Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Algebraic sum of intercepts made by the plane x+3y-4z+6=0 on the axes is

A
7

B
0

C
$$\frac{13}{2}$$

D
$$-\frac{13}{2}$$

Solution

Intercepts made by plane $$x+3y- 4z+6=0$$ on
$$x, y$$ and $$z-$$axis is found by putting $$(1,0,0), (0,1,0)$$ and $$(0,0,1)$$ in equation respectively
Then $$x-$$ intercept $$= -6$$
$$y-$$ intercept $$= -2$$
$$z-$$ intercept $$= \dfrac{3}{2}$$
Algebraic sum $$= -6-2+\dfrac{3}{2}$$
$$=-\dfrac{13}{2}$$
Question 2
1 / -0

If $$OA$$ is equally inclined to $$OX, OY$$ and $$OZ$$ and if $$A$$ is $$\sqrt 3 $$ units from the origin then $$A$$ is

A
$$\left( {3,3,3} \right)$$

B
$$\left( { - 1,\,1,\, - 1} \right)$$

C
$$\left( { - 1,\,1,\,1} \right)$$

D
$$\left( {1,\,1,\,1} \right)$$

Solution

By hit and tried method $$(1,1,1)=\sqrt {1^2+1^2+1^2}=\sqrt 3$$
and $$\cos \theta_1 =\cos \theta_2 =\cos \theta_3 =\dfrac {1}{\sqrt {3}}$$
Question 3
1 / -0

The distance between the points $$(-1,2,3)$$ and P is $$13$$. Then $$P=$$

A
$$(2,6,-9)$$

B
$$(-2,6,9)$$

C
$$(2,6,9)$$

D
$$(2,-6,9)$$

Solution

let $$A(-1,2,3)$$
suppose 'P' is (2,6,-9)
$$PA=\sqrt{(2+1)^{2}+(6-2)^{2}+(-9-3)^{2}}$$
$$=\sqrt{9+16+144}$$
$$=\sqrt{169}$$
$$=13$$
Hence, 'P' is (2,6,-9)
Question 4
1 / -0

The distance between the parallel planes given by the equations, $$\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})+3=0$$ and $$\vec{r}.(4\hat{i}-4\hat{j}+2\hat{k})+5=0$$ is-

A
$$1/2$$

B
$$1/3$$

C
$$1/4$$

D
$$1/6$$

Solution

Planes are $$2i-2j+k+3=0,4i-4j+2k+5=0,2i-2j+k+5/2=0$$
distance between them is $$=\cfrac{|c_1-c_2|}{\sqrt{a^2+b^2+c^2}}\\=\cfrac{|3-\cfrac{5}{2}|}{\sqrt{2^2+2^2+1^2}}=\cfrac{1}{6}$$
Question 5
1 / -0

The number of octants in which $$Z$$ coordinate is positive is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$1$$

Solution

$$Z$$- Coordinates are positive in Octant $$1, 2, 3$$ and $$4$$ .
Hence, then the number of octants are $$4$$.
Question 6
1 / -0

If $$\overline { a } ,\overline { b } $$ are the position vectors of $$A$$ and $$B$$ then one of the following points lie on $$\overline { AB } $$

A
$$\cfrac{2(\overline { a } +\overline { b } )}{3}$$

B
$$\cfrac{(\overline { a } -\overline { b } )}{3}$$

C
$$\cfrac{(\overline { a } +\overline { b } )}{3}$$

D
$$\cfrac{2\overline { a } +2\overline { b } }{3}$$

E
None of these

Solution

$$ \rightarrow $$ Point lying on $$ \overrightarrow{AB} $$ would divide $$ \overrightarrow{AB} $$
in some ratio, say $$ \lambda : 1 $$ internally
or externally thus it would be of
form $$ \frac{\lambda \bar{a}+\bar{b}}{\lambda +1} $$ (int) or $$ \frac{\lambda \bar{a}-\bar{b}}{\lambda -1} $$ (ext.)(section for.)
$$(A) \frac{2\bar{a}+2\bar{b}}{3} (\times ) 2+2\neq 4 $$ $$\left \{ In\, formula,\lambda +1 = \lambda +1 \,or\, \lambda -1 = \lambda -1 \right \} $$
$$ (B) \frac{\bar{a}+\bar{b}}{3} = \frac{\bar{a}-\bar{b}}{4-1} $$ $$\lambda $$ is 1 and $$ 4 (\times) $$
$$(C) \frac{\bar{a}+\bar{b}}{3} (\times) 1+1 \neq 3 $$
$$(D) \frac{2\bar{a}+2\bar{b}}{3} (\times) $$ same as (A)
$$ \Rightarrow $$ None option satisfy
Question 7
1 / -0

The distance between the points(4,3,7) and (1,-1,-5) is

A
7

B
12

C
13

D
25

Solution
Question 8
1 / -0

If $$L, M$$ are the feet of the perpendiculars from $$(2, 4, 5)$$ to the $$xy$$-plane, $$yz$$-plane respectively, then the distance $$LM$$ is:

A
$$\sqrt{41}$$

B
$$\sqrt{20}$$

C
$$\sqrt{29}$$

D
$$3\sqrt{5}$$

Solution

The coordinates of $$L$$ will be $$(2,4,0)$$, since it is the foot of the perpendicular from $$(2,4,5)$$ to $$xy$$ plane.
The coordinates of $$M$$ will be $$(0,4,5)$$, since it is the foot of the perpendicular from $$(2,4,5)$$ to $$yz$$ plane.
Applying distance formula, we get
$$\sqrt{(2-0)^{2}+(4-4)^{2}+(0-5)^{2}}$$
$$=\sqrt{4+25}$$
$$=\sqrt{29}$$
Question 9
1 / -0

The name of the figure formed by the points $$(0, 0, 0), (1, 0, 1)$$ and $$(0, 1, 1)$$ is

A
a straight line

B
an isosceles triangle

C
an equilateral triangle

D
a scalene triangle

Solution

The distance between the points $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2)$$ is $$\sqrt {(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$.
The distance between the origin and $$(x_1, y_1, z_1)$$ is $$\sqrt {x_1^2+y_1^2+z_1^2}$$.

lets take A $$(1,0,1)$$ and B$$(0,1,1)$$
and there is origin $$O(0,0,0)$$
length of line segment $$AO=\sqrt{2}=BO$$

length of line segment $$AB=\sqrt{1+1+0}=\sqrt{2}$$

since $$ AO=BO=AB \implies$$ All sides are equal $$\implies$$ equilateral triangle
Question 10
1 / -0

The name of the figure formed by the points $$(3, -5, 1), (-1, 0, 8)$$ and $$(7, -10, -6)$$ is

A
a triangle

B
a straight line

C
an isosceles triangle

D
an equilateral triangle

Solution

Let $$A=(3,-5,1), B = (-1,0,8)$$ and $$C=(7,-10,-6)$$
Now,
$$AB =\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}=\sqrt {90}=3\sqrt{10}$$,
$$BC =\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}=\sqrt{360}=6\sqrt{10}$$
and $$CA =\sqrt{(7-3)^2+(-10+5)^2+(-6-1)^2}=\sqrt{90}=3\sqrt{10}$$
Clearly $$BC = AB+CA$$
$$\therefore $$ given points lies on straight line.

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Introduction to Three Dimensional Geometry Test - 13

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Introduction to Three Dimensional Geometry Test - 13

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Question 1

7

0

$$\frac{13}{2}$$

$$-\frac{13}{2}$$

Solution

Question 2

$$\left( {3,3,3} \right)$$

$$\left( { - 1,\,1,\, - 1} \right)$$

$$\left( { - 1,\,1,\,1} \right)$$

$$\left( {1,\,1,\,1} \right)$$

Solution

Question 3

$$(2,6,-9)$$

$$(-2,6,9)$$

$$(2,6,9)$$

$$(2,-6,9)$$

Solution

Question 4

$$1/2$$

$$1/3$$

$$1/4$$

$$1/6$$

Solution

Question 5

$$2$$

$$3$$

$$4$$

$$1$$

Solution

Question 6

$$\cfrac{2(\overline { a } +\overline { b } )}{3}$$

$$\cfrac{(\overline { a } -\overline { b } )}{3}$$

$$\cfrac{(\overline { a } +\overline { b } )}{3}$$

$$\cfrac{2\overline { a } +2\overline { b } }{3}$$

None of these

Solution

Question 7

7

12

13

25

Solution

Question 8

$$\sqrt{41}$$

$$\sqrt{20}$$

$$\sqrt{29}$$

$$3\sqrt{5}$$

Solution

Question 9

a straight line

an isosceles triangle

an equilateral triangle

a scalene triangle

Solution

Question 10

a triangle

a straight line

an isosceles triangle

an equilateral triangle

Solution

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