Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 32

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Question 1
1 / -0

The ratio in which the plane $$x - 2y + 3z = 17$$ divides the line joining $$(-2, 4, 7)$$ and (3, -5, 8) is

A
$$1:2$$

B
$$3 : 1$$

C
$$3 : 10$$

D
$$10 : 1$$

Solution
Question 2
1 / -0

In the $$\Delta ABC$$ $$AB = \sqrt 2 ,AC = \sqrt {20} ,$$ B=$$(3,2,0)$$ and C=$$(0,1,4)$$ then the length of the median passing through A is

A
$$\dfrac{3}{2}$$

B
$$\dfrac{9}{2}$$

C
$$\dfrac{3}{{\sqrt 2 }}$$

D
$$\dfrac{{\sqrt 3 }}{2}$$

Solution

$$D=$$ mid point of $$BC$$ (since $$BD$$ is median)
$$D=\left(\dfrac{3+0}{2}, \dfrac{2+1}{2}, \dfrac{4+0}{2}\right)=\left(\dfrac{3}{2}, \dfrac{3}{2}, 2\right)$$
$$BD=1/2BC=\dfrac{1}{2}\sqrt{(3-0)^2+(2-1)^2+(0-4)^2}$$
$$=\dfrac{1}{2}\sqrt{9+1+16}$$
$$BD=\dfrac{1}{2}\sqrt{26}$$
using appollonius theorem :
$$AB^2+AC^2=2(AD^2+(BD)^2)$$
$$\Rightarrow AD^2=\dfrac{1}{2}[AB^2+AC^2]-(BD)^2$$
$$AD^2=\dfrac{1}{2}[(\sqrt{2})^2+(\sqrt{20})^2]0-(BD)^2$$
$$=\dfrac{1}{2}[22]-\dfrac{26}{4}=9/2$$
$$\therefore AD^2=\dfrac{18}{4}\Rightarrow AD=\dfrac{3}{2}\sqrt{2}=\dfrac{3}{\sqrt{2}}$$
$$\therefore$$ length of median through $$A=3/\sqrt{2}\ units$$.
Question 3
1 / -0

If a point C with y coordinate 2, lies on the line joining the points A(-1, -4, 5) and B(4, 6, -5), then find the coordinates of C.

A
(2, 2, -1)

B
(2, 1, 2)

C
(2, 0, 0)

D
(0, 2, 4)

Solution

Line joining $$A(-1,-4,5)$$ and $$B(4,6,-5)$$ is
$$\cfrac{x-4}{-1-4}=\cfrac{y-6}{-4-6}=\cfrac{z-(-5)}{5-(-5)}\\ \cfrac{x-4}{-5}=\cfrac{y-6}{-10}=\cfrac{z+5}{10}\\ \Rightarrow\cfrac{x-4}{1}=\cfrac{y-6}{2}=\cfrac{z+5}{-2}=r$$
$$\therefore$$ general c-ordinates of a point on this line is,
$$(x,y,z)=(r+4,2r+6,-2r-5)$$
y co-ordinates is given as$$2\Rightarrow y=2\\ \therefore 2r+6=2\\\Rightarrow r=-2$$
$$\therefore (x,y,z)=(2,2,-1)$$
Answer A
Question 4
1 / -0

The locus of a point P which moves such that $$PA^2-PB^2=2k^2$$ where A and B are $$(3, 4, 5)$$ and $$(-1, 3, -7)$$ respectively is

A
$$8x+2y+24z-9+2k^2=0$$

B
$$8x+2y+24z-2k^2=0$$

C
$$8x+2y+24z+9+2k^2=0$$

D
$$8x-2y+24z-2k^2=0$$

Solution

$$P{A}^{2}-P{B}^{2}=2{k}^{2}$$
$$\Rightarrow \left[{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z-5\right)}^{2}\right]-\left[{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z+7\right)}^{2}\right]=2{k}^{2}$$
$$\Rightarrow \left(x-3+x+1\right)\left(x-3-x-1\right)+\left(y-4+y-3\right)\left(y-4-y+3\right)+\left(z-5+z+7\right)\left(z-5-z-7\right)=2{k}^{2}$$
$$\Rightarrow \left(2x-2\right)\left(-4\right)+\left(2y-7\right)\left(-1\right)+\left(2z+1\right)\left(-12\right) =2{k}^{2}$$
$$\Rightarrow -8x+8-2y+7-24z-24=2{k}^{2}$$
$$\Rightarrow -8x-2y-24z-9=2{k}^{2}$$
$$\Rightarrow 8x+2y+24z+9+2{k}^{2}=0$$
Question 5
1 / -0

The perimeter of the triangle formed by the points $$(1,0,0),(0,1,0),(0,0,1)$$ is

A
$$\sqrt 2 $$

B
$$2\sqrt 2 $$

C
$$3\sqrt 2 $$

D
$$4\sqrt 2 $$

Solution

Given points $$A(1, 0, 0), B(0, 1, 0), C(0, 0, 1)$$ is
$$AB=\sqrt{1+1}=\sqrt{2}$$
$$BC=\sqrt{1+1}=\sqrt{2}$$
$$CA=\sqrt{1+1}=\sqrt{2}$$
Perimeter of the triangle is $$AB+BC+CA=\sqrt{2}+\sqrt{2}+\sqrt{2}=3\sqrt{2}$$.
Question 6
1 / -0

The distance between the points $$P(x,\,-1)$$ and $$Q(3,\,2)$$ is $$5$$ units. Find the value of $$x$$.

A
$$2,8$$

B
$$-2,9$$

C
$$1,8$$

D
$$-1,7$$

Solution

Distance=$$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}$$

P$$=(x_1,y_1)=(x,-1)$$

Q$$=(x_2,y_2)=(3,2)$$

$$25=(3-x)^2+(2+1)^2$$

$$25=(9+x^2-6x+9)$$

$$x^2-6x-7=0$$

$$x^2-7x+x-7=0$$

$$x(x-7)+1(x-7)=0$$

$$x=-1,7$$
Question 7
1 / -0

$$P$$ and $$Q$$ are points on the line joining $$A(-2,5)$$ and $$B(3,1)$$ such that $$AP=PQ=QB$$. Then, the distance of the midpoint of $$PQ$$ from the origin is

A
$$3$$

B
$$\frac {\sqrt 37}{4}$$

C
$$4$$

D
$$3.5$$

Solution
Question 8
1 / -0

If A=(1,-2,-1) B=(4,0,-3)C=(1,2,-1),D=(2,-4,-5) Find the distance between AB and CD lines.

A
40

B
0

C
80

D
20

Solution
Question 9
1 / -0

Find the co ordinates of points which trisect the line segment joining $$( 1 , - 2 )$$ and $$( - 3,4 )$$

A
$$\left( - \frac { 1 } { 3 } , 0 \right) \quad$$ and $$\left( - \frac { 5 } { 3 } , 2 \right)$$

B
$$\left( \frac { 1 } { 3 } , 0 \right)$$ and $$\left( - \frac { 5 } { 3 } , 2 \right)$$

C
$$\left( - \frac { 1 } { 3 } , 0 \right)$$ and $$\left( \frac { 5 } { 3 } , 2 \right)$$

D
None of the above

Solution
Question 10
1 / -0

If G is the centroid of $$\triangle ABC$$ and BC = 3, CA = 4, AB = 5 then BG =

A
$$\dfrac { \sqrt { 73 } }{ 3 } $$

B
$$\dfrac { \sqrt { 13 } }{ 3 } $$

C
$$\dfrac { \sqrt { 52 } }{ 3 } $$

D
$$\dfrac { \sqrt { 26 } }{ 3 } $$

Solution