Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

If two vertices of an equilateral triangle are $$(2, 1, 5)$$ and $$(3, 2, 3)$$, then its third vertex is:

A
$$(1, 2, 4)$$

B
$$(4, 0, 4)$$

C
$$(0, -4, 4)$$

D
$$(4, 4, 1)$$

Solution

Let the coordinates of third vertex be $$(x,y,z)$$
The distance of third vertex from above two vertices are same
So, After distance form between $$(2,1,5)\ and (x,y,z)$$ {and equate it with distance between $$(3,2,3)\ and (x,y,z)$$. After simplification we get
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }-4x-2y-10z+30={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }-6x-4y-6z+22$$
$$\Rightarrow x+y-2z+4=0$$
Among given options , only $$(4,0,4)$$ satisfies above equation
Therefore the correct option is $$B$$
Question 2
1 / -0

The point which is equidistant from the points $$(-1, 1, 3), (2, 1, 2), (0, 5, 6)$$ and $$(3,2, 2)$$ is

A
$$(-1, 3, 4)$$

B
$$(3, 1, 4)$$

C
$$(1, 3, 4)$$

D
$$(4,1, 3)$$

Solution

Let $$(x,y,z)$$ be the points equidistant from the given points by the distance $$d$$.
Then, we have the equations
$$(x+1)^2+(y-1)^2+(z-3)^2=d^2$$
$$(x-2)^2+(y-1)^2+(z-2)^2=d^2 $$
$$x^2+(y-5)^2+(z-6)^2=d^2$$
$$(x-3)^2+(y-2)^2+(z-2)^2=d^2$$
Using the equation 1 and 2 we get,
$$z=3x+1$$
And using equations 2 and 4 we get,
$$y=4-x$$.
Substituting these values in equation 1 and 3 we get,

$$x^2+2x+1+9-6x+x^2+9x^2-12x+4$$
$$=x^2+x^2+2x+1+9x^2-30x+25$$
$$\Rightarrow x=1$$.
Hence, the point is given by $$(x,4-x,3x+1)=(1,3,4)$$.
Question 3
1 / -0

The point which is equidistant from the points $$(a, 0, 0), (0, b, 0), (0, 0, c)$$ and $$(0, 0, 0)$$ is:

A
$$(a,b,c)$$

B
$$(\sqrt{{a}},\sqrt{{b}},\sqrt{{c}})$$

C
$$(2a,2b,2c)$$

D
$$\left (\dfrac {a}{2}, \dfrac {b}{2}, \dfrac {c}{2}\right)$$

Solution

Let the point $$(x,y,z)$$ be the equidistant from $$(a,0,0), (0,b,0), (0,0,c)$$ and $$(0,0,0)$$.
$$\Rightarrow (x-a)^2+(y-0)^2+(z-0)^2=(x-0)^2+(y-b)^2+(z-0)^2$$
$$=(x-0)^2+(y-0)^2+(z-c)^2=(x-0)^2+(y-0)^2+(z-0)^2$$
Solving above equations, we get $$x=\dfrac{a}{2}, y=\dfrac{b}{2}, z=\dfrac{c}{2}$$
Question 4
1 / -0

If the extremities of a diagonal of a square are $$(1, -2, 3)$$ and $$(2, -3, 5)$$, then the length of its side is:

A
$$\sqrt{6}$$

B
$$\sqrt{3}$$

C
$$\sqrt{5}$$

D
$$\sqrt{7}$$

Solution

The extremities of a diagonal of a square are given by $$(1,-2,3)$$ and $$(2,-3,5)$$.
The length of this diagonal will be given by $$ \sqrt {({1}^{2} + {1}^{2} + {2}^{2} )} = \sqrt{6 }$$
Length of a side $$ \times \sqrt{2} = $$ Length of diagonal
So, the length of the side of the square will be given by $$ = \sqrt {3 } $$
Question 5
1 / -0

If $$A, B$$ are the feet of the perpendiculars from $$(2, 4, 5)$$ to the $$x$$-axis, $$y$$-axis respectively, then the distance $$AB$$ is

A
$$2\sqrt{5}$$

B
$$\sqrt{29}$$

C
$$\sqrt{41}$$

D
$$3\sqrt{5}$$

Solution

Since the point is given by $$(2,4,5)$$ and $$A, B$$ are the projections on the $$x$$ and $$y$$ axes respectively,
$$A = (2,0,0) , B = (0,4,0)$$
The distance $$AB$$ is given by $$ \sqrt {({2}^{2} + {4}^{2})} = \sqrt{20} = 2\sqrt {5} $$
Question 6
1 / -0

If $$A(0, 4, 1), B(a, b, c), C(4, 5, 0), D(2, 6, 2)$$ are the consecutive vertices of a square, then the distance $$BD$$ is:

A
$$\sqrt{34}$$

B
$$6$$

C
$$18$$

D
$$3\sqrt{2}$$

Solution

Given vertices of square are $$A(0,4,1),B(a,b,c),C(4,5,0),D(2,6,2)$$.
The diagonals of a square are equal in lengths.
Length of diagonal $$BD =$$ Length of diagonal $$AC$$
$$BD = \sqrt {({4}^{2} + {1}^{2} + {1}^{2})} = 3\sqrt{2} $$
Question 7
1 / -0

If $$(p, q, r)$$ is equidistant from $$(1, 2, -3), (2, -3, 1)$$ and $$(-3, 1, 2)$$, then $$p +q + r$$ $$=$$

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$2$$

Solution

$$\textbf{Step 1: Check the type of the triangle formed by the given points.}$$

$$\text{The given points are }\mathrm{A(1,2,-3),B(2,-3,1)\ and\ C(-3,1,2).}$$

$$\text{We know, distance between }\mathrm{(x_1,y_1,z_1)\ and\ (x_2,y_2,z_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}$$

$$\text{Distance between A and B: }$$
$$\mathrm{AB=\sqrt{(2-1)^2+(-3-2)^2+(1+3)^2}}$$
$$\mathrm{=\sqrt{1+25+16}}$$
$$\mathrm{=\sqrt{42}}$$

$$\text{Distance between B and C: }$$
$$\mathrm{BC=\sqrt{(-3-2)^2+(1+3)^2+(2-1)^2}}$$
$$\mathrm{=\sqrt{25+16+1}}$$
$$\mathrm{=\sqrt{42}}$$

$$\text{Distance between C and A: }$$
$$\mathrm{CA=\sqrt{(-3-1)^2+(1-2)^2+(2+3)^2}}$$
$$\mathrm{=\sqrt{16+1+25}}$$
$$\mathrm{=\sqrt{42}}$$

$$\mathrm{\therefore AB=BC=CA }$$
$$\text{So, triangle ABC is an equilateral triangle.}$$

$$\textbf{Step 2: Find the required value using suitable property.}$$

$$\text{Given that, (p,q,r) is equidistant from the points A, B and C.}$$
$$\therefore\text{(p,q,r) is the circumcentre of triangle ABC.}$$
$$\text{We know, for equilateral triangles circumference and centroid are same point.}$$
$$\text{Centroid of a triangle with vertices } \mathrm{(x_1,y_1),(x_2,y_2)\ and\ (x_3,y_3)=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)}$$

$$\therefore\text{Centroid of triangle ABC } \mathrm{=\left(\dfrac{1+2-3}{3},\dfrac{2-3+1}{3},\dfrac{-3+1+2}{3}\right)=(0,0,0)}$$

$$\text{So, circumcentre of triangle ABC } \mathrm{=(0,0)}$$

$$\therefore\mathrm{(p,q,r)=(0,0,0)}$$

$$\mathrm{Thus,\ p+q+r=0+0+0=0}$$

$$\textbf{Hence, the correct option is C.}$$
Question 8
1 / -0

$$A = (1, -1, 2)$$ and $$B =$$ $$(2, 3, 7)$$ are two points. lf $$P,\ O$$ divide $$AB$$ in the ratios $$2:3, -2:3$$ respectively then $$P_x+Q_y=$$

A
$$\displaystyle \frac{-38}{5}$$

B
$$\displaystyle \frac{38}{5}$$

C
$$\displaystyle \frac{-2}{5}$$

D
$$\displaystyle \frac{-47}{6}$$

Solution

P divides line joining $$A(1,-1,2)$$ and $$B(2,3,7)$$ in the ratio $$2:3$$
$$\therefore P_x = \cfrac{2 \times 2 + 3 \times 1}{2 + 3} = \cfrac{7}{5}$$
Similarly, Q divides line joining $$A(1,-1,2)$$ and $$B(2,3,7)$$ in the ratio $$-2:3$$
$$\therefore Q_y = \cfrac{-2 \times 3 + 3 \times -1}{-2 + 3} = -9$$
$$\Rightarrow P_x + Q_y = -9 + \cfrac{7}{5} = \cfrac{-38}{5}$$
Question 9
1 / -0

$$\mathrm{If} \mathrm{A}=(1, 2, 3)$$ , $$\mathrm{B}=(2,3, 4)$$ and $$\mathrm{C}$$ is a point of trisection of$$\mathrm{A}\mathrm{B}$$ such that $$\displaystyle \mathrm{C}_{\mathrm{x}}+\mathrm{C}_{\mathrm{y}}=\frac{13}{3}$$ then $$\mathrm{C}_{\mathrm{z}}=$$

A
$$\displaystyle \dfrac{10}{3}$$

B
$$\displaystyle \dfrac{11}{3}$$

C
$$\displaystyle \dfrac{11}{2}$$

D
$$11$$

Solution

Given : $$A=(1,2,3), B=(2,3,4)$$

$$C$$ is a point of trisection of $$AB$$

$$\implies C=\dfrac{A+2B}{3}$$.............($$\because$$ this cuts the line in the ratio of $$1:2$$)

$$\implies C=\dfrac{(1,2,3)+2(2,3,4)}{3}$$

$$\implies C=\dfrac{(1+4,2+6,3+8)}{3}$$

$$\implies C=\left(\dfrac{5}{3},\dfrac{8}{3},\dfrac{11}{3}\right)$$

$$\therefore\ C_x=\dfrac{5}{3}, C_y=\dfrac{8}{3}, C_z=\dfrac{11}{3}$$

Now, $$C_x + C_y=\dfrac{5}{3}+\dfrac{8}{3}=\dfrac{13}{3}$$

Thus, $$C_z=\dfrac{11}{3}$$
Hence, option B is correct.
Question 10
1 / -0

$$A=(2, 4, 5)$$ and $$B=(3,5, -4)$$ are two points. lf the $$XY$$-plane, $$YZ$$-plane divide $$AB$$ in the ratio $$a:b$$ and $$ p:q$$ respectively, then $$\dfrac {a}{b}+\dfrac {p}{q}=$$

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

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

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

D
$$\displaystyle \dfrac{-22}{15}$$

Solution

Given points are $$ A(2,4,5)$$ and $$B(3,5,-4)$$.
Let the $$XY$$ plane divides the line $$AB$$ in the ratio $$\lambda:1$$.
Then the $$z$$ coordinate of $$XY$$ plane is zero.
$$0=5-4(\lambda)$$
$$\lambda=\dfrac {5}{4}=\dfrac {a}{b}$$
$$YZ$$ plane divides the line $$AB$$ such that x coordinate is zero.
$$0=3(\alpha)+2$$
$$\alpha=\dfrac {-2}{3}=p:q$$
Now, $$\dfrac {a}{b}+\dfrac {p}{q}=\dfrac {5}{4}-\dfrac {2}{3}=\dfrac {7}{12}$$
Hence, option C is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Three Dimensional Geometry Test - 14

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Introduction to Three Dimensional Geometry Test - 14

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SHARING IS CARING

Question 1

$$(1, 2, 4)$$

$$(4, 0, 4)$$

$$(0, -4, 4)$$

$$(4, 4, 1)$$

Solution

Question 2

$$(-1, 3, 4)$$

$$(3, 1, 4)$$

$$(1, 3, 4)$$

$$(4,1, 3)$$

Solution

Question 3

$$(a,b,c)$$

$$(\sqrt{{a}},\sqrt{{b}},\sqrt{{c}})$$

$$(2a,2b,2c)$$

$$\left (\dfrac {a}{2}, \dfrac {b}{2}, \dfrac {c}{2}\right)$$

Solution

Question 4

$$\sqrt{6}$$

$$\sqrt{3}$$

$$\sqrt{5}$$

$$\sqrt{7}$$

Solution

Question 5

$$2\sqrt{5}$$

$$\sqrt{29}$$

$$\sqrt{41}$$

$$3\sqrt{5}$$

Solution

Question 6

$$\sqrt{34}$$

$$6$$

$$18$$

$$3\sqrt{2}$$

Solution

Question 7

$$-1$$

$$1$$

$$0$$

$$2$$

Solution

Question 8

$$\displaystyle \frac{-38}{5}$$

$$\displaystyle \frac{38}{5}$$

$$\displaystyle \frac{-2}{5}$$

$$\displaystyle \frac{-47}{6}$$

Solution

Question 9

$$\displaystyle \dfrac{10}{3}$$

$$\displaystyle \dfrac{11}{3}$$

$$\displaystyle \dfrac{11}{2}$$

$$11$$

Solution

Question 10

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

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

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

$$\displaystyle \dfrac{-22}{15}$$

Solution

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