Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 31

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Question 1
1 / -0

The ratio in which the line joining $$(3,4,-7)$$ and $$(4,2,1)$$ is dividing the xy-plane

A
$$3:4$$

B
$$2:1$$

C
$$7:1$$

D
$$4:3$$

Solution

Let the given points be

$$A(3,4,-7),\,B(4,2,1)$$

Let a point on $$XY-plane $$ be $$P(x,y,0)$$ and the line $$AB$$ in the ratio $$k:1$$

then by section formula

$$0=\dfrac{k \times 1+1 \times -7}{k+1}$$

$$k-7 =0$$

$$k=7$$

Therefore, ratio is $$7:1$$

Hence, option $$C$$ is correct.
Question 2
1 / -0

The shortest distance of the point $$(1,2,3)$$ from $${x}^{2}+{y}^{2}=0$$ is

A
$$5$$

B
$$\sqrt{5}$$

C
$$2$$

D
$$\sqrt{14}$$

Solution

$${x}^{2}+{y}^{2}=0$$
$$\therefore$$ $$(x,y)\equiv (0,0)$$
Let point be $$\equiv (0,0,z)$$
$$d\equiv \sqrt { { (1-0) }^{ 2 }+{ (2-0) }^{ 2 }+{ (z-3) }^{ 2 } } =\sqrt { 1+4+{ (z-3) }^{ 2 } } \Rightarrow { (z-3) }^{ 2 }\ge 0$$
$${ d }_{ min }=\sqrt { 5 } $$
Question 3
1 / -0

A triangle $$ABC$$ is placed so that the mid-points of the sides are on the $$x,y,z$$ axes. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively $$\alpha, \beta, \gamma$$. Coordinates of the centroid of the triangle $$ABC$$ are

A
$$(-\alpha/ 3, \beta/ 3,\gamma/3)$$

B
$$(\alpha/3,-\beta/3,\gamma/3)$$

C
$$(\alpha/ 3, \beta/3, -\gamma/ 3)$$

D
$$(\alpha/3, \beta/ 3, \gamma/ 3)$$

Solution

$$\triangle ABC$$ has $$P$$ as mid-point of $$AB$$ on x-axis.
$$Q$$ as midpoint of $$BC$$ on y-axis.
$$R$$ as midpoint of $$AC$$ on z-axis.
$$\therefore A=({ x }_{ 1 }{ y }_{ 1 }{ z }_{ 1 })$$
$$\therefore B=({ x }_{ 2 }{ y }_{ 2 }{ z }_{ 2 })$$
$$\therefore C=({ x }_{ 3 }{ y }_{ 3 }{ z }_{ 3 })$$
$$\Rightarrow P=(\cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } ,\cfrac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } )\\ Q=(\cfrac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\cfrac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } ,\cfrac { { z }_{ 2 }+{ z }_{ 3 } }{ 2 } )\\ R=(\cfrac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } ,\cfrac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } ,\cfrac { { z }_{ 3 }+{ z }_{ 1 } }{ 2 } )\\ P=(\alpha ,0,0)\\ Q=(0,\beta ,0)\\ R=(0,0,\gamma )\\ \therefore \cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =\alpha \quad ,\quad \cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =0\quad ,\quad \cfrac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } =0\\ \cfrac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =0\quad ,\quad \cfrac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =\beta \quad ,\quad \cfrac { { z }_{ 2 }+{ z }_{ 3 } }{ 2 } =0\\ \cfrac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =0\quad ,\quad \cfrac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } =0\quad ,\quad \cfrac { { z }_{ 3 }+{ z }_{ 1 } }{ 2 } =\gamma \\ \therefore { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=\alpha ,\quad { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }=\beta ,\quad { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=\gamma $$
$$ \therefore$$ centroid $$ (G)=(\cfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } ,\cfrac { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }{ 3 } )\\ =(\cfrac { \alpha }{ 3 } ,\cfrac { \beta }{ 3 } ,\cfrac { \gamma }{ 3 } )$$
Question 4
1 / -0

The distance between two points $$(1,1)$$ and $$\left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{\left( {1 - t} \right)}^2}}}{{1 + {t^2}}}} \right)$$ is

A
4t

B
3t

C
1

D
none of these

Solution

Two point are $$A(1,1)$$ and $$B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$$

$$AB=\sqrt{\left(1-\dfrac{2t^{2}}{1+t^{2}}\right)^{2}+\left(1-\dfrac{(1-t^{2})}{1+t^{2}}\right)^{2}}$$

$$=\sqrt{\left(\dfrac{1+t^{2}-2t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{1+t^{2}-1-t^{2}+2t}{1+t^{2}}\right)^{2}}$$

$$=\sqrt{\left(\dfrac{1-t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{RT}{1+t^{2}}\right)^{2}}$$

$$=\sqrt{\dfrac{1+t^{4}-2t^{2}+4t^{2}}{(1+t^{2})^{2}}}$$

$$=\sqrt{\dfrac{(1+t^{2})^{2}}{(1+t^{2})^{2}}}$$

$$=1$$
Question 5
1 / -0

The points (-5,12), (-2,-3),(9,-10),(6,5) taken in order, form

A
Parallelogram

B
rectangle

C
rhombus

D
square

Solution

Given points $$A(-5, 12)\quad B(-2, -3), C(9, -10), D(6, 5)$$

Distance $$AB=\sqrt{(-5+2)^2+(12+3)^2}=\sqrt{9+225}=\sqrt{234}$$
Distance $$BC=\sqrt{(-2-9)^2+(-3+10)^2}=\sqrt{121+49}=\sqrt{170}$$

Distance $$CD=\sqrt{(9-6)^2+(-10-5)^2}=\sqrt{9+225}=\sqrt{234}$$
Distance $$AD=\sqrt{(-5-6)^2+(12-5)^2}=\sqrt{121+49}=\sqrt{170}$$

Distance $$AC=\sqrt{(-5-9)^2+(12+10)^2}=\sqrt{196+484}=\sqrt{680}$$
Distance $$BD=\sqrt{(-2-6)^2+(-3-5)^2}=\sqrt{64+64}=\sqrt{128}$$

These points forms a parallelogram, opposite pair of sides are equal and adjacent sides do not form right angles.
Question 6
1 / -0

The plane $$x = 0$$ divides the joinning of $$( - 2, 3, 4)$$ and $$(1, - 2, 3)$$ in the ratio :

A
$$2 : 1$$

B
$$1 : 2$$

C
$$3 : 2$$

D
$$ - 4 : 3$$

Solution

R.E.F image

Given:
place : $$ x = 0 $$ and two points $$ \rightarrow (-2,3,4) $$ and $$(1,-2,3) $$

let say a point $$(x,y,z) $$ in $$ x = 0 $$ place

So, $$ x = \dfrac{m+n(-2)}{m+n } = \dfrac{m-2n}{m+n} $$

$$ 0 = \dfrac{m-2n}{m+n}\Rightarrow m = 2n $$

So, $$ \dfrac{m}{n} = \dfrac{2}{1}\Rightarrow 2:1 $$ option -A
Question 7
1 / -0

The nearest point from the origin is

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

B
$$(-3,2,1)$$

C
$$(2,2,2)$$

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

Solution

Distance from the origin$$=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$$

Distance from $$\left(2,3,-1\right)$$ is $$\sqrt{{2}^{2}+{3}^{2}+{\left(-1\right)}^{2}}=\sqrt{4+9+1}=\sqrt{14}$$

Distance from $$\left(-3,2,1\right)$$ is $$\sqrt{{\left(-3\right)}^{2}+{2}^{2}+{1}^{2}}=\sqrt{9+4+1}=\sqrt{14}$$

Distance from $$\left(2,2,2\right)$$ is $$\sqrt{{2}^{2}+{2}^{2}+{2}^{2}}=\sqrt{8}$$

Distance from $$\left(1,2,-1\right)$$ is $$\sqrt{{1}^{2}+{2}^{2}+{\left(-1\right)}^{2}}=\sqrt{1+4+1}=\sqrt{6}$$

$$\sqrt{6}<\sqrt{8}<\sqrt{14}$$

Thus $$\left(1,2,-1\right)$$ is nearest point from the origin.
Question 8
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The vertices of a triangle are $$(2, 3, 5), (-1, 3, 2), (3, 5, -2)$$, then the angles are

A
$$30^{\circ}, 30^{\circ}, 120^{\circ}$$

B
$$\cos^{-1} \left (\dfrac {1}{\sqrt {5}}\right ), 90^{\circ}, \cos^{-1} \left (\dfrac {\sqrt {5}}{\sqrt {3}}\right )$$

C
$$30^{\circ}, 60^{\circ}, 90^{\circ}$$

D
$$\cos^{-1} \left (\dfrac {1}{\sqrt {3}}\right ), 90^{\circ}, \cos^{-1} \left (\sqrt{\dfrac { {2}}{ {3}}}\right )$$

Solution

Given vertices: $$A(2,3,5);B(-1,3,2);C(3,5,-2)$$
$$AB=\sqrt { 9+0+9 } =3\sqrt { 2 } $$
$$BC=\sqrt { 16+4+16 } =6$$
$$AC=\sqrt { 1+4+49 } =3\sqrt { 6 } $$
$$\cos { A } =\cfrac { { CA }^{ 2 }+{ BA }^{ 2 }-{ BC }^{ 2 } }{ 2CA\times BA } =\cfrac { 18+54-36 }{ 2\times 3\sqrt { 2 } \times 3\sqrt { 6 } } =\cfrac { 36 }{ 18\times 2\sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } } $$
$$\cos { B } =\cfrac { { AB }^{ 2 }+{ BC }^{ 2 }-{ AC }^{ 2 } }{ 2AB\times BC } =\cfrac { 18+36-54 }{ 2.3\sqrt { 2 } \times 6 } =0$$
$$\cos { C } =\cfrac { 54+36-18 }{ 2\times 3\sqrt { 6 } \times 6 } =\cfrac { 72 }{ 36\sqrt { 6 } } =\cfrac { 2 }{ \sqrt { 6 } } =\sqrt { \cfrac { 2 }{ 3 } } $$
$$\therefore A=\cos ^{ -1 }{ \left( \cfrac { 1 }{ \sqrt { 3 } } \right) } ;B=90°;C=\cos ^{ -1 }{ \left( \sqrt { \cfrac { 2 }{ 3 } } \right) } $$
Question 9
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The plane passing through the point $$\left(-2,-2,2\right)$$ and containing the line joining the points $$\left(1,1,1\right)$$ and $$\left(1,-1,2\right)$$ makes intercepts on the coordinates axes, the sum whose lengths is ?

A
$$3$$

B
$$4$$

C
$$6$$

D
$$12$$

Solution
Question 10
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The points $$(3,\ 2,\ 0),\ (5,\ 3,\ 2)$$ and $$(-9,\ 6,\ -3)$$, are the vertices of a triangle $$ABC.AD$$ is the internal bisector of $$\angle\ BAC$$ which meets $$BC$$ at $$D$$. Then the co-ordinates of $$D$$, are

A
$$\left[ {\dfrac{{17}}{{16}},\ \dfrac{{57}}{{16}},\ \dfrac{{19}}{8}} \right]$$

B
$$\left[ {\dfrac{{19}}{{8}},\ \dfrac{{57}}{{16}},\ \dfrac{{17}}{16}} \right]$$

C
$$\left[0,\ 0,\ {\dfrac{{17}}{{16}}}\right]$$

D
$$\left[{\dfrac{{17}}{{16}}},\ 0,\ 0\right]$$

Solution

$$AD$$ is bisector of $$ \angle BAC $$

$$ \Rightarrow $$ Ratio at $$D$$ is $$ c:b$$ where

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

$$ = \sqrt{4+1+4} = \sqrt{9} = 3 $$

$$ b = AC = \sqrt{(3+9)^{2}+(2-6)^{2}+(0+3)^{2}} $$

$$ = \sqrt{144+16+9} = \sqrt{169} = 13 $$

for point $$D$$

$$ x = \dfrac{c(-9)+b(6)}{c+b} = \dfrac{3(-9)+13(5)}{3+13} = \dfrac{38}{16} = \dfrac{19}{8} $$

$$ y = \dfrac{c(6)+b(3)}{c+b} = \dfrac{3(6)+13(3)}{3+13} = \dfrac{57}{16} $$

$$ z = \dfrac{c(-3)+b(2)}{c+b} = \dfrac{3(-3)+13(2)}{3+13} = \dfrac{17}{16} $$

Hence, point $$D$$ is $$ [\dfrac{19}{8},\dfrac{57}{16},\dfrac{17}{6}] $$

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Re-Attempt Weekly Quiz Competition

Introduction to Three Dimensional Geometry Test - 31

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Introduction to Three Dimensional Geometry Test - 31

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SHARING IS CARING

Question 1

$$3:4$$

$$2:1$$

$$7:1$$

$$4:3$$

Solution

Question 2

$$5$$

$$\sqrt{5}$$

$$2$$

$$\sqrt{14}$$

Solution

Question 3

$$(-\alpha/ 3, \beta/ 3,\gamma/3)$$

$$(\alpha/3,-\beta/3,\gamma/3)$$

$$(\alpha/ 3, \beta/3, -\gamma/ 3)$$

$$(\alpha/3, \beta/ 3, \gamma/ 3)$$

Solution

Question 4

4t

3t

1

none of these

Solution

Question 5

Parallelogram

rectangle

rhombus

square

Solution

Question 6

$$2 : 1$$

$$1 : 2$$

$$3 : 2$$

$$ - 4 : 3$$

Solution

Question 7

$$(2,3,-1)$$

$$(-3,2,1)$$

$$(2,2,2)$$

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

Solution

Question 8

$$30^{\circ}, 30^{\circ}, 120^{\circ}$$

$$\cos^{-1} \left (\dfrac {1}{\sqrt {5}}\right ), 90^{\circ}, \cos^{-1} \left (\dfrac {\sqrt {5}}{\sqrt {3}}\right )$$

$$30^{\circ}, 60^{\circ}, 90^{\circ}$$

$$\cos^{-1} \left (\dfrac {1}{\sqrt {3}}\right ), 90^{\circ}, \cos^{-1} \left (\sqrt{\dfrac { {2}}{ {3}}}\right )$$

Solution

Question 9

$$3$$

$$4$$

$$6$$

$$12$$

Solution

Question 10

$$\left[ {\dfrac{{17}}{{16}},\ \dfrac{{57}}{{16}},\ \dfrac{{19}}{8}} \right]$$

$$\left[ {\dfrac{{19}}{{8}},\ \dfrac{{57}}{{16}},\ \dfrac{{17}}{16}} \right]$$

$$\left[0,\ 0,\ {\dfrac{{17}}{{16}}}\right]$$

$$\left[{\dfrac{{17}}{{16}}},\ 0,\ 0\right]$$

Solution

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