MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 9

The ratio in which the plane

$x - 2y + 3z = 17$ divides the line joining

$(-2, 4, 7)$ and (3, -5, 8) is

0%
$1:2$
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$3 : 1$
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$3 : 10$
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$10 : 1$

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

0%
$\dfrac{3}{2}$
0%
$\dfrac{9}{2}$
0%
$\dfrac{3}{{\sqrt 2 }}$
0%
$\dfrac{{\sqrt 3 }}{2}$

Explanation

$D=$ 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

$A=3/\sqrt{2}\ units$ .

1355085_1199559_ans_27fdc62b88e44569bfb806cc6d32d5c4.png

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.

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

Explanation

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

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

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

Explanation

$P{A}^{2}-P{B}^{2}=2{k}^{2}$

$\Rightarrow \left[{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z-5\right)}^{2}\right]-\left[{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z+7\right)}^{2}\right]=2{k}^{2}$

$\Rightarrow \left(x-3+x+1\right)\left(x-3-x-1\right)+\left(y-4+y-3\right)\left(y-4-y+3\right)+\left(z-5+z+7\right)\left(z-5-z-7\right)=2{k}^{2}$

$\Rightarrow \left(2x-2\right)\left(-4\right)+\left(2y-7\right)\left(-1\right)+\left(2z+1\right)\left(-12\right) =2{k}^{2}$

$\Rightarrow -8x+8-2y+7-24z-24=2{k}^{2}$

$\Rightarrow -8x-2y-24z-9=2{k}^{2}$

$\Rightarrow 8x+2y+24z+9+2{k}^{2}=0$

The perimeter of the triangle formed by the points

$(1,0,0),(0,1,0),(0,0,1)$ is

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$\sqrt 2$
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$2\sqrt 2$
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$3\sqrt 2$
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$4\sqrt 2$

Explanation

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

$AB+BC+CA=\sqrt{2}+\sqrt{2}+\sqrt{2}=3\sqrt{2}$ .

$ABCD$ is a tetrahedron. The principal medians (four in all ) and the lateral medians ( three in all ) are concurrent

0%
True
0%
False

$P$ 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

0%
$3$
0%
$\frac {\sqrt 37}{4}$
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$4$
0%
$3.5$

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.

0%
40
0%
0
0%
80
0%
20

The distance between the points

$P(x,\,-1)$ and

$Q(3,\,2)$ is

$5$ units. Find the value of

$x$ .

0%
$2,8$
0%
$-2,9$
0%
$1,8$
0%
$-1,7$

Explanation

Distance=

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

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

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

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

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

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

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

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

$x=-1,7$

Find the co ordinates of points which trisect the line segment joining

$( 1 , - 2 )$ and

$( - 3,4 )$

0%
$\left( - \frac { 1 } { 3 } , 0 \right) \quad$ and $\left( - \frac { 5 } { 3 } , 2 \right)$
0%
$\left( \frac { 1 } { 3 } , 0 \right)$ and $\left( - \frac { 5 } { 3 } , 2 \right)$
0%
$\left( - \frac { 1 } { 3 } , 0 \right)$ and $\left( \frac { 5 } { 3 } , 2 \right)$
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None of the above

If G is the centroid of

$\triangle ABC$ and BC = 3, CA = 4, AB = 5 then BG =

0%
$\dfrac { \sqrt { 73 } }{ 3 }$
0%
$\dfrac { \sqrt { 13 } }{ 3 }$
0%
$\dfrac { \sqrt { 52 } }{ 3 }$
0%
$\dfrac { \sqrt { 26 } }{ 3 }$

If R divides the line segment joining p(2,3,4) and Q(4,5,6) in the ratio -3 : 2 , then the value of the parameter which represents R is

0%
3
0%
2
0%
1
0%
-1

Explanation

${\textbf{Step1: Find coordinates of the point which divide two points internally.}}$

${\text{Given points are P(2,3,4) & Q(4,5,6)}}$

${\text{ratio is}}$

$= - \dfrac{3}{2} = \dfrac{m}{n}$

$\text{ We know that, the coordinates of the point}$

$P(x, y, z)$

$\text{dividing the line segment}$

$\text{joining the points}$

$A(x_1, y_1, z_1)$

$\text{and}$

$B(x_2, y_2, z_2)$

$\text{in the ratio m : n internally is}$

$\text{} = \left( {\dfrac{{m{x_2} + nx_1}}{{m + n}},\dfrac{{m{y_2} + ny_1}}{{m + n}},\dfrac{{m{z_2} + nz_1}}{{m + n}}} \right)$

${\text{Let, }}$

${\text{R be}}$

$\left( {{x_{3,}}{y_{3,}}{z_3}} \right)$

$\text{R} = \left( {\dfrac{{m{x_2} + nx_1}}{{m + n}},\dfrac{{m{y_2} + ny_1}}{{m + n}},\dfrac{{m{z_2} + nz_1}}{{m + n}}} \right)$

$= \left( {\dfrac{{ - 3(4) + 2(2)}}{{ - 3 + 2}},\dfrac{{ - 3(5) + 2(3)}}{{ - 3 + 2}},\dfrac{{ - 3(6) + 2(4)}}{{ - 3 + 2}}} \right)$

$= \left( {\dfrac{-8}{{ - 1}},\dfrac{-9}{{ - 1}}, \dfrac{{-10}}{{ - 1}},} \right)$

$= (8,9,10)$

${\textbf{Thus, the value of parameter which represent R is (8,9,10). }}$

$( 3,4 )$ and

$( 6,5 )$ are the extremities of a diagonal of a parallelogram and

$( 2,1 )$ is is third vertex, then its fourth vertex is _______.

0%
$( - 1,0 )$
0%
$( - 1,1 )$
0%
$( 0,-1 )$
0%
$( 7,8 )$

Explanation

${\textbf{Step -1: Form equation}}$

${\text{Let A }= (3,4), \text{C } = (6,5) \text{as they are end points of the diagonal and B }= (2,1)}$

${\text{Let the fourth vertex D }= (x,y)}$

${\text{We know that the diagonals of a parallelogram bisect each other}}{\text{. }}$

${\text{So, the midpoint of AC is same as the mid point of BD}}{\text{.}}$

${\text{Mid point of two points }}({x_1},{y_1}) {\text{and }}({x_2},{y_2})$

${\text{which}}\;{\text{ is calculated by the formula}};(\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2})$

${\textbf{Step -2: Find fourth point of vertex}}$

${\text{So, midpoint of AC = Mid point of BD}}$

$\Rightarrow \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$

$\Rightarrow \left( {\dfrac{{3 + 6}}{2},\dfrac{{4 + 5}}{2}} \right) = \left( {\dfrac{{2 + x}}{2},\dfrac{{1 + y}}{2}} \right)$

${\text{So }}\dfrac{{3 + 6}}{2} = \dfrac{{2 + x}}{2}\;{\text{and }}\dfrac{{4 + 5}}{2} = \dfrac{{1 + y}}{2}$

${\text{So, }}\dfrac{9}{2} = \dfrac{{2 + x}}{2}$

$\Rightarrow 9 = 2 + x$

$\Rightarrow x = 7$

${\text{and }}\dfrac{9}{2} = \dfrac{{1 + y}}{2}$

$\Rightarrow 9 = 1 + y$

$\Rightarrow y = 8$

${\text{So fourth vertex is (7,8)}}$

${\textbf{Hence the correct answer is option D}}{\text{.}}$

If the point

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

$C(-1, 4, -2)$ are collinear then

$(C : AB)$

0%
$1 : 2$
0%
$-2 : 1$
0%
$-1 : 2$
0%
$4 : 0$

Centroid of triangle formed by foot of perpendiculars from (-3, -6, -9) on coordinate axes is:-

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

$( - 6 , - 4 ) , ( 3,5 ) , ( - 2,1 )$ are the vertices of a parallelogram, then remaining vertex can be:

0%
$( 0 , - 1 )$
0%
$( 7,10 )$
0%
$( - 1,0 )$
0%
$( - 11 , - 8 )$

The distance of the point (1,3) from the line 2x-3y+9=0 measured along a line x-y+1=0 is

0%
$\sqrt 2$
0%
$\sqrt 5$
0%
$2\sqrt 2$
0%
1

Explanation

Slope of line

$2x-3y+9=0$ is

$m_1=\frac{2}{3}$

Slope of line

$x-y+1=0$ is

$m_2=1$

Now, angle between the two lines is

$tan\theta=\frac{m_2-m_1}{1+m_1m_2}=\frac{1-\frac{2}{3}}{1+1\times\frac{2}{3}}=\frac{\frac{1}{3}}{\frac{5}{3}}=\frac{1}{5}$

So,

$sin\theta=\frac{1}{\sqrt{26}}$

Now, Perpendicular distance of

$(1,3)$ from

$2x-3y+9=0$ is

$p=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

$=\frac{|2(1)+(-3)(3)+9|}{\sqrt{(2)^2+(-3)^2}}$

$=\frac{|2-9+9|}{\sqrt{4+9}}$

$=\frac{2}{\sqrt{13}}$

Now, distance along the given line

$=\frac{p}{sin\theta}=\frac{2}{\sqrt{13}}\times\sqrt{26}=2\sqrt{2}$ units

So, the correct option is

$\text{C}$ .

$L_1$ is the line of intersection of the plane

$2x-2y+3z-2=0, x-y+z+1=0$ and

$L_2$ is the line of intersection of the plane

$x+2y-z-3=0, 3x-y+2z-1=0$ , then the distance of origin from from the plane containing the lines

$L_1$ +

$L_2$ is :

0%
$\dfrac{1}{\sqrt{2}}$
0%
$\dfrac{1}{4\sqrt{2}}$
0%
$\dfrac{1}{2\sqrt{2}}$
0%
none of these

Which one of the following 3D shapes does not have a vertex?

0%
Sphere
0%
Pyramid
0%
Prism
0%
Cone

is a point on the line segment joining the points A(2,-3,4) and B(8,0,10). if the value of y-coordinate of C is-2, then the z-coordinate of c is ___________________.

0%
4
0%
6
0%
-4
0%
5

Find the ratio on which the plane 3x + 4y - 5z = 1 divides the line joining the points (-2, 4, -6) and (3, -5, 8).

0%
3 : 4
0%
3 : 5
0%
4 : 3
0%
2 : 3

The centroid of triangle

$A(3,4,5);B(6,7,2);C(0,-5,2)$ is

0%
$(3,2,3)$
0%
$(5,2,1)$
0%
$(2,5,1)$
0%
$(3,4,1)$

Explanation

Given points

$A(3,4,5);B(6,7,2);C(0,-5,2)$

Centroid is given as

$\left(\dfrac{3+6+0}{3},\dfrac {4+7-5}3,\dfrac {5+2+2}{3}\right)\\\left(\dfrac 9 3,\dfrac 63 ,\dfrac 93\right)=(3,2,3)$

The shortest distances of a point from x-axis, y-axis and z-axis are 2,3,6 respectively. The distance of this point from the origin is

0%
$\frac{7}{\sqrt{2}}$
0%
7
0%
11
0%
$\frac{49}{2}$

If R divides the line segment joining P(2, 3, 4) and Q (4, 5, 6) in the ratio -3 : 2, then the value of the parameter which represents R is

0%
$=(8,9,10)$
0%
$=(10,9,8)$
0%
$=(10,8,9)$
0%
$=(9,10,8)$

Explanation

Given,

$P(2,3,4),Q(4,5,6)$

Given Ratio

$=-\dfrac{3}{2}=\dfrac{m}{n}$

Let

$R$ be

$(x_3,y_3,z_3)$

$R=\left ( \dfrac{mx_2+nx}{m+n},\dfrac{my_2+ny}{m+n}+\dfrac{mz_2+nz}{m+n}\right )$

$=\left ( \dfrac{-3(4)+2(2)}{-3+2},\dfrac{-3(5)+2(3)}{-3+2}+\dfrac{-3(6)+2(4)}{-3+2}\right )$

$=\left ( -\dfrac{8}{-1},-\dfrac{9}{-1},-\dfrac{10}{-1} \right )$

$=(8,9,10)$

If (2, 3, -1) is the midpoint of AB where A=(-1,5,3) then B =

0%
(5, 1, -5)
0%
(5, -1, 5)
0%
(-5, 1, 5)
0%
NONE OF THESE

Explanation

Given,

$(2,3,-1)$ is mid point

then it will divide

$AB$ in

$1:1$

let,

$B$ be

$(x,y,z)$

so, according to section formula

$(\dfrac{1x+1(-1)}{1+1},\dfrac{1y+1(5)}{1+1},\dfrac{1z+1(3)}{1+1})=(2,3,-1)$

$(\dfrac{x-1}{2},\dfrac{y+5}{2},\dfrac{z+3}{2})=(2,3,-1)$

$\implies{\dfrac{x-1}{2}=2}$

$\implies{x-1=4}$

$\implies{x=5}$

Also,

$\dfrac{y+5}{2}=3$

$\implies{y+5=6}$

$\implies{y=6-5=1}$

Also,

$\dfrac{z+3}{2}=-1$

$\implies{z+3=-2}$

$\implies{z=-2-3=-5}$

hence,

$B=(5,1,-5)$

The coordinates of the point where the line through

$(3, -4, -5)$ and

$(2, -3, 1)$ crosses the plane passing through three points

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

$(4, -1, 0)$ is

0%
$(1, 2, 7)$
0%
$(-1, 2, -7)$
0%
$(1, -2, 7)$
0%
None of these

Locate the point

$(3, 2, -1)$ in

$3\text{D}$ octant.

0%
$1st$ Octant
0%
$2nd$ Octant
0%
$7th$ Octant
0%
$8th$ Octant

If the distance of the point P(4, 3, 5) from the Y-axis is

$\lambda$ , then the value of

${ 7\lambda }^{ 2 }$ is

0%
$287$
0%
$7\sqrt { 41 }$
0%
$63$
0%
$21$

Explanation

$\Rightarrow$

$A=(x_1,y_1,z_1)=(0,3,0)$ and

$P=(x_2,y_2,z_2)=(4,3,5)$

Here,

$PA=\lambda$

$\Rightarrow$

$AP(\lambda)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

$=\sqrt{(4-0)^2+(3-3)^2+(5-0)^2}$

$=\sqrt{16+0+25}$

$=\sqrt{41}$

$\Rightarrow$

$7\lambda^2=7(\sqrt{41})^2$

$=7\times 41$

$=287$

1458848_1428108_ans_ac95c722c3704ec186b690d82c38d213.png

Let

$A(3, 0, -1), B(2, 10, 6)$ and

$C(1, 2, 1)$ be the vertices of a triangle and

$M$ be the midpoint of

$AC$ . If

$G$ divides

$BM$ in the ratio,

$2 : 1$ , then

$\cos (\angle GOA)$ (

$O$ being the origin) is equal to:

0%
$\dfrac{1}{\sqrt{30}}$
0%
$\dfrac{1}{6\sqrt{10}}$
0%
$\dfrac{1}{\sqrt{15}}$
0%
$\dfrac{1}{2\sqrt{15}}$

Explanation

$G$ is the centroid of

$\Delta ABC$

$G \equiv (2, 4, 2)$

$\overrightarrow{OG} = 2\hat{i} + 4\hat{i} + 2\hat{k}$

$\overrightarrow{OA} = 3\hat{i} - \hat{k}$

$\cos (\angle GOA) = \dfrac{\overrightarrow{OG} \cdot \overrightarrow{OA}}{|\overrightarrow{OG}| |\overrightarrow{OA}|} = \dfrac{1}{\sqrt{15}}$

$30$ consider at three dimensional figure represented by

$xyz^2=2$ , then its minimum distance from origin is?

0%
$2$
0%
$4$
0%
Both A & B
0%
None of the above

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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