Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

A $$= (1, 1, 4)$$ and B $$= (5,-3, 4)$$ are two points. If the points $$P$$, $$Q$$ are on the line AB such that AP $$=$$ PQ $$=$$ QB then PQ $$=$$

A
$$2\sqrt{2}$$

B
$$4$$

C
$$\sqrt{\displaystyle \frac{32}{9}}$$

D
$$\sqrt{2}$$

Solution

$$AB=\sqrt {(1-5)^2+(1+3)^2+(4-4)^2}$$
$$AB = \sqrt {(-4)^2+4^2}$$
$$AB = \sqrt {32}$$

Given that $$AP=PQ=QB$$
But, $$AB=AP+PQ+QB$$
Hence, $$AB=3 \times PQ$$
So, $$PQ = \cfrac {\sqrt {32}}{3} = \sqrt {\cfrac {32}{9}}$$
Question 2
1 / -0

If P $$(3, 2, -4)$$ , Q $$(5, 4, -6)$$ and R $$(9, 8, -10)$$ are collinear, then R divides PQ in the ratio

A
$$3 : 2$$ internally

B
$$3 : 2$$ externally

C
$$2 : 1$$ internally

D
$$2 : 1$$ externally

Solution

Let R divides PQ in ratio m:n.
By Section Formula
coordinates of R are:
$$R=(\dfrac{5m+3n}{m+n},\dfrac{4m+2n}{m+n},\dfrac{-6m-4n}{m+n})$$

we have coordinates of R=(9,8,-10)
Therefore,
$$9=\dfrac{5m+3n}{m+n}$$
$$9m+9n=5m+3n$$
$$-6n=4m$$
$$\dfrac{m}{n}=\dfrac{-3}{2}$$
Therefore R divides PQ in 3:2 ratio and negative sign indicates,it cuts Externally.
Therefore option (B) is Correct.
Question 3
1 / -0

The distance from the origin to the centroid of the tetrahedron formed by the points $$(0, 0, 0), (3, 0, 0), (0, 4, 0), (0, 0, 5)$$ is

A
$$\displaystyle \frac{\sqrt{\mathrm{3}+\mathrm{4}+\mathrm{5}}}{4}$$

B
$$\displaystyle \frac{\sqrt{\mathrm{3}+\mathrm{4}+\mathrm{5}}}{3}$$

C
$$\displaystyle \frac{\sqrt{\mathrm{3}^{2}+\mathrm{4}^{2}+\mathrm{5}^{2}}}{16}$$

D
$$\displaystyle \frac{\sqrt{\mathrm{3}^{2}+\mathrm{4}^{2}+\mathrm{5}^{2}}}{4}$$

Solution

$$G=(\dfrac{a}{4},\cfrac{b}{4},\cfrac{c}{4})$$
$$\therefore OG=\sqrt{(\cfrac{a^2}{16}+\cfrac{b^2}{16}+\cfrac{c^2}{16})}$$
Thus, here $$a=3,b=4,c=5$$
substituting in above equation we get
$$OG=\sqrt{\cfrac{3^2+4^2+5^2}{4}}$$
Question 4
1 / -0

The position vectors of the four angular point of a tetrahedron $$OABC$$ are $$(0, 0, 0)$$; $$(0, 0, 2)$$; $$(0, 4, 0)$$ and $$(6, 0, 0)$$ respectively. Find the coordinates of cenroid

A
$$\left (2, \displaystyle \frac {4}{3}, \displaystyle \frac {2}{3}\right )$$

B
$$\left (\displaystyle \frac {6}{4}, 1, \displaystyle \frac {2}{4}\right )$$

C
$$(0, 0, 0)$$

D
none of these

Solution

Angular points of tetrahedron OABC are $$(0,0,0),(0,0,2),(0,4,0),(6,0,0)$$
To find the coordinates of the centroid of the tetrahedron whose vertices are $$(x_1, y_1, z_1 ), (x_2, y_2, z_2 ), (x_3, y_3, z_3 )$$ and $$(x_4 , y_4 , z_4 ).$$
Hence, the centroid is
$$(\cfrac{x_1+x_2+x_3+x_4}{4}),(\cfrac{y_1+y_2+y_3+y_4}{4}),(\cfrac{z_1+z_2+z_3+z_4}{4})$$
Now, substituting the values we get
$$(\cfrac{0+0+0+6}{4}),(\cfrac{0+0+4+0}{4}),(\cfrac{0+2+0+0}{4})$$
$$\therefore $$ the coordinates of the centroid are : $$(\cfrac{6}{4},1,\cfrac{2}{4})$$
Question 5
1 / -0

Let $$A= \left ( 1,2,3 \right )B= \left ( -1,-2,-1 \right )C= \left ( 2,3,2 \right )$$ and $$ D= \left ( 4,7,6 \right )$$. Then $$ABCD$$ is a

A
rectangle

B
square

C
parallelogram

D
none of these

Solution

AB$${=}$$ $$\sqrt{{(-1-1)}^{2}+{(-2-2)}^{2}+{(-1-3)}^{2}}$$
AB$${=}$$ $$\sqrt{{(-2)}^{2}+{(-4)}^{2}+{(-4)}^{2}}$$
AB$${=}$$ $$\sqrt{36}$$
AB$${=}$$ 6
Similarly you find that BC$${=}$$ $$\sqrt{43}$$ CD$${=}$$ 6 and DA$${=}$$ $$\sqrt{43}$$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC$${=}$$ $$\sqrt{{(2-1)}^{2}+{(3-2)}^{2}+{(2-3)}^{2}}$$
AC$${=}$$ $$\sqrt{3}$$
similarly BD$${=}$$ $$\sqrt{155}$$
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
hence it is parallelogram
Question 6
1 / -0

If $$A= \left ( 5,-1,1 \right ),B= \left ( 7,-4,7 \right ),C= \left ( 1,-6,10 \right ),D= \left ( -1,-3,4 \right )$$. Then $$ABCD$$ is a

A
square

B
rectangle

C
rhombus

D
none of these

Solution

AB$${=}$$ $$\sqrt{{(7-5)}^{2}+{(-4+1)}^{2}+{(7-1)}^{2}}$$
AB$${=}$$ $$\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$$
AB$${=}$$ $$\sqrt{49}$$
AB$${=}$$ 7
Similarly you find that BC$${=}$$ $$\sqrt{49}$$ CD$${=}$$ 7 and DA$${=}$$7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC$${=}$$ $$\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$$
AC$${=}$$ $$\sqrt{122}$$
similarly BD$${=}$$ $$\sqrt{74}$$
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus
Question 7
1 / -0

The equation of the plane which is parallel to the $$xy-$$plane is

A
$$x=y$$

B
$$z=c$$

C
$$y=c$$

D
$$z=xy$$

Solution

The equation of a plane which is parallel to the $$xy-$$ plane is
$$z=c$$
Question 8
1 / -0

The distance of the point $$(1,-2,3)$$ from the plane $$x-y+z=5$$ measured parallel to the line $$\displaystyle \frac{x}{2}=\displaystyle \frac{y}{3}=\displaystyle \frac{z}{-6}$$ is

A
$$1$$

B
$$19$$

C
$$155$$

D
$$\sqrt{271}$$

Solution

Equation of the line through $$(1,-2,3)$$ parallel to the line $$\displaystyle \frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z}{ -6 } $$ is
$$\displaystyle \frac { x-1 }{ 2 } =\frac { y+2 }{ 3 } =\frac { z-3 }{ -6 } =r$$ (say) ...(1)
Then any point on (1) is $$(2r+1,3r-2,-6r+3)$$
If this point lies on the plane $$x-y+z=5$$ then $$(2r+1)-(3r-2)+(-6r+3)=5$$
$$\displaystyle \Rightarrow -7r+6=5\Rightarrow r=\frac { 1 }{ 7 } $$
Hence, the point is $$\displaystyle \left( \frac { 9 }{ 7 } ,\frac { -11 }{ 7 } ,\frac { 15 }{ 7 } \right) $$
Distance between $$(1,-2,3)$$ and $$\displaystyle \left( \frac { 9 }{ 7 } ,\frac { -11 }{ 7 } ,\frac { 15 }{ 7 } \right) $$
$$\displaystyle =\sqrt { \frac { 4 }{ 49 } +\frac { 9 }{ 49 } +\frac { 36 }{ 49 } } =\sqrt { \frac { 49 }{ 49 } } =1$$
Question 9
1 / -0

If $$C_1:{x^2+y^2}-20x+64=0$$ and $$C_2:{x^2+y^2}+30x+144=0$$. Then the length of the shortest line segment $$PQ$$ which touches $$C_1$$ at $$P$$ and to $$C_2$$ at $$Q$$ is

A
$$10$$

B
$$15$$

C
$$22$$

D
$$27$$

Solution

Given $$C_1 : x^2+y^2-20x+64=0 \Rightarrow (x-10)^2+y^2=36$$
and $$C_2 : x^2+y^2+30x+144=0 \Rightarrow (x+15)^2+y^2=81$$
So centre and radius of $$C_1$$ and $$C_2$$ are $$(10,0)$$, $$(-15,0)$$ and $$6,9$$ respectively.
Then, distance between $$C_1$$ and $$C_2$$ is $$\sqrt{(15+10)^{2}+(0-0)^{2}}=25$$.
$$PQ$$ touches $$C_{1}$$ at $$P$$ and $$C_{2}$$ at $$Q$$.
Then, shortest length of $$PQ$$ is $$=25-(9+6)=25-15=10$$
Question 10
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
$$\dfrac {1}{4}$$

C
$$3$$

D
$$\dfrac {1}{3}$$

Solution

$${\textbf{Step - 1: Using section formula.}}$$
$${\text{ Let given points (1,-1,5) and (2,3,4) be}}$$ $$(x_1,y_1,z_1)$$ $${\text{and}}$$ $$(x_2,y_2,z_2)$$
$${\text{The given ratios are}}$$ $$\lambda : 1$$ $${\text{be}}$$ $$m:n$$
$${\text{Section formula are:}}$$
$$\mathbf{\left( {\dfrac{{mx_2 + nx_1}}{{m + n}},\dfrac{{my_2 + ny_1}}{{m + n}},\dfrac{{mz_2 + nz_1}}{{m + n}}} \right)}$$
$$\text{putting values we get, } \left( {\dfrac{{2\lambda + 1}}{{\lambda + 1}},\dfrac{{3\lambda - 1}}{{\lambda + 1}},\dfrac{{4\lambda + 5}}{{\lambda + 1}}} \right)$$

$${\textbf{Step - 2: Taking y co-ordinate equals to zero}}$$
$${\text{Since, this points lies in XOZ plane then its y co-ordinate should be zero}}$$
$$ \Rightarrow$$ $$\dfrac{{3\lambda - 1}}{{\lambda + 1}} = 0$$
$$ \Rightarrow$$ $${{3\lambda - 1}}= 0$$
$$ \Rightarrow$$ $$3\lambda= 1$$
$$\Rightarrow$$ $$\lambda = \dfrac{1}{3}$$
$${\textbf{Hence, the correct option is (D)}}$$ $$\mathbf{\lambda = \dfrac{1}{3}}$$

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Re-Attempt Weekly Quiz Competition

Introduction to Three Dimensional Geometry Test - 21

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Introduction to Three Dimensional Geometry Test - 21

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SHARING IS CARING

Question 1

$$2\sqrt{2}$$

$$4$$

$$\sqrt{\displaystyle \frac{32}{9}}$$

$$\sqrt{2}$$

Solution

Question 2

$$3 : 2$$ internally

$$3 : 2$$ externally

$$2 : 1$$ internally

$$2 : 1$$ externally

Solution

Question 3

$$\displaystyle \frac{\sqrt{\mathrm{3}+\mathrm{4}+\mathrm{5}}}{4}$$

$$\displaystyle \frac{\sqrt{\mathrm{3}+\mathrm{4}+\mathrm{5}}}{3}$$

$$\displaystyle \frac{\sqrt{\mathrm{3}^{2}+\mathrm{4}^{2}+\mathrm{5}^{2}}}{16}$$

$$\displaystyle \frac{\sqrt{\mathrm{3}^{2}+\mathrm{4}^{2}+\mathrm{5}^{2}}}{4}$$

Solution

Question 4

$$\left (2, \displaystyle \frac {4}{3}, \displaystyle \frac {2}{3}\right )$$

$$\left (\displaystyle \frac {6}{4}, 1, \displaystyle \frac {2}{4}\right )$$

$$(0, 0, 0)$$

none of these

Solution

Question 5

rectangle

square

parallelogram

none of these

Solution

Question 6

square

rectangle

rhombus

none of these

Solution

Question 7

$$x=y$$

$$z=c$$

$$y=c$$

$$z=xy$$

Solution

Question 8

$$1$$

$$19$$

$$155$$

$$\sqrt{271}$$

Solution

Question 9

$$10$$

$$15$$

$$22$$

$$27$$

Solution

Question 10

$$-3$$

$$\dfrac {1}{4}$$

$$3$$

$$\dfrac {1}{3}$$

Solution

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