MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 1

Distance between

$A(4,5,6)$ from origin

$O$ is

0%
$25\sqrt3$
0%
$\sqrt {77}$
0%
$3\sqrt5$
0%
Data Insufficient

Explanation

Origin is

$O(0,0,0)$ and given point is

$A(4,5,6)$

So, distance

$=$

$\sqrt {(4-0)^2+(5-0)^2+(6-0)^2}$

$=\sqrt {4^2 + 5^2 + 6^2} = \sqrt {77}$

In the given figure, co-ordinates of the midpoint of

$AB$ are

0%
$(0,2)$
0%
$(0,3)$
0%
$(1,2)$
0%
$(3,1)$

Explanation

Here we are given two points A(0, 0) and B(6, 2)
Since the midpoint falls midway between A and
B, it has x-co-ordinates of A and B
x-co-ordinates of midpoint of

$\displaystyle AB=\frac{0+6}{2}=3$
and y-co-ordinates of Midpoint of

$\displaystyle AB=\frac{0+2}{2}=1$
Thus the midpoint of AB has co-ordinates (3, 1)

Name three undefined terms.

0%
Point
0%
Line
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Plane
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All of the above

State the following statement is True or False

If two distinct lines are intersecting each other in a plane then they cannot have more than one point in common.

0%
True
0%
False

Explanation

suppose lines

$l1$ and

$l2$ intersect at two disticnt points say P and Q.Then

$l1$ contains points P and Q.

Also,

$l2$ contains points P and Q.

So two lines

$l1$ and

$l2$ pass through two distinct points P and Q.

But only one line can pass through two different points. (axiom 3)

Hence, two lines cannot have more than one point in common.

$\therefore$ Option A is correct.

Distance between the points

$(12,4,7)$ and

$(10,5,3)$ is

0%
$\sqrt{21}$
0%
$\sqrt{5}$
0%
$\sqrt{17}$
0%
none of these

Explanation

Consider the problem,

Let the given points

$A(12,4,7)$ and

$B(10,5,3)$

So, distance between

$A$ and

$B$ by distance formula.

$AB=\sqrt{(10-12)^2+(5-4)^2+(3-7)^2}=\sqrt{(-2)^2+1^2+(-4)^2}$

$=\sqrt{4+1+16}=\sqrt{21}$

So, distance between the points

$(12,4,7)$ and

$(10,5,3)$ is

$\sqrt{21}$ sq. units.

The three planes divides the space into

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four parts
0%
six parts
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eight parts
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sixteen parts

A point on XOZ-plane divides the join of

$(5, -3, -2)$ and

$(1, 2, -2)$ at

0%
$\left(\displaystyle\frac{13}{5}, 0, -2\right)$
0%
$\left(\displaystyle\frac{13}{5}, 0, 2\right)$
0%
$(5, 0, 2)$
0%
$(5, 0, -2)$

Explanation

$\begin{array}{l} The\, \, line\, \, through\, \, the\, \, po{ { int } }\, \, \left( { 5,-3,-2 } \right) \\ and\left( { 1,2,-2 } \right) \\ \vec { r } =\hat { i } +2\hat { j } -2\hat { k } +\lambda \left( { 4\hat { i } -5\hat { j } } \right) \\ \vec { r } =\left( { 1+4\lambda } \right) \hat { i } +\left( { 2-5\lambda } \right) \hat { j } -2\vec { k } \\ equation\, \, of\, \, XOZ\, \, plane \\ y=0 \\ \therefore 2-5\lambda =0 \\ \therefore \lambda =\dfrac { 2 }{ 5 } \\ \therefore \vec { r } =\dfrac { { 13 } }{ 5 } \hat { i } -2\hat { k } \\ \therefore \left( { \dfrac { { 13 } }{ 5 } ,0,-2 } \right) \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$

In three dimensions, the coordinate axes of a rectangular cartesian coordinate system are

0%
three mutually parallel lines
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three mutually perpendicular lines
0%
two mutually perpendicular lines and any two parallel
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None of these

If G is centroid of

$\triangle ABC$ , then

0%
$\overrightarrow {G}=\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow{c}$
0%
$\displaystyle \overrightarrow {G} = \frac {\overrightarrow{a} + \overrightarrow {b} +\overrightarrow{c}}{2}$
0%
$3\overrightarrow {G}=\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow {c}$
0%
$\displaystyle 3\overrightarrow {G}=\frac{\overrightarrow {a}+\overrightarrow{b}+\overrightarrow{c}}{2}$

Explanation

We have,

In a $\Delta ABC$

Let

$\overrightarrow{A}=\overrightarrow{a}$

$\overrightarrow{B}=\overrightarrow{b}$

$\overrightarrow{C}=\overrightarrow{c}$

Then,

We know that,

$G\,=\dfrac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$

$3\overrightarrow{G}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$

Hence, this is the answer.

Find the distance between

$(12,3,4)$ and

$(4,5,2)$

0%
$\sqrt {72}$
0%
$\sqrt {62}$
0%
$\sqrt {64}$
0%
None of these

Explanation

Consider the problem,

Let the given points

$A(12,3,4)$ and

$B(4,5,2)$

So, distance between

$A$ and

$B$ by distance formula.

$AB=\sqrt{(4-12)^2+(5-3)^2+(2-4)^2}=\sqrt{(-8)^2+2^2+(-2)^2}$

$=\sqrt{64+4+4}=\sqrt{72}$

So, distance between the points

$(12,3,4)$ and

$(4,5,2)$ is

$\sqrt{72}$ sq. units.

A point at which all the three perpendicular coordinate axes meets is known as

0%
Meeting point
0%
Origin
0%
Triple point
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None of these

Explanation

The three perpendicular coordinate axes meets at one of the point which divides the plane into eight quadrant.

The

$1^{st}$ quadrant has all positive points,

$2^{nd}$ has

$x$ -ve and remaining

$2$ +ve points and so on.

Only

$(0,0,0)$ is not included in any quadrant and is the intersection point.

Thus, the three axes meet at

$(0,0,0)$ from where the eight quadrants originate.

Hence, the point is known as origin.

The coordinates of any point, which lies on

$x$ axis are

0%
$(0,x,0)$
0%
$(x,0,0)$
0%
$(x,x,0)$
0%
$(x,x,x)$

Explanation

In 3-dimensional plane, the point which lies on

$x$ -axis does not have any part in y and z axes.

At that point, the value of

$y$ and

$z$ will be

$0$ .

Hence, coordinate of any point which lies on

$x$ axis are

$(x,0,0)$ .

Point

$D$ has coordinates as

$(3,4,5)$ . Referring to the given figure, find the coordinates of point

$C$ .

0%
$(4,0,0)$
0%
$(0,3,4)$
0%
$(0,3,0)$
0%
$(0,4,0)$

Explanation

It is given that

$D=(3,4,5)$ . Then, obviously,

$E=(0,4,5)$ . Since the point C lies on y axis, hence the corresponding point C will be (from the coordinates of E we found earlier )

$C=(0,4,0)$ .

Point

$D$ has coordinates as

$(3,4,5)$ . Referring to the given figure, find the coordinates of point

$B$ .

0%
$(3,0,4)$
0%
$(4,3,0)$
0%
$(4,0,3)$
0%
$(3,4,0)$

Explanation

From the given figure, D has x-coordinate

$=3$ , y-coordinate

$=4$ and z-coordinate

$=5$ .

The point B lies in xy-plane. So, z-coordinate will be

$0$ .

B is in the same line of point D.

$\therefore$ x and y coordinates will be same as of point D.

Hence, coordinates of point B is

$(3,4,0)$ .

Point

$D$ has coordinates as

$(3,4,5)$ . Find the coordinates of the point

$F$ .

0%
$(0,4,0)$
0%
$(0,0,4)$
0%
$(0,0,5)$
0%
$(0,5,0)$

Explanation

From the given figure, D has x-coordinate

$=3$ , y-coordinate

$=4$ and z-coordinate

$=5$ .

Lines GF and DE are parallel and has the same height.

$\therefore$ All four points D,E,F and G has z-coordinate same i.e.

$5$ .

Point F lie on z-axis

$\implies$ x-coordinate and y-coordinate are

$0$ .

Hence, coordinates of F is

$(0,0,5)$ .

The coordinate of any point, which lies in

$xy$ plane , is

0%
$(x,0,y)$
0%
$(x,x,0)$
0%
$(x, 0, x)$
0%
$(y,0,x)$

Explanation

Given that the point lies in

$xy$ plane

$xy$ plane , the coordinate of

$z$ will be zero

$(x,x,0)$ represents a point which lies in

$xy$ plane

Therefore option

$B$ is correct

Point

$D$ has coordinates as

$(3,4,5)$ . Find the coordinates of point

$G$ .

0%
$(0,3,5)$
0%
$(3,0,4)$
0%
$(3,5,4)$
0%
$(3,0,5)$

Explanation

From the given figure, D has x-coordinate

$=3$ , y-coordinate

$=4$ and z-coordinate

$=5$ .

The points A,B,D and G form a rectangle whose x-coordinate is same i.e.

$3$ .

So, G has x-coordinate as

$3$ .

G lie on the same height as D i.e. z-coordinate will be same.

$\therefore$ G has z-coordinate

$=5$ .

Since, G lie on XZ plane

$\implies y-$ coordinate is

$0$ .

Hence, coordinates of point G is

$(3,0,5)$ .

In which octant, does

$(-4,2,3)$ lie?

0%
III
0%
VIII
0%
VI
0%
II

Explanation

The three perpendicular coordinate axes meet at one of the points which divide the plane into eight quadrants.

The

$1^{st}$ quadrant has all positive points,

$2^{nd}$ has

$x$ -ve and remaining

$2$ are +ve points and so on.

Thus, the point

$(-4,2,3)$ lies in

$II$ octant.

If point

$p$ lies in first octant, then the sign of

$x-$ coordinate will always be

0%
$+$
0%
$-$
0%
$x$ coordinate is always $0$
0%
$x$ coordinate can be $+$ or $-$

Explanation

In the first octant, the values of x,y and z axes are positive.

Any point which lies in first octant has all their coordinate values as positive.

Since point

$p$ lies in first octant, so, p will have all its coordinates as positive.

Hence, sign of

$x$ -coordinate will always be

$+$ .

Point

$D$ has coordinates as

$(3,4,5)$ . Referring to the given figure, find the coordinates of point

$A$ .

0%
$(4,0,0)$
0%
$(3,4,0)$
0%
$(3,0,0)$
0%
$(4,3,0)$

Explanation

From the given figure, D has x-coordinate

$=3$ , y-coordinate

$=4$ and z-coordinate

$=5$ .

The points A,B,D and G form a rectangle whose x-coordinate is same i.e.

$3$ .

Point A lies on x-axis and thus, y and z coordinates will be

$0$ .

Hence, coordinates of point A is

$(3,0,0)$ .

$(-1,-5,-7)$ lies in Octant

0%
I
0%
VII
0%
V
0%
III

Explanation

Here all the three

$x,y,z$ coordinate are negative of the given point.

Therefore, it will lie in the seventh Octant.

A rectangular wooden prism is made up of three pieces, each consisting of four cubes of wood glued together. Which of the following pieces has the same shape as the black piece?

If the

$zx$ -plane divides the line segment joining

$(1,-1,5)$ and

$(2,3,4)$ in the ratio

$p:1$ , then

$p+1=$

0%
$\dfrac{1}{3}$
0%
$1:3$
0%
$\dfrac{3}{4}$
0%
$\dfrac{4}{3}$

Explanation

Given points are

$(1,-1,5)$ and

$(2,3,4)$

since ZX-plane divides the line segment in the ratio

$p:1,$ y-coordinate will be

$0$

the y-coordinate of the point dividing the line segment will be

$\dfrac{3{p}-1}{p+1}=0$

$\ p=\dfrac{1}{3}$

$p+1=\dfrac{1}{3}+1=\dfrac{4}{3}$

The point

$(0 , -2 , 5)$ lies on the

0%
z axis
0%
x axis
0%
xy plane
0%
yz plane
0%
xz plane

The ratio in which the plane

$2x+3y-2z+7=0$ divides the line segment joining the points

$(-1, 1, 3)$ ,

$(2, 3, 5)$ is

0%
$3:5$
0%
$7:5$
0%
$9:11$
0%
$1:5$ externally

Explanation

Let the plane divide the line

$AB$ at

$C$ in the ratio

$k:1$

The coordinayes of

$C$

$= \left( {\dfrac{{2k - 1}}{{k + 1}},\dfrac{{3k + 1}}{{k + 1}},\dfrac{{5k + 3}}{{k + 1}}} \right)$

Since this point lies on the plane

$2x+3y-2z+7=0$

Therefore

$2\left( {\dfrac{{2k - 1}}{{k + 1}}} \right) + 3\left( {\dfrac{{3k + 1}}{{k + 1}}} \right) - 2\left( {\dfrac{{5k + 3}}{{k + 1}}} \right) + 7 = 0$

$\Rightarrow 4k-2+9k+3-10k-6+7k+7=0$

$\Rightarrow 10k+2=0$

$\Rightarrow k= \dfrac{-2}{10}= \dfrac{-1}{5}$

If the divides the line in the ratio

$1:5$ externally

The points

$(2, 5)$ and

$(5, 1)$ are the two opposite vertices of a rectangle. If the other two vertices are points on the straight line

$y = 2x + k$ , then the value of k is

0%
4
0%
3
0%
-4
0%
-3
0%
1

Explanation

Points $(2,5)$ and $(5,1)$ form a diagonal of the rectangle.

Hence, the mid point of these points will lie on the other diagonal.

Mid Point $= \left( \dfrac{2+5}{2}, \dfrac{5+1}{2} \right) = \left( \dfrac{7}{2}, 3 \right)$

Equation of the other diagonal is $y = 2x + k$

Therefore, $3 = 2 \times \dfrac{7}{2} +k \Rightarrow k = -4$

An ordered triplet corresponds to ___________ in three dimensional space.

0%
three points
0%
a unique point
0%
a point in each octant
0%
infinite number of points

Explanation

It is fundamental fact that:

The ordered triplet

$(x,y,z)$ represents an unique point in three dimensional space

The ratio in which

$xy-$ plane divides the line joining the points

$(1, 0, -3)$ and

$(1, -5, 7)$ is given by

0%
$7:3$
0%
$3:7$
0%
$3:4$
0%
$4:7$

Explanation

Let

$xy-plane$ divide the line joining the given points in the ratio

$k:1$ and the point of intersection is

$x,y,0$

then,

$\dfrac{7k-3}{k+1}=0$

$7k-3=0$

$k=\dfrac{3}{7}$

Therefore,

The ratio is

$3:7$

The distance between the origin and the centroid of the tetrahedron whose vertices are

$(0, 0, 0)$ ,

$(a, 0, 0), (0, b, 0), (0, 0, c)$ is?

0%
$\sqrt{a^2+b^2+c^2}$
0%
$\dfrac{\sqrt{a^2+b^2+c^2}}{2}$
0%
$\dfrac{\sqrt{a^2+b^2+c^2}}{4}$
0%
$4\sqrt{a^2+b^2+c^2}$

Explanation

Given,

Centroid of tatrahedron =

$\left( {\frac{a}{4},\,\frac{b}{4}\,,\frac{c}{4}} \right)$

distance between origin & centroid of tetrahedron..................

$\left( {\frac{a}{4},\,\frac{b}{4}\,,\frac{c}{4}} \right)$ (0,0,0)

$\begin{array}{l} =\sqrt { { { \left( { \frac { a }{ 4 } } \right) }^{ 2 } }+{ { \left( { \frac { b }{ 4 } } \right) }^{ 2 } }+{ { \left( { \frac { c }{ 4 } } \right) }^{ 2 } } } \\ \\ =\frac { { \sqrt { { a_{ 2 } }+{ b_{ 2 } }+{ c_{ 2 } } } } }{ 4 } \end{array}$

So that the correct option is C.

1149910_1076188_ans_440ef707a7ce478e8d80509904a4317e.png

The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is

0%
$\sqrt{17}$
0%
$\sqrt{29}$
0%
$\sqrt{34}$
0%
$\sqrt{41}$

Explanation

$P(1,2,3),$ Plane :

$x-y+z=5$

$F$ is foot of perpendicular form

$P$ to plane and

$I$ is image,then

$PF=FI$

$\therefore$ If

$(x,y,z)=(r+1,-r+2,r+3)$ are foot of perpendicular.

$\Rightarrow (r+1)-(-r+2)+r+3=5\quad \quad \Rightarrow r=1\\ \therefore F=(2,1,4)\\ \therefore I=(3,0,5)$

$\therefore$ distance of

$I$ from origin

$=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 34 }$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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