Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 30

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Question 1
1 / -0

If $$A(2, 2, -3), B(5, 6, 9), C(2, 7, 9)$$ be the vertices of a triangle. The internal bisector of the angle $$A$$ meets $$BC$$ at the point $$D$$, then find the coordinates of $$D$$.

A
$$(\dfrac{13}{2},\dfrac{7}{2},9)$$

B
$$(\dfrac{7}{2},\dfrac{13}{2},9)$$

C
$$(\dfrac{9}{2},\dfrac{7}{2},9)$$

D
$$(\dfrac{13}{2},\dfrac{9}{2},9)$$

Solution

We have,

Point

$$ A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 2,2,-3 \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 5,6,9 \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( 2,7,9 \right) $$

Let the coordinates of point $$D\left( x,y,z \right).$$

So,

According to question,

$$ AB=\sqrt{{{\left( 5-2 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}+{{\left( 9-3 \right)}^{2}}} $$

$$ AB=\sqrt{9+16+149}=\sqrt{{{13}^{2}}}=13 $$

$$ AC=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 7-2 \right)}^{2}}+{{\left( 9+3 \right)}^{2}}} $$

$$ AC=\sqrt{0+25+144}=\sqrt{169}=13 $$

Thus $$ABC$$ is isosceles triangle with $$AB=AC$$

So, angle bisector $$AD$$ bisects $$BC$$

$$D\equiv \left( \dfrac{5+2}{2},\dfrac{6+7}{2},\dfrac{9+9}{2} \right)\equiv \left( \dfrac{7}{2},\dfrac{13}{2},9 \right)$$
Hence, this is the ansewr.
Question 2
1 / -0

If the lines $$\frac{x - 0}{1} =\frac{y+1}{2}=\frac{z-1}{-1}$$ and $$\frac{x+1}{k}=\frac{y-3}{-2}=\frac{z-2}{1}$$ are at right angles, then the value of k is

A
$$5$$

B
$$0$$

C
$$3$$

D
$$-1$$

Solution

If $$({ l }_{ 1 }{ m }_{ 1 }{ n }_{ 1 })$$ and $$({ l }_{ 2 }{ m }_{ 2 }{ n }_{ 2 })$$ are directions of two $$\bot$$ lines then,
$${ l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+n_{ 1 }{ n }_{ 2 }=0\\ k-4-1=0\\ k=5$$
Question 3
1 / -0

The graph of the equation $$y^{2}+z^{2}=0$$ in three dimensional space is

A
x- axis

B
y- axis

C
z- axis

D
yz-plane

Solution

Consider the problem

$${y^2} + {z^2} = 0$$

$$x=0$$ and $$z=0$$

Therefore,
The graph of the equation

$${y^2} + {z^2} = 0$$ is $$x-axis$$.

Hence, the correct option is $$x-axis$$.
Question 4
1 / -0

The area of triangle whose vertices are $$(1, 2, 3), (2, 5, -1)$$ and $$(-1, 1, 2)$$ is

A
$$150\ sq. units$$

B
$$145\ sq. units$$

C
$$\sqrt {155}/2\ sq. units$$

D
$$155/2\ sq. units$$

Solution

Let the vertices of triangle are

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

Then,

$$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =\hat { i } +3\hat { j } -4\hat { k } \\ \\ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =-2\hat { i } -\hat { j } -\hat { k } \end{array}$$

Then,

$$\begin{array}{l} \overrightarrow { AB } \times \overrightarrow { AC } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 1 & 3 & { -4 } \\ { -2 } & { -1 } & { -1 } \end{array} } \right| \\ \\ =-7\hat { i } +9\hat { j } +5\hat { k } \\ \\ \left| { \overrightarrow { AB } \times \overrightarrow { AC } } \right| =\sqrt { { { \left( { -7 } \right) }^{ 2 } }+{ { \left( 9 \right) }^{ 2 } }+{ { \left( 5 \right) }^{ 2 } } } \\ \\ =\sqrt { 49+8+25 } \\ \\ =\sqrt { 155 } \end{array}$$

Area of triangle $$ABC=\frac{1}{2}|\vec {AB} \times \vec {AC}|$$

$$=\frac{1}{2} \times \sqrt {155}$$

$$=\frac{\sqrt {155}}{2}$$ sq. units
Question 5
1 / -0

A tangent to the curve $$y = f(x)$$ at $$p(x, y)$$ meets $$x - axis$$ at $$A$$ and $$y-axis$$ at $$B$$. If $$\overline {AP} : \overline {BP} = 1 : 3$$ and $$f(1) = 1$$ then the curve also passes through the point.

A
$$\left (\dfrac {1}{2}, 4\right )$$

B
$$\left (\dfrac {1}{3}, 24\right )$$

C
$$\left (2, \dfrac {1}{8}\right )$$

D
$$\left (3, \dfrac {1}{28}\right )$$

Solution

$$\dfrac{(y - y_1)}{(x - x_1)} = f'(x_1)$$

$$y - y_1 = f'(x_1)(x - x_1)$$

$$y = 0, \dfrac{-y}{f'(x_1)} = x - x_1$$

$$\Rightarrow x = x_1 = \dfrac{y_1}{f'(x_1)}$$

$$A = \left(x_1 - \dfrac{y_1}{r'(x_1)}, 0\right)$$

$$x = 0, y - y_1 = f'(x_1). (-x_1)$$

$$\Rightarrow y = y_1 - x_1 f'(x_1)$$

$$B = (0, y_1 - x_1 f'(x_1))$$

P divides $$AB$$ in ration $$1 : 3$$

$$x_1 = \dfrac{3\left[x_1 - \dfrac{y_1}{f'(x_1)} \right]}{4} , y_1 = \dfrac{[y_1 - x_1 f'(x_1)]}{4}$$

$$4y_1 = y_1 x_1 f'(x_1)$$

$$f'(x_1) = \dfrac{-3y_1}{x_1}$$

$$f'(x) = \left(\dfrac{-3y}{x} \right)$$

$$\dfrac{dy}{dx} = \dfrac{-3y}{x}$$

$$\dfrac{dy}{y} = \dfrac{-3dx}{x}$$

$$\ln y = -3 \ln x + c$$

$$y = kx^{-3}$$

$$y(1) = 1$$

$$\Rightarrow k = 1 \Rightarrow y = \dfrac{1}{x^3} $$ put $$x = 2 , y = \dfrac{1}{8} \left(2, \dfrac{1}{8} \right)$$
Question 6
1 / -0

If the distance between a point $$P$$ and the point $$(1, 1, 1)$$ on the line $$\dfrac {x - 1}{3} = \dfrac {y - 1}{4} = \dfrac {z - 1}{12}$$ is $$13$$, then the coordinates of $$P$$ are

A
$$(3, 4, 12)$$

B
$$\left (\dfrac {3}{13}, \dfrac {4}{13}, \dfrac {12}{13}\right )$$

C
$$(4, 5, 13)$$

D
$$(40, 53, 157)$$

Solution

Let the given points be $$A(1,1,1)$$
Consider,
$$\dfrac{{x - 1}}{3} = \dfrac{{y - 1}}{4} = \dfrac{{z - 1}}{{12}} = \lambda $$

$$\begin{array}{l} x=3\lambda +1 \\ \\ y=4\lambda + 1\\ \\ z=12\lambda +1 \end{array}$$

General point on the line is
$$3\lambda + 1,\,4\lambda + 1,\,12\lambda + 1$$

Given that,
$$AP=13$$

$$\sqrt {{{\left( {3\lambda + 1 - 1} \right)}^2} + {{\left( {\,4\lambda + 1 - 1} \right)}^2} + {{\left( {12\lambda + 1 - 1} \right)}^2}} = 13$$

$$13\lambda =13$$

$$\lambda =1$$

$$\begin{array}{l} 3\lambda +1=4 \\ \\ 4\lambda +1=5 \\ \\ 12\lambda +1=13 \end{array}$$

Therefore, required point $$P$$ is $$(4,5,13)$$.
Hence the correct option is $$C$$.
Question 7
1 / -0

The values of a for which $$(8, -7, a), (5, 2, 4)$$ and $$(6, -1, 2)$$ are collinear, is given by?

A
$$2$$

B
$$-2$$

C
$$-1$$

D
$$1$$

Solution

$$\begin{matrix} then\, a=? \\ The\, equation\, lineAB\, is\, \\ \Rightarrow \dfrac { { x-8 } }{ { 5-8 } } =\dfrac { { y+7 } }{ { 2+7 } } =\dfrac { { z-a } }{ { 4-a } } ....\left( 1 \right) \\ po{ int }\, c\, lies\, in\, the\, line\, AB \\ po{ int }\left( { 6,-1,2 } \right) satisfy\, eq\left( 1 \right) \\ \Rightarrow \dfrac { { -2 } }{ { -3 } } =\dfrac { { -1+7 } }{ 9 } =\dfrac { { 2-a } }{ { 4-a } } \\ \Rightarrow \dfrac { 2 }{ 3 } =\dfrac { { 2-a } }{ { 4-a } } \\ \Rightarrow 8-2a=6-3a \\ \Rightarrow 3a-2a=6-8 \\ a=-2 \\ \end{matrix}$$
Question 8
1 / -0

The vertices of a triangle are $$)2, 3, 5), (-1, 3, 2), (3, 5, -2)$$, then the angles are

A
$$30^o, 30^o, 30^o$$

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

C
$$30^o, 60^o, 90^o$$

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

Solution

$$\overrightarrow { AB } =-3\hat { i } -3\hat { k } \\ \overrightarrow { BC } =4\hat { i } +2\hat { j } -4\hat { k } \\ \overrightarrow { CA } =-\hat { i } -2\hat { j } +7\hat { k } \\ \cos { A } =\cfrac { \left| \overrightarrow { AB } .\overrightarrow { CA } \right| }{ \left| \overrightarrow { AB } \right| .\left| \overrightarrow { CA } \right| } =\cfrac { 1 }{ \sqrt { 3 } } \\ \Rightarrow A=\cos ^{ -1 }{ (\cfrac { 1 }{ \sqrt { 3 } } ) } \\ \cos { B } =\cfrac { \left| \overrightarrow { AB } .\overrightarrow { BC } \right| }{ \left| \overrightarrow { AB } \right| .\left| \overrightarrow { BC } \right| } =0\\ \Rightarrow B={ 90 }^{ \circ }\\ \cos { C } =\cfrac { \left| \overrightarrow { CA } .\overrightarrow { BC } \right| }{ \left| \overrightarrow { CA } \right| .\left| \overrightarrow { BC } \right| } =\cfrac { \sqrt { 2 } }{ \sqrt { 3 } } \\ \Rightarrow C=\cos ^{ -1 }{ (\cfrac { \sqrt { 2 } }{ \sqrt { 3 } } ) } $$
Question 9
1 / -0

The ratio in which the line joining points $$(2,4,5)$$ and $$(3,5,-4)$$ divide YZ -plane is

A
$$-2:3$$

B
$$2:3$$

C
$$-3:2$$

D
$$3:2$$

Solution

Let the ratio be $$\lambda:1$$
The dividing plane is $$YZ$$ plane.
So, the co-ordinate of $$x$$ after dividing the line $$= 0$$
Now According to internal division rule of line segment
$$x=\dfrac{3\times \lambda - 1\times 2}{\lambda + 1}$$
$$\Rightarrow \dfrac{3\lambda - 2}{\lambda + 1}=0$$
$$\Rightarrow 3\lambda - 2=0 \ \Rightarrow \lambda=\dfrac{2}{3}$$.
So. Ratio $$=2:3$$
Question 10
1 / -0

The points $$(10,7,0)$$, $$(6,6-1)$$ and $$(6,9,-4)$$ form a

A
Right -angled triangle

B
Isosceles triangle

C
Both $$(1)$$ & $$(2)$$

D
Equilateral triangle

Solution

$$P(10, 7, 0),\ Q(6, 6, -1),\ R(6, 9, -4)$$
$$PQ=\sqrt {4^2+(-1)^2 +(-1)^2}$$
$$=\sqrt {16+1+1}=3\sqrt 3$$
$$QR=\sqrt {0^2+3^2+-3^2}$$
$$PR=\sqrt {4^2+2^2+(-4)^2}$$
$$=\sqrt {16+4+16}=6$$
$$PQ^2+QR^2=(3\sqrt 2)^2+(3\sqrt 2)^2=6^2=PR^2$$
$$\therefore \ \triangle PQR$$ is a right angle $$D$$ of $$Q,\ PQ=QR$$
$$\triangle PQR$$ is a isoscels traingle
$$(C)$$ Both $$(1)$$ & $$(2)$$

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Introduction to Three Dimensional Geometry Test - 30

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Introduction to Three Dimensional Geometry Test - 30

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Question 1

$$(\dfrac{13}{2},\dfrac{7}{2},9)$$

$$(\dfrac{7}{2},\dfrac{13}{2},9)$$

$$(\dfrac{9}{2},\dfrac{7}{2},9)$$

$$(\dfrac{13}{2},\dfrac{9}{2},9)$$

Solution

Question 2

$$5$$

$$0$$

$$3$$

$$-1$$

Solution

Question 3

x- axis

y- axis

z- axis

yz-plane

Solution

Question 4

$$150\ sq. units$$

$$145\ sq. units$$

$$\sqrt {155}/2\ sq. units$$

$$155/2\ sq. units$$

Solution

Question 5

$$\left (\dfrac {1}{2}, 4\right )$$

$$\left (\dfrac {1}{3}, 24\right )$$

$$\left (2, \dfrac {1}{8}\right )$$

$$\left (3, \dfrac {1}{28}\right )$$

Solution

Question 6

$$(3, 4, 12)$$

$$\left (\dfrac {3}{13}, \dfrac {4}{13}, \dfrac {12}{13}\right )$$

$$(4, 5, 13)$$

$$(40, 53, 157)$$

Solution

Question 7

$$2$$

$$-2$$

$$-1$$

$$1$$

Solution

Question 8

$$30^o, 30^o, 30^o$$

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

$$30^o, 60^o, 90^o$$

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

Solution

Question 9

$$-2:3$$

$$2:3$$

$$-3:2$$

$$3:2$$

Solution

Question 10

Right -angled triangle

Isosceles triangle

Both $$(1)$$ & $$(2)$$

Equilateral triangle

Solution

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