Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 27

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Question 1
1 / -0

Let $$O$$ be the origin and $$A$$ be the point $$(64, 0)$$. If $$P$$ and $$Q$$ divide $$OA$$ in the ratio $$1 : 2 : 3$$, then the point $$P$$ is

A
$$\left (\dfrac {32}{3}, 0\right )$$

B
$$(32, 0)$$

C
$$\left (\dfrac {64}{3}, 0\right )$$

D
$$(16, 0)$$

E
$$\left (\dfrac {16}{3}, 0\right )$$

Solution

Now, by internal section formula,
$$P\equiv \left (\dfrac {1\times 64 + 5\times 10}{1 + 5} \cdot \dfrac {1\times 0 + 5\times 0}{1 + 5}\right )$$
$$\equiv \left (\dfrac {64}{6}, 0\right ) = \left (\dfrac {32}{3}, 0\right )$$.
Question 2
1 / -0

The shortest distance between z-axis and the line
$$x+y+2z-3=0=2x+3y+4z-4$$, is _____________

A
$$1$$

B
$$2$$

C
$$4$$

D
$$3$$

Solution

$$x+y+2z-3=0=2x+3y+4z-4$$ at $$z-axis, x=y=0$$
$$x+y=3-2z$$ and $$2x+3y=4(1-z)$$
solving $$x$$ and $$y$$ in function $$z$$,
$$2x+3(3-2z-x)=4(1-z)$$
$$\implies 2x+9-6z-3x=4-4z$$
$$\implies x=9-6z-4+4z$$
$$\implies x=5-2z\quad equation (2)$$
$$\implies (5-2z)+y=3-2z$$
$$\implies y=3-2z-5+2z$$
$$y=-2\quad equation (2)$$
So, $$\sqrt {x^2+y^2}=\sqrt {(-2)^2+(5-2z)^2}$$
at, $$z=\cfrac {5}{2}$$, this would be minimum.
Question 3
1 / -0

If $$z_{1}$$ and $$z_{2}$$ are $$z$$ co-ordinates of the points of trisection of the segment joining the points $$A(2, 1, 4), B(-1, 3, 6)$$ then $$z_{1} + z_{2} =$$

A
$$1$$

B
$$4$$

C
$$5$$

D
$$10$$

Solution

For $$z_{1}$$, the ratio of line segment is $$1:2$$

For $$z_{2}$$, the ratio of line segment is $$2:1$$

By internal division formula,

$$Z=z_{1}+z_{1}$$

$$=\dfrac{1(6)+2(4)}{1+2}+\dfrac{2(6)+1(4)}{2+1}$$

$$=\dfrac{6+8}{3}+\dfrac{12+4}{3}$$

$$=\dfrac{14+16}{3}=\dfrac{30}{3}=10$$

Hence, $$z_{1}+z_{2}=10$$
Question 4
1 / -0

The point which divides the line joining the points $$ (1, 3, 4)$$ and $$(4,3,1)$$ internally in the ratio $$ 2:1 , $$ is

A
$$ (2, -3, 3 ) $$

B
$$ (2, 3, 3) $$

C
$$ \left( \dfrac {5}{2} , 3 , \dfrac {5}{2} \right) $$

D
$$ ( -3, 3, 2 ) $$

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

Solution

Therefore, $$ R (x,y,z) = \left( \dfrac {2 \times 4 + 1 \times 1 }{2 +1} , \dfrac {2 \times 3 + 1 \times 3 }{2+1} , \dfrac { 2 \times 1 + 1 \times 4 }{2+1} \right) $$
$$ = \left( \dfrac{9}{3}, \dfrac{9}{3}, \dfrac{6}{3} \right) = (3,3,2) $$
Question 5
1 / -0

The distance between the X-axis and the point $$(3, 12, 5)$$ is

A
3

B
13

C
14

D
12

E
5

Solution

The distance between the $$x$$-axis and point $$(3, 12, 5) =$$ Distance between $$PQ$$
$$= \sqrt{(3 - 3)^2 + (12 - 0)^2 + (5 - 0)^2}$$
$$= \sqrt{0^2 + 12^2 + 5^2} = \sqrt{144 + 25}$$
$$= \sqrt{169} = 13$$
Question 6
1 / -0

The coordinates of the foot of the perpendicular drawn from the point $$A(1,0,3)$$ to the join of the points $$B(4,7,1)$$ and $$C(3,5,3)$$ are

A
$$(5,7,17)$$

B
$$\left( \cfrac { -5 }{ 7 } ,\cfrac { 7 }{ 3 } ,\cfrac { -17 }{ 3 } \right) $$

C
$$\left( \cfrac { 5 }{ 7 } ,\cfrac { -7 }{ 3 } ,\cfrac { 17 }{ 3 } \right) $$

D
$$\left( \cfrac { 5 }{ 7 } ,\cfrac { 7 }{ 3 } ,\cfrac { 17 }{ 3 } \right) $$

Solution

The cartesian equation of the line passing through given 2-point form

$$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$$

$$\dfrac{x-3}{1}=\dfrac{y-5}{2}=\dfrac{z-3}{-2}$$

let P be the coordinates of foot

$$P\equiv (\lambda +3,2\lambda+5,-2\lambda +3 )$$

DR of $$PA\equiv (\lambda+2,2\lambda+5,-2\lambda)$$

So,

$$(\lambda+2)1+(2\lambda+5)2+(-2\lambda)(-2)=0$$

$$9\lambda +12=0$$

$$\lambda =-\dfrac{4}{3}$$

$$P\equiv \left ( \dfrac{5}{3},\dfrac{7}{3},\dfrac{17}{3} \right )$$
Question 7
1 / -0

In the triangle with vertices $$A(1, -1, 2), B(5, -6, 2)$$ and $$C(1,3,-1)$$ find the altitude $$n=|BD|$$.

A
$$5$$

B
$$10$$

C
$$5\sqrt{2}$$

D
$$\displaystyle\frac{10}{\sqrt{2}}$$

Solution

Given $$A(1,-1,2) , B(5,-6,2) , C(1,3,-1)$$

Let the coordinates of $$D$$ be $$D(h,k,l)$$
the equation of line $$AC$$ is $$\dfrac{x-1}{0}=\dfrac{y+1}{4}=\dfrac{z-2}{-3}$$
So the point of line $$AC$$ will be in the form of $$( 1,4t-1,2-3t)$$
Since $$D$$ lies on $$AC$$ , we have $$h=1,k=4t-1,l=2-3t$$

The drs of $$BD$$ are $$h-5,k+6,l-2$$ and the drs of $$AC$$ are $$0,4,-3$$
Since $$BD$$ and $$AC$$ are perpendicular, we have $$4(k+6)-3(l-2)=0$$
$$\Rightarrow 4(4t+5)-3(-3t)=0$$

$$\Rightarrow t=-\dfrac{4}{5}$$
So, coordinates of $$D$$ are $$D\left(1,-\dfrac{21}{5},\dfrac{22}{5}\right)$$

So, the value of $$n=\sqrt { 16+\dfrac { 81 }{ 25 } +\dfrac { 144 }{ 25 } } =\sqrt { 25 } =5$$
Question 8
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 $$\left( {{a \over b},{p \over q}} \right)$$ may be

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

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

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

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

Solution

Let the co-ordinate of xy plane be $$(x, y, 0)$$
According to the section formula
$$0 = \dfrac{-4a + 5b}{a + b}$$ {for z co-cordinates}
$$\Rightarrow 5b = 4a$$
$$\dfrac{a}{b} = \dfrac{5}{4}$$
Let the co-ordinate of yz-plane be (0, y, z)
According to the section formula,
$$0 = \dfrac{2q + 3p}{p + q}$$
$$\therefore \dfrac{p}{q} = \dfrac{-2}{3}$$
$$(\dfrac{a}{b} + \dfrac{p}{q}) = (\dfrac{5}{4} - \dfrac{2}{3}) = (\dfrac{15 - 8}{12}) = \dfrac{7}{12}$$
Option C should be correct
Question 9
1 / -0

A swimmer can swim $$2$$ km in $$15$$ minutes in a lake and in a river he can swim a distance of $$4$$ km in $$20$$ minutes along the stream. If a paper boat is put in the river, then the distance covered by it in $$\displaystyle $$2$$ \, \frac{1}{2}$$2 hours will be

A
$$18$$ km

B
$$12$$ km

C
$$8$$ km

D
$$10$$ km

Solution

Speed of the man in still water $$\cfrac{2}{15/60}$$ = $$8$$ km/hr
Speed of the man in downstream = $$\cfrac{4}{20/60}$$ = $$12$$ km/hr
Speed of stream = $$4$$ km/hr
Distance covered by paper boat in $$2$$ $$\frac{1}{2}$$ hours = $$\cfrac{5}{2} \, \times$$ $$4$$ = $$10$$ km
Question 10
1 / -0

In a $$\triangle {ABC}$$, side $$AB$$ has the equation $$2x+3y=29$$ and the side $$AC$$ has the equation $$x+2y=16$$. If the mid point of $$BC$$ is $$(5,6)$$, then the equation of $$BC$$ is

A
$$2x+y=7$$

B
$$x+y=1$$

C
$$2x-y=17$$

D
None of these

Solution

Let co-ordinates of $$B$$ be $$(x_1, y_1)$$ & $$C$$ be $$(x_2, y_2)$$

$$\therefore$$ $$(5,6)$$ is the mid point,

so, $$\dfrac{x_1 + x_2}{2} = 5, \dfrac{y_1 + y_2}{2} = 6$$

$$\Rightarrow x_1 + x_2 = 10, y_1 + y_2 = 12$$

$$B(x_1, y_1)$$ lies on the line $$2x + 3y = 29$$

$$\therefore 2x_1 + 3y_1 = 29$$ ----(1)

$$C(x_2, y_2)$$ lies on the line $$x + 2y = 16,$$

$$\therefore x_2 + 2y_2 = 16$$ ----(2)

$$\therefore$$ putting $$x_1, y_1$$ in the form of $$x_2, y_2$$ in (1)

$$2(10 - x_2) + 3(12 - y_2) = 29$$ {$$x_1 = 10 - x_2, y_1 = 12 - y_2$$}

$$\Rightarrow 20 - 2x_2 + 36 - 3y_2 = 29$$

$$\Rightarrow 2x_2 + 3y_2 = 27$$ ----(3)

on subtracting $$(3)$$ and $$(2)$$ $$\times$$ $$2$$

$$-y_2 = -5$$

$$y_2 = 5$$

Putting $$y_2 \, in (2)$$

$$x_2 + 2(5) = 16$$

$$x_2 = 6$$

$$x_1 = 10 - x_2$$

$$= 4$$

$$y_1 = 12 - 5 = 7$$

Equation :

$$\dfrac{x - x_1}{x_2 - x_1} = \dfrac{y - y_1}{y_2 - y_1}$$

$$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$$

$$\Rightarrow -x + 4 = y - 7$$

$$\Rightarrow x + y = 11$$

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Introduction to Three Dimensional Geometry Test - 27

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Introduction to Three Dimensional Geometry Test - 27

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Question 1

$$\left (\dfrac {32}{3}, 0\right )$$

$$(32, 0)$$

$$\left (\dfrac {64}{3}, 0\right )$$

$$(16, 0)$$

$$\left (\dfrac {16}{3}, 0\right )$$

Solution

Question 2

$$1$$

$$2$$

$$4$$

$$3$$

Solution

Question 3

$$1$$

$$4$$

$$5$$

$$10$$

Solution

Question 4

$$ (2, -3, 3 ) $$

$$ (2, 3, 3) $$

$$ \left( \dfrac {5}{2} , 3 , \dfrac {5}{2} \right) $$

$$ ( -3, 3, 2 ) $$

$$ ( 3,3,2) $$

Solution

Question 5

3

13

14

12

5

Solution

Question 6

$$(5,7,17)$$

$$\left( \cfrac { -5 }{ 7 } ,\cfrac { 7 }{ 3 } ,\cfrac { -17 }{ 3 } \right) $$

$$\left( \cfrac { 5 }{ 7 } ,\cfrac { -7 }{ 3 } ,\cfrac { 17 }{ 3 } \right) $$

$$\left( \cfrac { 5 }{ 7 } ,\cfrac { 7 }{ 3 } ,\cfrac { 17 }{ 3 } \right) $$

Solution

Question 7

$$5$$

$$10$$

$$5\sqrt{2}$$

$$\displaystyle\frac{10}{\sqrt{2}}$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$18$$ km

$$12$$ km

$$8$$ km

$$10$$ km

Solution

Question 10

$$2x+y=7$$

$$x+y=1$$

$$2x-y=17$$

None of these

Solution

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