+c\hat k$$ and $$x\hat i+y\hat j+z\hat k$$ is given by
$$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $$
Given $$4\hat i+3\hat j-6\hat k$$, $$-2\hat i+\hat j-\hat k$$
Distance $$=\sqrt { \left( 4+2 \right) ^{ 2 }+\left( 3-1 \right) ^{ 2 }+\left( -6+1 \right) ^{ 2 } } $$
$$=\sqrt { 65 } $$
Question 9
1 / -0

Find the point which divides the lines joining $$A(2,3,5)$$ and $$B(-6,5,8)$$ in the ratio $$2:3$$ externally

A
$$(18,-1,-1)$$

B
$$(12,-1,-1)$$

C
$$(3,4,4)$$

D
$$(5,8,11)$$

Solution

The given points are

$$A(x_1,y_1,z_1)=A(2,3,5)$$

$$B(x_2,y_2,z_2)=B(-6,5,8)$$

Let $$P(x,y,z)$$ divides AB in the ratio

$$m:n=2:3$$ externally

Then by using section Formula,

$$P(x,y,z)=\left(\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n},\dfrac{mz_2-nz_1}{m-n}\right)$$

$$P(x,y,z)=\left(\dfrac{2\times (-6)-3\times 2}{2-3},\dfrac{2\times 5-3\times 3}{2-3},\dfrac{2\times 8-3\times 5}{2-3}\right)$$

$$P(x,y,z)=\left(\dfrac{-12-6}{-1},\dfrac{10-9}{-1},\dfrac{16-15}{-1}\right)$$

$$P(x,y,z)=\left(\dfrac{-18}{-1},\dfrac{1}{-1},\dfrac{1}{-1}\right)$$

$$P(x,y,z)=(18,-1,-1)$$
Question 10
1 / -0

The ordinate of the point which divides the lines joining the origin and the point $$(1,2) $$ externally in the ratio of $$3:2$$ is

A
$$-2$$

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

C
$$ \displaystyle \frac{2}{5} $$

D
$$6$$

Solution

Co-ordinates of the required point will be
$$\displaystyle y=\frac{m_{1}y_{2}-m_{2}y_{1}}{m_{1}-m_{2}}=\frac{3\times 2-2\times 0}{3-2}=6$$

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Re-Attempt Weekly Quiz Competition

Introduction to Three Dimensional Geometry Test - 24

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Introduction to Three Dimensional Geometry Test - 24

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SHARING IS CARING

Question 1

$$3:2$$

$$2:3$$

$$1:2$$

$$1:3$$

Solution

Question 2

$$\displaystyle 3\sqrt{14}$$

$$\displaystyle \sqrt{54}$$

$$\displaystyle 3\sqrt{19}$$

$$\displaystyle \sqrt{57}$$

Solution

Question 3

Parallelogram

Rectangle

Rhombus

Square

Solution

Question 4

$$1:2$$

$$3:2$$

$$2:3$$

$$1:3$$

Solution

Question 5

$$\displaystyle \sqrt{118}$$

$$8$$

$$\displaystyle \sqrt{26}$$

none of these

Solution

Question 6

$$- 3$$

$$-1 / 3$$

$$3$$

$$1 / 3$$

Solution

Question 7

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

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

$$\displaystyle 5\sqrt{2}.$$

$$\displaystyle 2\sqrt{5}.$$

Solution

Question 8

$$\displaystyle \sqrt{65}$$

$$\displaystyle \sqrt{69}$$

$$13$$

none of these

Solution

Question 9

$$(18,-1,-1)$$

$$(12,-1,-1)$$

$$(3,4,4)$$

$$(5,8,11)$$

Solution

Question 10

$$-2$$

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

$$ \displaystyle \frac{2}{5} $$

$$6$$

Solution

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