MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 3

If the extremities of a diagonal of a square are

$(1, -2, 3)$ and

$(2, -3, 5)$ , then area of the square is

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$6$
0%
$3$
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$\displaystyle \dfrac{3}{2}$
0%
$\sqrt{3}$

Explanation

Let the extremities of the diagonal of a square be

$(1,-2,3)$ and

$B(2,-3,5)$ .
Then diagonal

$AB$ is given by

${({1}^{2} + {1}^{2} + {2}^{2})}^{0.5}$ =

$\sqrt{6}$
Hence, length of the side

$=\text{length of diagonal}\div\sqrt2= \sqrt {3}$
So, area of square will be

$\sqrt{3} \times \sqrt{3} = 3$

$A=(2, 4, 5)$ and

$B=(3,5, -4)$ are two points. lf the

$XY$ -plane,

$YZ$ -plane divide

$AB$ in the ratio

$a:b$ and

$p:q$ respectively, then

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

0%
$\displaystyle \dfrac{23}{12}$
0%
$\displaystyle \dfrac{-7}{12}$
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$\displaystyle \dfrac{7}{12}$
0%
$\displaystyle \dfrac{-22}{15}$

Explanation

Given points are

$A(2,4,5)$ and

$B(3,5,-4)$ .
Let the

$XY$ plane divides the line

$AB$ in the ratio

$\lambda:1$ .
Then the

$z$ coordinate of

$XY$ plane is zero.

$0=5-4(\lambda)$

$\lambda=\dfrac {5}{4}=\dfrac {a}{b}$

$YZ$ plane divides the line

$AB$ such that x coordinate is zero.

$0=3(\alpha)+2$

$\alpha=\dfrac {-2}{3}=p:q$
Now,

$\dfrac {a}{b}+\dfrac {p}{q}=\dfrac {5}{4}-\dfrac {2}{3}=\dfrac {7}{12}$

Hence, option C is correct.

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

$AB$ is produced upto

$C$ such that

$2AB = BC$ , then

$C =$

0%
$(5, 4, 6)$
0%
$(6, 2, 4)$
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$(4, 5, 6)$
0%
$(6, 4, 5)$

Explanation

Let the point

$C$ be

$(i,j,k)$ .
Since,

$B$ divides

$AC$ in the ratio

$1:2$
Coordinates of

$B$ should be

$\left ( \dfrac{2+i}{3} , \dfrac{4+j}{3} , \dfrac{k+6}{3}\right)$
Comparing the values given already for

$B$ , we get,

$i = 4$ ,

$j = 5$ and

$k = 6$

If the points

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

$B(1, 1, 1)$ and

$C(-1, 4, -2)$ are collinear, then the ratio in which

$C$ divides

$AB$ is

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

Explanation

Let

$C$ divide

$AB$ in the ratio

$x:y$ .
Let us compare the

$x$ -coordinate of

$C$ by using section formula:

$-1 = \dfrac {x\times 1 + y\times3}{x+y}$

$\Rightarrow x+3y=-x-y$

$\Rightarrow x = -2y$
Hence, point

$C$ divides

$AB$ in the ratio

$-2 :1$ . As the ratio is negative, it means

$C$ divides

$AB$ externally

Two opposite vertices of a square are

$(2, -3, 4)$ and

$(4, 1, -2)$ . The length of the side of the square is

0%
$\sqrt{58}$
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$2\sqrt{7}$
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$\sqrt{14}$
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$\sqrt{7}$

Explanation

Length of diagonal

$=$ distance between the opposite vertices of the square

$\\=\sqrt{(2-4)^2+(-3-1)^2+(4+2)^2}=\sqrt{56}=d$ (say)

Hence, length of sides of square

$=\dfrac{d}{\sqrt 2}=\sqrt{28}=2\sqrt 7$

$A = (2, -3, 1), B = (3, -4, 6)$ and

$C$ is a point of trisection of

$AB$ , then

${C}_{{y}}=$

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

Explanation

Given,

$C$ is a point of trisection of

$AB$ .
Hence,

$C$ either divides

$AB$ in the ratio

$2:1$ or

$1:2$ .
Case 1:

$C$ divides in the ratio

$2:1$ .
The coordinates of

$C$ will be

$\left(\dfrac{8}{3}, -\dfrac{11}{3}, \dfrac{13}{3}\right)$
Case 2:

$C$ divides in the ratio

$1:2$ .
The coordinates of

$C$ will be

$\left(\dfrac{7}{3}, -\dfrac{10}{3}, \dfrac{8}{3}\right)$

Hence, either

$C_y = -\dfrac{11}{3}$ or

$-\dfrac{10}{3}$

The point

$P$ is on the

$y$ -axis. If

$P$ is equidistant from

$(1,2, 3)$ and

$(2,3, 4)$ , then

$P_{y}=$

0%
$\displaystyle \dfrac{15}{2}$
0%
$15$
0%
$30$
0%
$\displaystyle \dfrac{3}{2}$

Explanation

Let the point be

$(0,y,0)$
Since, it is equidistant from the

$2$ points.

$\Rightarrow (1+ (y-2)^{2} + 9) = (4 + (y-3)^{2} + 16)$

$\Rightarrow (10+y^2-4y+4)= (20+y^2-6y+9)$

$\Rightarrow y^2-4y+14=y^2-6y+29$

$\Rightarrow 2y=15$

$\Rightarrow y =$

$\displaystyle \dfrac{15}{2}$

$A = (1, -2, 3)$ ,

$B =$ (2, 1, 3),

$C =$ (4, 2, 1) and

$G= (-1,3, 5)$ is the centroid of the tetrahedron

$ABCD$ . Then the fourth coordinate is

0%
$(11,11,13)$
0%
$(-11,11,45)$
0%
$(-11,11,13)$
0%
$(11,13,11)$

Explanation

Given :

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

$G=(-1,3,5)$

Let

$D=(a,b,c)$

$G$ is centroid of the tetrahedron

$ABCD$ then,

$G =\dfrac{A+B+C+D}{4}$

$\implies G=\dfrac{(1,-2,3)+(2,1,3)+(4,2,1)+(a,b,c)}{4}$

$\implies 4G=(1+2+4+a,-2+1+2+b,3+3+1+c)$

$\implies (-4, 12, 20)=(7+a, 1+b, 7+c)$

$\implies 7+a=-4, 1+b=12, 7+c=20$

$\implies a=-11, b=11, c=13$

Hence,

$D=(-11,11,13)$ is the fourth coordinate.

$XOZ$ plane divides the join of

$(2,3,1)$ and

$(6,7,1)$ in the ratio

0%
$3: 7$
0%
$2 : 7$
0%
$-3 : 7$
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$-2 : 7$

Explanation

Let the plane divide the line in the ratio

$p : 1$ .
A point that divides the line joining these

$2$ points in the ratio

$p : 1$ is given by,

$\left ( \dfrac{6p+2}{p+1} , \dfrac{7p+3}{p+1}, \dfrac{p+1}{p+1} \right)$
Since, this point has to lie on the

$zx$ plane, so

$7p +3 =0$

$\dfrac{p}{1} =$

$\dfrac{-3}{7}$

The shortest distance of

$(a,b,c)$ from

$x$ -axis is

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

Explanation

The shortest distance of a point is always the perpendicular distance.
The projection of the point

$(a,b,c)$ on the

$x$ -axis is

$(a,0,0)$ .
This distance is given by

$\sqrt {{{b}^{2} + {c}^{2}} }$ .

The distance between the circumcentre and the ortho centre of the triangle formed by the points

$(2, 1, 5), (3, 2, 3)$ and

$(4, 0, 4)$ is

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$\sqrt{6}$
0%
$\displaystyle \dfrac{\sqrt{6}}{2}$
0%
$2\sqrt{6}$
0%
$0$

Explanation

let the points

$A(2,1,5) B(3,2,3)$ and

$C(4,0,4)$
Now using distance formula in

$3D$ , we have

$AB=\sqrt{6}$ ,

$BC=\sqrt{6}$ and

$AC=\sqrt{6}$
Therefore, it is a equilateral triangle and as we know that the circumcentre and orthocentre in equilateral triangle coincide each other.

Therefore distance between them is

$0$ .

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%
$\displaystyle \dfrac{1}{3}$
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$1$
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$\displaystyle \dfrac{3}{4}$
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$\displaystyle \dfrac{4}{3}$

Explanation

Let the points be given by

$(1,-1,5)$ and

$(2,3,4)$ .
A point that divides the line joining these

$2$ points in the ratio

$p : 1$ is given by

$\left ( \dfrac{2p+1}{p+1} , \dfrac{3p-1}{p+1}, \dfrac{4p+5}{p+1} \right)$ .
Since, this point has to lie on the

$zx$ plane, so

$3p - 1 =0$

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

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

The ratio in which

$yz$ -plane divides the line segment joining

$(-3, 4, 2), (2, 1, 3)$ is

0%
$-4 : 1$
0%
$3 : 2$
0%
$-2 : 3$
0%
$1: 4$

Explanation

Let the plane divide the line in the ratio

$p:1$ .
A point that divides the line joining these

$2$ points in the ratio

$p : 1$ is given by

$\left ( \dfrac{2p-3}{p+1} , \dfrac{p+4}{p+1}, \dfrac{3p+2}{p+1} \right)$
Since, this point has to lie on the

$zy$ plane, so

$2p -3 =0$

$p = \dfrac{3}{2}$

$(4, 2, p)$ is the centroid of the tetrahedron formed by the points

$(k, 2, -1), (4, 1, 1), (6,2, 5)$ and

$(3, 3, 3)$ then

$k + p=$

0%
$\displaystyle \frac{17}{3}$
0%
$1$
0%
$\displaystyle \frac{5}{3}$
0%
$5$

Explanation

Centroid of the tetrahedron formed by the points

$(k, 2, -1), (4, 1, 1), (6,2, 5)$ and

$(3, 3, 3)$ is,

$=\left(\dfrac{k+4+6+3}{4},\dfrac{2+1+2+3}{4},\dfrac{-1+1+5+3}{4} \right)=\left( \dfrac{k+ 13}{4},2,2\right)=(4,2,p)$ (given)

$\therefore \dfrac{k+13}{4}=4\Rightarrow k=3$

and

$p=2$

Hence

$k+p=3+2=5$

The circum centre of the triangle formed by the points

$(2, 5, 1), (1, 4, -3)$ and

$(-2, 7, -3)$ is

0%
$(6,0,1)$
0%
$(0,6,-1)$
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$(-1,6,2)$
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$(6,1,-2)$

Explanation

Let the points

$A(2,5,1) B(1,4,-3)$ and

$C(-2,7,-3)$
Now using distance formula in

$3D$ , we have

$AB {=}$

$\sqrt{18}$ ,

$BC {=}$

$\sqrt{18}$ , and

$AC {=}$

$\sqrt{36}$
Since,

${AB}^{2}+{BC}^{2}$

${=}$

${AC}^{2}$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e

$AC$ .
Therefore by using section formula (1:1), circumcentre

${=}(0,6,-1)$

The distance from the origin to the centroid of the tetrahedron formed by the points

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

0%
$\displaystyle \dfrac{\sqrt{a+{b}+{c}}}{4}$
0%
$\displaystyle \dfrac{\sqrt{{a}+{b}+{c}}}{3}$
0%
$\displaystyle \dfrac{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}{16}$
0%
$\displaystyle \dfrac{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}{4}$

Explanation

Let centroid of the tetrahedron having vertices

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

$(0,0,c)$ be

$G$ .

So, coordinate of

$G$ is given by,

$G =\left(\dfrac{0+a+0+0}{4},\dfrac{0+0+b+0}{4}, \dfrac{0+0+0+c}{4}\right)=\left(\dfrac{a}{4},\dfrac{b}{4},\dfrac{c}{4} \right)$

Hence, distance of

$G$ from origin is

$=OG=\sqrt{\left(\dfrac{a}{4}-0\right)^2+\left(\dfrac{b}{4}-0\right)^2+\left(\dfrac{c}{4}-0\right)^2}=\dfrac{\sqrt{a^2+b^2+c^2}}{4}$

The circum radius of the triangle formed by the points

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

$(-3, 1, 2)$ is:

0%
$\sqrt{14}$
0%
$14$
0%
$\sqrt{13}$
0%
$0$

Explanation

Let

$A(1,2,-3) , B(2,-3,1) , C(-3,1,2)$

$|AB | = \sqrt{1^2+ 5^2+4^2} = \sqrt{42}$

$|BC| = \sqrt {5^2+4^2+1^2} = \sqrt{42}$

$|AC| = \sqrt { 4^2+1^2+5^2} = \sqrt{42}$
It is an equilateral triangle. Hence, the value of

$R$ will be

$\displaystyle \dfrac{a}{\sqrt{3}}$

$= \sqrt {14}$

$G(1, 1, -2)$ is the centroid of the triangle

$ABC$ and

$D$ is the mid point of

$BC$ . If

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

$D =$

0%
$\left (\displaystyle \dfrac{1}{2},1, \dfrac{-5}{2}\right)$
0%
$(5,1,2)$
0%
$(-5,-1,-2)$
0%
$(2,1, -1)$

Explanation

Let the coordinates of

$D$ be

$(p,q,r)$ .
Since, the centroid divides the line joining

$AD$ in the ratio

$2:1$ , the coordinates of centroid should be,

$\left ( \dfrac{2p-1}{3} , \dfrac{2q+1}{3} , \dfrac{2r-4}{3} \right)$
Comparing it with the coordinates of the centroid given,

$D(2,1,-1)$ .

If the centroid of tetrahedron

$OABC$ where

$A,B,C$ are given by

$(a,2,3), (1,b,2)$ and

$(2,1,c)$ respectively is

$(1,2,-2)$ , then distance of

$P(a,b,c)$ from origin is

0%
$\sqrt{195}$
0%
$\sqrt{14}$
0%
$\sqrt{\dfrac{107}{14}}$
0%
$\sqrt{13}$

Explanation

The Centroid of tetrahedron

$OABC$ with vertices

$O(0,0,0), A(a,2,3), B(1,b,2)$ and

$C(2,1,c)$ is

$G\left(\dfrac{a+3}{4},\dfrac{b+3}{4},\dfrac{5+c}{4}\right)$ .
Comparing the coordinates of the Centroid

$G$ with the given coordinates of centroid

$(1,2,-2)$ , we get

$a = 1, b= 5, c = -13$

$\therefore$ distance of

$P(1,5,-13)$ from origin

$O(0,0,0)$ is

$D: \sqrt{1^2 + 5^2 + (-13)^2} = \sqrt{195}$

The circum radius of the triangle formed by the points

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

$(3, -4, -4)$ is

0%
$\displaystyle \dfrac{\sqrt{6}}{2}$
0%
$\displaystyle \dfrac{\sqrt{35}}{2}$
0%
$\displaystyle \dfrac{\sqrt{41}}{2}$
0%
$\sqrt{41}$

Explanation

Let the points

$A(2,-1,1) B(1,-3,-5)$ and

$C(3,-4,-4)$
Now using distance formula in

$3D$ , we have

$AB=\sqrt{41}$ ,

$BC=\sqrt{6}$ and

$AC=\sqrt{35}$
Since,

${BC}^{2}+{AC}^{2}$

${=}$

${AB}^{2}$
Hence, it is right angle triangle and as we know that the circum radius of right angled triangle is equal to half of hypotenuse.
Thus, radius

$=\dfrac {\sqrt {41}}{2}$

$\mathrm{A}= (-1,6, 6)$ ,

$\mathrm{B}=(-4,9, 6)$ ,

$\displaystyle \mathrm{G}=\frac{1}{3}(-5,22,22)$ and

$\mathrm{G}$ is the centroid of the

$\Delta \mathrm{A}\mathrm{B}\mathrm{C}$ then the name of the triangle

$\mathrm{A}\mathrm{B}\mathrm{C}$ is

0%
an isosceles triangle
0%
a right angled triangle
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an equilateral triangle
0%
a right-angled isosceles triangle

Explanation

The centroid of a triangle if given by

$G=((\dfrac{x_{1}+x_{2}+x_{3}}{3}),\dfrac{y_{1}+y_{2}+y_{3}}{3}),\dfrac{z_{1}+z_{2}+z_{3}}{3}))$
Let the coordinates of C be

$(x,y,z)$ ; then

$G=(\dfrac{-5}{3},\dfrac{22}{3},\dfrac{22}{3})=(\dfrac{x-1-4}{3},\dfrac{y+6+9}{3},\dfrac{z+6+6}{3})$
Comparing LHS and RHS gives us

$x-5=-5$ or

$x=0$ ; similarly

$y+15=22$ or

$y=7$ . and

$z+12=22$ or

$z=10$ .
Hence

$C=(0,7,10)$ .
Thus the coordinates of the vertices of the triangle are

$A(-1,6,6)$

$B=(-4,9,6)$ and

$C(0,7,10)$ .
Now

$AB=3\sqrt{2}$

$BC=6$

$AC=3\sqrt{2}$
Now

$AB^{2}+AC^{2}=(3\sqrt{2})^{2}+(3\sqrt{2})^{2}=18+18=36$

$=BC^{2}$ .
Hence it is a right angled isosceles triangle right angled at A

If the orthocentre, circumcentre of a triangle are

$(-3, 5, 2), (6, 2, 5)$ respectively then the centroid of the triangle is

0%
$(3,3, 4)$
0%
$\left (\displaystyle \dfrac{3}{2},\dfrac{7}{2},\dfrac{9}{2}\right)$
0%
$(9,9,12)$
0%
$\left (\displaystyle \dfrac{9}{2}\dfrac{-3}{2},\dfrac{3}{2}\right)$

Explanation

Since, the centroid divides the line joining the orthocentre and circumcentre in the ratio $2:1$

The coordinates of the centroid will be,

( $\dfrac{9}{3} , \dfrac{9}{3}, \dfrac{12}{3}$ )

$= (3,3,4)$

The extremities of a diagonal of a rectangular parallelopiped whose faces are parallel to the reference planes are

$(-2, 4, 6)$ and

$(3, 16, 6)$ . The length of the base diagonal is

0%
$7$
0%
$10$
0%
$11$
0%
$13$

Explanation

$AC=$ base diagonal

where,

$A(-2,4,6),C(3,16,6)$

$=\sqrt{(3+2)^2+(16-4)^2+(6-6)^2}$

$=\sqrt{5^2+12^2}$

$=13$

In the tetrahedron

$ABCD,\ A= (1, 2, -3)$ and

$G(-3,4, 5)$ is the centroid of the tetrahedron. If

$P$ is the centroid of the

$\Delta BCD$ , then

$AP=$

0%
$\displaystyle \dfrac{8\sqrt{21}}{3}$
0%
$\displaystyle \dfrac{4\sqrt{21}}{3}$
0%
$4\sqrt{21}$
0%
$\displaystyle \dfrac{\sqrt{21}}{3}$

Explanation

Given, $A=(1,2,-3), G(-3,4,5)$

Therefore, $AG=\sqrt { { (-3-1) }^{ 2 }+{ (4-2) }^{ 2 }+{ (5-(-3)) }^{ 2 } }$

and $AG=\sqrt { 84 } =2\sqrt { 21 }$

$P$ is the centroid of $\triangle BCD$

So, $G$ divides $AP$ in $3:1$ .

Let $AG=3x$ , then $GP=x$

$3x=2\sqrt { 21 } \\ x=\dfrac { 2\sqrt { 21 } }{ 3 }$

Now $AP=AG+GP$

$\Rightarrow AP=3x+x$

$\Rightarrow AP=4x$

$\Rightarrow AP=4\left( \dfrac { 2\sqrt { 21 } }{ 3 } \right) =\dfrac { 8\sqrt { 21 } }{ 3 }$

So, option A is correct.

In the

$\Delta$ ABC , A

$=$ (1, 3, -2) and G (-1, 4, 2) is the centroid of the triangle. If D is the mid point of BC then AD

$=$

0%
$\displaystyle \dfrac{\sqrt{21}}{2}$
0%
$\displaystyle \dfrac{3\sqrt{21}}{2}$
0%
$\sqrt{21}$
0%
$\dfrac{63}{2}$

Explanation

First, we calculate the distance AG,
It is

${(4+1+16)}^{0.5} = {21}^{0.5}$
From the property of the centroid that it divides the line joining AD in the ratio 2:1,
The distance

$AD = \frac{3}{2}$ AG
Hence,

$AD = \frac{3}{2} {21}^{0.5}$

The harmonic conjugate of

$(2, 3, 4)$ with respect to the points

$(3, -2, 2)$ and

$(6, -17, -4)$ is

0%
$(\dfrac{18}{5}, -5,\dfrac{4}{5})$
0%
$(11, -16,2)$
0%
$\left (\displaystyle \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}\right)$
0%
$(0, 0, 0)$

Explanation

Let us assume that point

$(2,3,4)$ divides

$(3,-2,2)$ and

$(6,-17,-4)$ in the ratio

$\lambda:1$

We have

$\frac{6\lambda+3}{1+\lambda}=2$

$\Rightarrow \lambda=-\frac{1}{4}$

Therefore point

$(2,3,4)$ divides given two points in the ratio

$1:4$ externally

Let the point which divides the given two points in the ratio

$1:4$ internally be

$(x,y,z)$

We have

$x=\frac{6+12}{5}=\frac{18}{5}$ ,

$y=\frac{-17-8}{5}=-5$ and

$z=\frac{-4+8}{5}=\frac{4}{5}$

Therefore option

$A$ is correct

$A = (2, 3, 0)$ and

$B = (2,1, 2)$ are two points. If the points

$P, Q$ are on the line

$AB$ such that

$AP= PQ = QB$ , then

$PQ=$

0%
$2\sqrt{2}$
0%
$6\sqrt{2}$
0%
$\sqrt{\dfrac{8}{9}}$
0%
$\sqrt{2}$

Explanation

$AB =\sqrt{(2-2)^2+(3-1)^2+(0-2)^2}=\sqrt 8$

Since

$AP=PQ=QB$

$\therefore PQ =\dfrac{AB}{3}=\dfrac{\sqrt 8}{3}=\sqrt{\dfrac{8}{9}}$

Then the correct matching is

List - I	List - II
A: The coordinates of the mid point of the line joining $(-1,-1,1)$ and $(-1,1,-1)$	1) $(-2,1,1)$
B: The coordinates of the point which divides the line segment joining $($ 2,3,1 $)$ and $(5, 0,4)$ in the ratio1:2	2) $(-1,0,0)$
$\mathrm{C}$ : The points and $P(2,1, -3)$ are three vertices of a parallelogram $\mathrm{P}\mathrm{Q}\mathrm{R}\mathrm{S}$ , the fourth vertex	3) $(\displaystyle \frac{13}{3},\frac{-11}{3},6)$
$\mathrm{D}$ : The vertices of a triangle are $(7,-4,7)$ , $(1,-6,10)$ and $(5, -1,1)$ . centroid of the triangle	4) $($ 3, 2, 2 $)$

0%
$A-2,\ B-4,\ C-1,\ D-3$
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$A-1,\ B-2,\ C-3,\ D-4$
0%
$A-2,\ B-3,\ C-1,\ D-4$
0%
$A-1,\ B-4,\ C-3,\ D-2$

Explanation

A: The co-ordinate of the mid point of the line joining

$(-1,-1,1)$ and

$(-1,1,-1)$

from mid point formula

$\left( \dfrac {-1-1}{2},\dfrac{-1+1}{2},\dfrac{1-1}{2}\right)$

$=(-1,0,0)$

$A\rightarrow 2$

B: co-ordinates of the point which divides the line segment Joining

$(2,3,1)$ and

$(5,0,4)$ in ratio

$1:2$

so from section formula

$\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2},\dfrac{m_1z_2+m_2z_1}{m_1+m_2}\right)$

Here

$m_1=1,m_2=2$

$=\left(\dfrac {1\times +2\times 2}{3},\dfrac{1(0)+2(3)}{3},\dfrac{1(4)+2(1)}{3}\right)$

$=(3,2,2)$

$B\rightarrow 4$

C: The point

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

$(2,1,-3)$ and

$A(x,y,z)$ vertices of a parallelogram

$PQRS$ then fourth vertex

we know that the diagonals of parallelogram bisect each other have they mid point of

$PR=$ mid point of

$QS$ .... [Ref. image]

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

$\Rightarrow \dfrac{x+2}{2}=\dfrac{7}{2} \Rightarrow x=5$

$\Rightarrow \dfrac{y+3}{2}=\dfrac{-1}{2} \Rightarrow y=-2$

$\Rightarrow \dfrac{z+1}{2}= \dfrac{1}{2} \Rightarrow z=0$

D: The vertixes of a triangle

$ax$

$(7,-4,7),(1,-6,10)$ and

$(5,-1,1)$ Centroid of triangle

Then centroid

$=\left(\dfrac{7+1+5}{3},\dfrac{-4-6-1}{3},\dfrac{7+10+1}{3}\right)$

$=\left(\dfrac{13}{3},\dfrac{-11}{3},6\right)$

$D \rightarrow 3$

Hence correct option is A

If the points

$A,B,C,D$ are collinear and

$C,D$ divide

$AB$ in the ratios

$2:3, -2:3$ respectively, then the ratio in which

$A$ divides

$CD$ is

0%
$5:1$
0%
$2:3$
0%
$3:2$
0%
$1:5$

Explanation

$C$ divide

$AB$ in the ratio of

$2:3$

$\Rightarrow AC=2x,CB=3x$

$D$ divide

$AB$ in the ratio of

$-2:3$

$\Rightarrow DA=2y,DB=3y$

$\Rightarrow AB=DB-DA=y$

$\Rightarrow 5x=y\Rightarrow x=\dfrac { y }{ 5 }$

$\Rightarrow CA:AD=2x:2y=\dfrac { x }{ y } = \cfrac{\dfrac y5}{y} = \cfrac15$

$\Rightarrow$

$A$ divides

$CD$ in the ratio

$1:5$

$P (1,1,1 )$ and

$Q(\lambda, \lambda, \lambda)$ are two points in the space such that

$PQ=\sqrt{27}$ , then the value(s) of

$\lambda$ can be

0%
$-4$
0%
$-2,4$
0%
$2$
0%
$4,3$

Explanation

The square of the distance between the

$2$ points is

${3\times{(\lambda - 1)}^{2}}$
This has to be equal to

$27$ .
Hence,

$\lambda - 1$

$= 3,-3$

$\lambda = -2,4$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 3 - 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 3 - MCQExams.com

0:0:2

Practice Class 11 Engineering Maths Quiz Questions and Answers

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