Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

Ratio in which the $$xy-$$plane divides the join of $$(1, 2, 3)$$ and $$(4, 2, 1)$$ is

A
$$3:1$$ internally

B
$$3:1$$ externally

C
$$1:2$$ internally

D
$$2:1$$ externally

Solution

Points $$A(1,2,3),B(4,2,1)$$
$$x-y$$ plane $$\Rightarrow z=0$$
$$\therefore$$ lines containing $$A,B$$ is $$\cfrac { x-1 }{ 3 } =\cfrac { y-2 }{ 0 } =\cfrac { z-3 }{ -2 } $$
point of intersection of line and $$xy$$ plane is $$(\cfrac { 11 }{ 2 } ,2,0)$$
$$\therefore (\cfrac { 4\lambda +1 }{ \lambda +1 } ,\cfrac { 2\lambda +2 }{ \lambda +1 } ,\cfrac { \lambda +3 }{ \lambda +1 } )=(\cfrac { 11 }{ 2 } ,2,0)\\ \Rightarrow \lambda =-3$$
Answer $$(B)$$
Question 2
1 / -0

The equation of the set of points which are equidistant from the points $$(1, 2, 3)$$ and $$(3, 2, -1)$$.

A
$$x-2z=0$$

B
$$2x-z=0$$

C
$$2x+y=0$$

D
$$x-2y=0$$

Solution

Let the given points be A($$1,2,3$$) and B($$3,2,-1$$) and let the point equidistant from A and B be P($$x,y,z$$)
then $$PA=PB$$
$$\sqrt {{{(x - 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}} = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} + {{(z + 1)}^2}} $$
Squaring both sides
$${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$$
$${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$$
$$-2x-6z+10=-6x+2z+10$$
$$4x-8z=0$$
$$x-2z=0$$
Question 3
1 / -0

$$A(3, 2, 0), B(5, 3, 2), C(-9, 6, -3)$$ are three points forming a triangle. If $$AD$$, the bisector of $$\angle BAC$$ meets $$BC$$ in $$D$$ then coordinates of $$D$$ are

A
$$\left (-\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$$

B
$$\left (\dfrac {19}{8}, -\dfrac {57}{16}, \dfrac {17}{16}\right )$$

C
$$\left (\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$$

D
None of these

Solution

According to question,
$$\begin{array}{l} AB=\sqrt { 4+1+4 } =3 \\ AC=\sqrt { 144+16+9 } =13 \\ BD:DC=AB:AC=3:13 \\ D=\left( { \dfrac { { 3(-9)+13(5) } }{ { 3+13 } } ,\dfrac { { 3(6)+13(3) } }{ { 3+13 } } ,\dfrac { { 3(-3)+13(2) } }{ { 3+13 } } } \right) \\ \, \, =\left( { \dfrac { { -38 } }{ { 16 } } ,\dfrac { { 57 } }{ { 16 } } ,\dfrac { { 17 } }{ { 16 } } } \right) \\ \, \, \, \, \, \, \, \, \, \therefore \, \, \, \left( { \dfrac { { -19 } }{ 8 } ,\dfrac { { 57 } }{ { 16 } } ,\dfrac { { 17 } }{ { 16 } } } \right) \\ So,the\, \, correct\, \, option\, \, is\, \, A. \end{array}$$
Question 4
1 / -0

$$A=\left(2,4,5\right)$$ and $$B=\left(3,5,-4\right)$$ are two points. If the $$xy$$-plane, $$yz$$-plane divide $$AB$$ in the ratios $$a:b,p:q$$ respectively then $$\dfrac{a}{b}+\dfrac{p}{q}$$=

A
$$\dfrac{7}{15}$$

B
$$\dfrac{-7}{12}$$

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

D
$$\dfrac{22}{25}$$

Solution

$$A(2,4,5),B(3,5,-4)$$
xy plane $$\Rightarrow z=0$$
yz plane $$\Rightarrow x=0$$
line through $$A$$ & $$B$$ $$\Rightarrow \cfrac { x-2 }{ 1 } =\cfrac { y-4 }{ 1 } =\cfrac { z-5 }{ -9 } =t$$
$$\therefore$$ let $$P(t+2,t+4,-9t+5)$$ be point of intersection of line with xy plane.
$$\therefore P=(\cfrac { 23 }{ 9 } ,\cfrac { 41 }{ 9 } ,0)$$
Similarly,for $$Q(t+2,t+4,9t+5)$$ be point of intersection of line with yz plane $$\Rightarrow t=-2$$.
$$Q(0,2,23)$$
(i)let $$P$$ divide $$AB$$ in $$1:\lambda (a:b)$$
$$\therefore (\cfrac { 23 }{ 9 } ,\cfrac { 41 }{ 9 } ,0)=(\cfrac { 2\lambda +3 }{ \lambda +1 } ,\cfrac { 4\lambda +5 }{ \lambda +1 } ,\cfrac { 5\lambda -4 }{ \lambda +1 } )\\ \Rightarrow \lambda =\cfrac { 4 }{ 5 } \\ \therefore \cfrac { a }{ b } =\cfrac { 5 }{ 4 } $$
(ii)let $$Q$$ divide $$AB$$ in $$\lambda:1 (p:q)$$
$$\therefore (0,2,23)=(\cfrac { 2+3\lambda }{ \lambda +1 } ,\cfrac { 4+5\lambda }{ \lambda +1 } ,\cfrac { 5-4\lambda }{ \lambda +1 } )\\ \Rightarrow \lambda =\cfrac { -2 }{ 3 } \\ \therefore \cfrac { p }{ q } =\cfrac { -2 }{ 3 } \\ \therefore \cfrac { a }{ b } +\cfrac { p }{ q } =\cfrac { 5 }{ 4 } -\cfrac { 2 }{ 3 } =\cfrac { 7 }{ 12 } $$
Question 5
1 / -0

If $$z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$$, then

A
$$\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$$

B
$$\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$$

C
$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$$

D
$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$$

Solution

if $$z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)$$ then.
As $$z=|z|e^{i arq (z)}$$
$$\therefore z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)=e^{i\left(\pi/6\right)}\quad [\because e^{i\theta}=\cos\theta+i\sin \theta]$$
$$\Rightarrow |z|=1, arq=\dfrac{\pi}{6}$$
Question 6
1 / -0

The x-coordinate of a point on the line joining the points $$P(2,2,1)$$ and $$Q(5,1,-2)$$ is $$4$$. Find its z-coordinate.

A
$$-1$$

B
$$-2$$

C
$$1$$

D
$$2$$

Solution

$$P(2,2,1),Q(5,1,-2)$$

$$\therefore$$ Equation of line through $$P$$ & $$Q$$,

$$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \\ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r$$

$$\therefore P$$ be point of line

$$\Rightarrow P\equiv (-3r+2,r+2,3r+1)$$

$$ \therefore -3r+2=4$$ ($$\because $$ since x-coordinate is $$4$$)

$$\Rightarrow r=\cfrac { -2 }{ 3 } $$

$$\therefore$$ z-coordinate $$=3r+1=-1$$
Question 7
1 / -0

Solve the following differential equation.
$$\dfrac{dy}{dx}=x-1$$.

A
$$y=x^2+x$$

B
$$y=x^2$$

C
$$y=x^2-x$$

D
None of the above

Solution

Given,
$$\dfrac{dy}{dx}=x-1$$.

Integrating on both sides

$$\displaystyle \int \dfrac {dy}{dx}=\int x-1\ dx$$

$$y=x^2-x +c$$
Question 8
1 / -0

The distance between (5,1,3) and the line x=3, y=7+t, z=1+t is

A
4

B
2

C
6

D
8

Solution

$$\begin{array}{l} x=3\, \, \, ,y=7+t\, \, ,z=1+t \\ A\left( { 3,7+t,1+t } \right) \\ 0.\left( { 3-5 } \right) +1\left( { 7+t-1 } \right) +1\left( { 1+t-3 } \right) =0 \\ 6+t+t-2=0 \\ 2t=-4 \\ t=-2 \\ A:\left( { 3,5,-1 } \right) \\ dis\tan ce=\sqrt { 4+16+16 } \\ =6 \\ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
Question 9
1 / -0

The perimeter of triangle with vertices at $$(1,0,0) , (0,1,0) and (0,0,1)$$ is :

A
$$3$$

B
$$2$$

C
$$2\sqrt {2}$$

D
$$3\sqrt {2}$$

Solution

REF.Image
$$AB=\sqrt{1+1+0}=\sqrt{2}$$
$$BC=\sqrt{0+1+1}=\sqrt{2}$$
$$AC=\sqrt{1+0+1}=\sqrt{2}$$
Hence, perimeter of $$\triangle ABC$$ is given by
$$AB+BC+AC= 3\sqrt{2}$$
Question 10
1 / -0

If the points $$A(9, 8, -10), B(3, 2, -4)$$ and $$C(5, 4, -6)$$ be collinear, then the point $$C$$ divides the line $$AB$$ in the ratio

A
$$2:1$$

B
$$3:1$$

C
$$1:2$$

D
$$-1:2$$

Solution

REF.Image
Let the ratio be K:1
Then, using section formula
we have
$$C(5,4,-6)=\frac{K(3,2,-4)+(9,8,-10)}{K+1}$$
$$\Rightarrow 5=\frac{3K+9}{K+1}$$
$$\Rightarrow 5K= 3K+4$$
$$\Rightarrow 2K=4$$
$$K=2$$
$$\Rightarrow $$ C divides AB in ratio 2:1

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Introduction to Three Dimensional Geometry Test - 12

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Introduction to Three Dimensional Geometry Test - 12

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Question 1

$$3:1$$ internally

$$3:1$$ externally

$$1:2$$ internally

$$2:1$$ externally

Solution

Question 2

$$x-2z=0$$

$$2x-z=0$$

$$2x+y=0$$

$$x-2y=0$$

Solution

Question 3

$$\left (-\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$$

$$\left (\dfrac {19}{8}, -\dfrac {57}{16}, \dfrac {17}{16}\right )$$

$$\left (\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$$

None of these

Solution

Question 4

$$\dfrac{7}{15}$$

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

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

$$\dfrac{22}{25}$$

Solution

Question 5

$$\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$$

$$\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$$

$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$$

$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$$

Solution

Question 6

$$-1$$

$$-2$$

$$1$$

$$2$$

Solution

Question 7

$$y=x^2+x$$

$$y=x^2$$

$$y=x^2-x$$

None of the above

Solution

Question 8

4

2

6

8

Solution

Question 9

$$3$$

$$2$$

$$2\sqrt {2}$$

$$3\sqrt {2}$$

Solution

Question 10

$$2:1$$

$$3:1$$

$$1:2$$

$$-1:2$$

Solution

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