If $$z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$$, then

$$\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$$

$$\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$$

$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$$

$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$$

MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 2

Ratio in which the

$xy-$ plane divides the join of

$(1, 2, 3)$ and

$(4, 2, 1)$ is

0%
$3:1$ internally
0%
$3:1$ externally
0%
$1:2$ internally
0%
$2:1$ externally

Explanation

Points

$A(1,2,3),B(4,2,1)$

$x-y$ plane

$\Rightarrow z=0$

$\therefore$ lines containing

$A,B$ is

$\cfrac { x-1 }{ 3 } =\cfrac { y-2 }{ 0 } =\cfrac { z-3 }{ -2 }$

point of intersection of line and

$xy$ plane is

$(\cfrac { 11 }{ 2 } ,2,0)$

$\therefore (\cfrac { 4\lambda +1 }{ \lambda +1 } ,\cfrac { 2\lambda +2 }{ \lambda +1 } ,\cfrac { \lambda +3 }{ \lambda +1 } )=(\cfrac { 11 }{ 2 } ,2,0)\\ \Rightarrow \lambda =-3$

Answer

$(B)$

1074333_1095499_ans_a1416ac9fff8486e8384bfe0d3e9b899.png

The equation of the set of points which are equidistant from the points

$(1, 2, 3)$ and

$(3, 2, -1)$ .

0%
$x-2z=0$
0%
$2x-z=0$
0%
$2x+y=0$
0%
$x-2y=0$

Explanation

Let the given points be A(

$1,2,3$ ) and B(

$3,2,-1$ ) and let the point equidistant from A and B be P(

$x,y,z$ )

then

$PA=PB$

$\sqrt {{{(x - 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}} = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} + {{(z + 1)}^2}}$

Squaring both sides

${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$

${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$

$-2x-6z+10=-6x+2z+10$

$4x-8z=0$

$x-2z=0$

$A(3, 2, 0), B(5, 3, 2), C(-9, 6, -3)$ are three points forming a triangle. If

$AD$ , the bisector of

$\angle BAC$ meets

$BC$ in

$D$ then coordinates of

$D$ are

0%
$\left (-\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$
0%
$\left (\dfrac {19}{8}, -\dfrac {57}{16}, \dfrac {17}{16}\right )$
0%
$\left (\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$
0%
None of these

Explanation

According to question,

$\begin{array}{l} AB=\sqrt { 4+1+4 } =3 \\ AC=\sqrt { 144+16+9 } =13 \\ BD:DC=AB:AC=3:13 \\ D=\left( { \dfrac { { 3(-9)+13(5) } }{ { 3+13 } } ,\dfrac { { 3(6)+13(3) } }{ { 3+13 } } ,\dfrac { { 3(-3)+13(2) } }{ { 3+13 } } } \right) \\ \, \, =\left( { \dfrac { { -38 } }{ { 16 } } ,\dfrac { { 57 } }{ { 16 } } ,\dfrac { { 17 } }{ { 16 } } } \right) \\ \, \, \, \, \, \, \, \, \, \therefore \, \, \, \left( { \dfrac { { -19 } }{ 8 } ,\dfrac { { 57 } }{ { 16 } } ,\dfrac { { 17 } }{ { 16 } } } \right) \\ So,the\, \, correct\, \, option\, \, is\, \, A. \end{array}$

$A=\left(2,4,5\right)$ and

$B=\left(3,5,-4\right)$ are two points. If the

$xy$ -plane,

$yz$ -plane divide

$AB$ in the ratios

$a:b,p:q$ respectively then

$\dfrac{a}{b}+\dfrac{p}{q}$ =

0%
$\dfrac{7}{15}$
0%
$\dfrac{-7}{12}$
0%
$\dfrac{7}{12}$
0%
$\dfrac{22}{25}$

Explanation

$A(2,4,5),B(3,5,-4)$

xy plane

$\Rightarrow z=0$

yz plane

$\Rightarrow x=0$

line through

$A$ &

$B$

$\Rightarrow \cfrac { x-2 }{ 1 } =\cfrac { y-4 }{ 1 } =\cfrac { z-5 }{ -9 } =t$

$\therefore$ let

$P(t+2,t+4,-9t+5)$ be point of intersection of line with xy plane.

$\therefore P=(\cfrac { 23 }{ 9 } ,\cfrac { 41 }{ 9 } ,0)$

Similarly,for

$Q(t+2,t+4,9t+5)$ be point of intersection of line with yz plane

$\Rightarrow t=-2$ .

$Q(0,2,23)$

(i)let

$P$ divide

$AB$ in

$1:\lambda (a:b)$

$\therefore (\cfrac { 23 }{ 9 } ,\cfrac { 41 }{ 9 } ,0)=(\cfrac { 2\lambda +3 }{ \lambda +1 } ,\cfrac { 4\lambda +5 }{ \lambda +1 } ,\cfrac { 5\lambda -4 }{ \lambda +1 } )\\ \Rightarrow \lambda =\cfrac { 4 }{ 5 } \\ \therefore \cfrac { a }{ b } =\cfrac { 5 }{ 4 }$

(ii)let

$Q$ divide

$AB$ in

$\lambda:1 (p:q)$

$\therefore (0,2,23)=(\cfrac { 2+3\lambda }{ \lambda +1 } ,\cfrac { 4+5\lambda }{ \lambda +1 } ,\cfrac { 5-4\lambda }{ \lambda +1 } )\\ \Rightarrow \lambda =\cfrac { -2 }{ 3 } \\ \therefore \cfrac { p }{ q } =\cfrac { -2 }{ 3 } \\ \therefore \cfrac { a }{ b } +\cfrac { p }{ q } =\cfrac { 5 }{ 4 } -\cfrac { 2 }{ 3 } =\cfrac { 7 }{ 12 }$

$z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$ , then

0%
$\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$
0%
$\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$
0%
$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$
0%
$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$

Explanation

$z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)$ then.

$z=|z|e^{i arq (z)}$

$\therefore z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)=e^{i\left(\pi/6\right)}\quad [\because e^{i\theta}=\cos\theta+i\sin \theta]$

$\Rightarrow |z|=1, arq=\dfrac{\pi}{6}$

The x-coordinate of a point on the line joining the points

$P(2,2,1)$ and

$Q(5,1,-2)$ is

$4$ . Find its z-coordinate.

0%
$-1$
0%
$-2$
0%
$1$
0%
$2$

Explanation

$P(2,2,1),Q(5,1,-2)$

$\therefore$ Equation of line through

$P$ &

$Q$ ,

$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \\ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r$

$\therefore P$ be point of line

$\Rightarrow P\equiv (-3r+2,r+2,3r+1)$

$\therefore -3r+2=4$ (

$\because$ since x-coordinate is

$4$ )

$\Rightarrow r=\cfrac { -2 }{ 3 }$

$\therefore$ z-coordinate

$=3r+1=-1$

Solve the following differential equation.

$\dfrac{dy}{dx}=x-1$ .

0%
$y=x^2+x$
0%
$y=x^2$
0%
$y=x^2-x$
0%
None of the above

Explanation

Given,

$\dfrac{dy}{dx}=x-1$ .

Integrating on both sides

$\displaystyle \int \dfrac {dy}{dx}=\int x-1\ dx$

$y=x^2-x +c$

The distance between (5,1,3) and the line x=3, y=7+t, z=1+t is

0%
4
0%
2
0%
6
0%
8

Explanation

$\begin{array}{l} x=3\, \, \, ,y=7+t\, \, ,z=1+t \\ A\left( { 3,7+t,1+t } \right) \\ 0.\left( { 3-5 } \right) +1\left( { 7+t-1 } \right) +1\left( { 1+t-3 } \right) =0 \\ 6+t+t-2=0 \\ 2t=-4 \\ t=-2 \\ A:\left( { 3,5,-1 } \right) \\ dis\tan ce=\sqrt { 4+16+16 } \\ =6 \\ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

The perimeter of triangle with vertices at

$(1,0,0) , (0,1,0) and (0,0,1)$ is :

0%
$3$
0%
$2$
0%
$2\sqrt {2}$
0%
$3\sqrt {2}$

Explanation

REF.Image

$AB=\sqrt{1+1+0}=\sqrt{2}$

$BC=\sqrt{0+1+1}=\sqrt{2}$

$AC=\sqrt{1+0+1}=\sqrt{2}$

Hence, perimeter of

$\triangle ABC$ is given by

$AB+BC+AC= 3\sqrt{2}$

1068862_1181473_ans_39cef25120f24b65854764f6a31c89cd.png

If the points

$A(9, 8, -10), B(3, 2, -4)$ and

$C(5, 4, -6)$ be collinear, then the point

$C$ divides the line

$AB$ in the ratio

0%
$2:1$
0%
$3:1$
0%
$1:2$
0%
$-1:2$

Explanation

REF.Image

Let the ratio be K:1

Then, using section formula

we have

$C(5,4,-6)=\frac{K(3,2,-4)+(9,8,-10)}{K+1}$

$\Rightarrow 5=\frac{3K+9}{K+1}$

$\Rightarrow 5K= 3K+4$

$\Rightarrow 2K=4$

$K=2$

$\Rightarrow$ C divides AB in ratio 2:1

1118838_1202108_ans_8bd38d2778184388979a53afda31a58a.jpg

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

0%
7
0%
0
0%
$\frac{13}{2}$
0%
$-\frac{13}{2}$

Explanation

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

$= -6-2+\dfrac{3}{2}$

$=-\dfrac{13}{2}$

$OA$ is equally inclined to

$OX, OY$ and

$OZ$ and if

$A$ is

$\sqrt 3$ units from the origin then

$A$ is

0%
$\left( {3,3,3} \right)$
0%
$\left( { - 1,\,1,\, - 1} \right)$
0%
$\left( { - 1,\,1,\,1} \right)$
0%
$\left( {1,\,1,\,1} \right)$

Explanation

By hit and tried method

$(1,1,1)=\sqrt {1^2+1^2+1^2}=\sqrt 3$

and

$\cos \theta_1 =\cos \theta_2 =\cos \theta_3 =\dfrac {1}{\sqrt {3}}$

The distance between the points

$(-1,2,3)$ and P is

$13$ . Then

$P=$

0%
$(2,6,-9)$
0%
$(-2,6,9)$
0%
$(2,6,9)$
0%
$(2,-6,9)$

Explanation

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)

1068689_1181094_ans_aeee9ea04f7946e898bd073f4f8c30ea.png

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-

0%
$1/2$
0%
$1/3$
0%
$1/4$
0%
$1/6$

Explanation

Planes are

$2i-2j+k+3=0,4i-4j+2k+5=0,2i-2j+k+5/2=0$

distance between them is

$=\cfrac{|c_1-c_2|}{\sqrt{a^2+b^2+c^2}}\\=\cfrac{|3-\cfrac{5}{2}|}{\sqrt{2^2+2^2+1^2}}=\cfrac{1}{6}$

The number of octants in which

$Z$ coordinate is positive is

0%
$2$
0%
$3$
0%
$4$
0%
$1$

Explanation

$Z$ - Coordinates are positive in Octant

$1, 2, 3$ and

$4$ .

Hence, then the number of octants are

$4$ .

1183887_1240453_ans_9d7c1e42c6a146f286105efd300c4edf.png

$\overline { a } ,\overline { b }$ are the position vectors of

$A$ and

$B$ then one of the following points lie on

$\overline { AB }$

0%
$\cfrac{2(\overline { a } +\overline { b } )}{3}$
0%
$\cfrac{(\overline { a } -\overline { b } )}{3}$
0%
$\cfrac{(\overline { a } +\overline { b } )}{3}$
0%
$\cfrac{2\overline { a } +2\overline { b } }{3}$
0%
None of these

Explanation

$\rightarrow$ 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

1118165_1187820_ans_f444e6cf52524d9ba3d1b2f78f9f2165.jpg

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

0%
7
0%
12
0%
13
0%
25

$L, M$ are the feet of the perpendiculars from

$(2, 4, 5)$ to the

$xy$ -plane,

$yz$ -plane respectively, then the distance

$LM$ is:

0%
$\sqrt{41}$
0%
$\sqrt{20}$
0%
$\sqrt{29}$
0%
$3\sqrt{5}$

Explanation

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

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

$=\sqrt{4+25}$

$=\sqrt{29}$

The name of the figure formed by the points

$(0, 0, 0), (1, 0, 1)$ and

$(0, 1, 1)$ is

0%
a straight line
0%
an isosceles triangle
0%
an equilateral triangle
0%
a scalene triangle

Explanation

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

The name of the figure formed by the points

$(3, -5, 1), (-1, 0, 8)$ and

$(7, -10, -6)$ is

0%
a triangle
0%
a straight line
0%
an isosceles triangle
0%
an equilateral triangle

Explanation

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.

If two vertices of an equilateral triangle are

$(2, 1, 5)$ and

$(3, 2, 3)$ , then its third vertex is:

0%
$(1, 2, 4)$
0%
$(4, 0, 4)$
0%
$(0, -4, 4)$
0%
$(4, 4, 1)$

Explanation

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

$B$

The point which is equidistant from the points

$(a, 0, 0), (0, b, 0), (0, 0, c)$ and

$(0, 0, 0)$ is:

0%
$(a,b,c)$
0%
$(\sqrt{{a}},\sqrt{{b}},\sqrt{{c}})$
0%
$(2a,2b,2c)$
0%
$\left (\dfrac {a}{2}, \dfrac {b}{2}, \dfrac {c}{2}\right)$

Explanation

${\textbf{Step - 1: Writing equations using distance formula}}$

${\text{Let the point which is equidistant from A(a,0,0), B(0,b,0), C(0,0,c) and O(0,0,0) be P(x, y, z)}}$

$\therefore {\text{ AP = BP = CP = OP}}$

$\Rightarrow {\text{ }}\sqrt {{{\left( {{\text{x - a}}} \right)}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - b}}} \right)}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\left( {{\text{z - c}}} \right)}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}}$

${\text{Squaring all sides,}}$

$\Rightarrow {\text{ }}{\left( {{\text{x - a}}} \right)^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{\left( {{\text{y - b}}} \right)^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{\left( {{\text{z - c}}} \right)^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}$

$\Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ - 2ax + }}{{\text{a}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ - 2by + }}{{\text{b}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ - 2zc + }}{{\text{c}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}$

$\Rightarrow {\text{ }}{{\text{a}}^{\text{2}}}{\text{ - 2ax = }}{{\text{b}}^{\text{2}}}{\text{ - 2by = }}{{\text{c}}^{\text{2}}}{\text{ - 2cz = 0}}$

${\textbf{Step - 2: Solving for x, y and z}}$

$\because {\text{ }}{{\text{a}}^{\text{2}}}{\text{ - 2ax = }}{{\text{b}}^{\text{2}}}{\text{ - 2by = }}{{\text{c}}^{\text{2}}}{\text{ - 2cz = 0}}$

$\therefore {\text{ }}{{\text{a}}^{\text{2}}}{\text{ - 2ax = 0}}$

$\Rightarrow {\text{ 2ax = }}{{\text{a}}^{\text{2}}}$

$\Rightarrow {\text{ x = }}\dfrac{{\text{a}}}{{\text{2}}}$

${\text{Similarly, }}{{\text{b}}^{\text{2}}}{\text{ - 2by = 0}}$

$\Rightarrow {\text{ 2by = }}{{\text{b}}^{\text{2}}}$

$\Rightarrow {\text{ y = }}\dfrac{{\text{b}}}{{\text{2}}}$

${\text{Similarly, }}{{\text{c}}^{\text{2}}}{\text{ - 2cz = 0}}$

$\Rightarrow {\text{ 2cz = }}{{\text{c}}^{\text{2}}}$

$\Rightarrow {\text{ z = }}\dfrac{{\text{c}}}{{\text{2}}}$

${\text{So the point is }}$ $\left( {\dfrac{{\text{a}}}{{\text{2}}}{\text{, }}\dfrac{{\text{b}}}{{\text{2}}}{\text{, }}\dfrac{{\text{c}}}{{\text{2}}}} \right)$

$\textbf{Hence option D is correct}$

The point which is equidistant from the points

$(-1, 1, 3), (2, 1, 2), (0, 5, 6)$ and

$(3,2, 2)$ is

0%
$(-1, 3, 4)$
0%
$(3, 1, 4)$
0%
$(1, 3, 4)$
0%
$(4,1, 3)$

Explanation

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

$(x,4-x,3x+1)=(1,3,4)$ .

If the extremities of a diagonal of a square are

$(1, -2, 3)$ and

$(2, -3, 5)$ , then the length of its side is:

0%
$\sqrt{6}$
0%
$\sqrt{3}$
0%
$\sqrt{5}$
0%
$\sqrt{7}$

Explanation

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

$= \sqrt {3 }$

$A, B$ are the feet of the perpendiculars from

$(2, 4, 5)$ to the

$x$ -axis,

$y$ -axis respectively, then the distance

$AB$ is

0%
$2\sqrt{5}$
0%
$\sqrt{29}$
0%
$\sqrt{41}$
0%
$3\sqrt{5}$

Explanation

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

$\sqrt {({2}^{2} + {4}^{2})} = \sqrt{20} = 2\sqrt {5}$

$(p, q, r)$ is equidistant from

$(1, 2, -3), (2, -3, 1)$ and

$(-3, 1, 2)$ , then

$p +q + r$

$=$

0%
$-1$
0%
$1$
0%
$0$
0%
$2$

Explanation

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

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

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

$\text{Distance between A and B: }$

$\mathrm{AB=\sqrt{(2-1)^2+(-3-2)^2+(1+3)^2}}$

$\mathrm{=\sqrt{1+25+16}}$

$\mathrm{=\sqrt{42}}$

$\text{Distance between B and C: }$

$\mathrm{BC=\sqrt{(-3-2)^2+(1+3)^2+(2-1)^2}}$

$\mathrm{=\sqrt{25+16+1}}$

$\mathrm{=\sqrt{42}}$

$\text{Distance between C and A: }$

$\mathrm{CA=\sqrt{(-3-1)^2+(1-2)^2+(2+3)^2}}$

$\mathrm{=\sqrt{16+1+25}}$

$\mathrm{=\sqrt{42}}$

$\mathrm{\therefore AB=BC=CA }$

$\text{So, triangle ABC is an equilateral triangle.}$

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

$\text{Given that, (p,q,r) is equidistant from the points A, B and C.}$

$\therefore\text{(p,q,r) is the circumcentre of triangle ABC.}$

$\text{We know, for equilateral triangles circumference and centroid are same point.}$

$\text{Centroid of a triangle with vertices } \mathrm{(x_1,y_1),(x_2,y_2)\ and\ (x_3,y_3)=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)}$

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

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

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

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

$\textbf{Hence, the correct option is C.}$

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

$BD$ is:

0%
$\sqrt{34}$
0%
$6$
0%
$18$
0%
$3\sqrt{2}$

Explanation

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

$AC$

$BD = \sqrt {({4}^{2} + {1}^{2} + {1}^{2})} = 3\sqrt{2}$

$A = (1, -1, 2)$ and

$B =$

$(2, 3, 7)$ are two points. lf

$P,\ O$ divide

$AB$ in the ratios

$2:3, -2:3$ respectively then

$P_x+Q_y=$

0%
$\displaystyle \frac{-38}{5}$
0%
$\displaystyle \frac{38}{5}$
0%
$\displaystyle \frac{-2}{5}$
0%
$\displaystyle \frac{-47}{6}$

Explanation

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

$-2:3$

$\therefore Q_y = \cfrac{-2 \times 3 + 3 \times -1}{-2 + 3} = -9$

$\Rightarrow P_x + Q_y = -9 + \cfrac{7}{5} = \cfrac{-38}{5}$

$\mathrm{If} \mathrm{A}=(1, 2, 3)$ ,

$\mathrm{B}=(2,3, 4)$ 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

$\mathrm{C}_{\mathrm{z}}=$

0%
$\displaystyle \dfrac{10}{3}$
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$\displaystyle \dfrac{11}{3}$
0%
$\displaystyle \dfrac{11}{2}$
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$11$

Explanation

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.

$(1, 1, a)$ is the centroid of the triangle formed by the points

$(1, 2, -3)$ ,

$(\mathrm{b}, 0, 1)$ and

$(-1, 1, -4)$ then

$a-b$

$=$

0%
$-5$
0%
$-7$
0%
$5$
0%
$1$

Explanation

The coordinates of the vertices of the triangle are given by $(1,2,-3) , (b,0,1), (-1,1,-4)$

Accordingly the coordinates of the centroid of this triangle will be given by

( $\dfrac{b}{3}, 1 , -2$ )

Hence, $\dfrac{b}{3}$ $= 1$ Or, $b= 3$

and $a = -2$

So $a- b = -5$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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