MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 5

$P= (0, 0, 0), Q = (3, 6, 9)$ and

$R$ is a point of trisection of

$PQ$ , then

$R_y=$

0%
$\displaystyle \frac{4}{3}$
0%
$2$
0%
$3$
0%
$9$

Explanation

$R$ is a point of trisection of

$PQ$ , then it divides the line

$PQ$ in the ratio of

$2:1$ or

$1:2$
Case 1:

$R$ divides

$PQ$ in

$2:1$ ratio,
Then coordinates of

$R$ are

$\left(\dfrac{2\times3+1\times0}{3}, \dfrac{2\times6+1\times0}{3}, \dfrac{2\times9+1\times0}{3}\right) = (2,4.6)$
Case 2:

$R$ divides

$PQ$ in

$1:2$ ratio,
Then coordinates of

$R$ are

$\left(\dfrac{1\times3+2\times0}{3}, \dfrac{1\times6+2\times0}{3}, \dfrac{1\times9+2\times0}{3}\right) = (1,2,3)$

$\therefore R_y = 2$ or

$4$
Hence, option B is correct.

If the distance of a point

$(a,a,a)$ from the origin is

$\sqrt { 108 }$ , then the value of

$a$ is

0%
$9$
0%
$6$
0%
$-9$
0%
$-6$

Explanation

Distance between

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$\sqrt { { ({ x }_{ 1 }-{ x }_{ 2 }) }^{ 2 }+{ ({ y }_{ 1 }-{ y }_{ 2 }) }^{ 2 } }$

Distance between

$(0,0,0)$ and

$(a,a,a )$ is

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

$\sqrt { 3\times { (a) }^{ 2 } } =\sqrt { 108 } =6\sqrt { 3 } \\ \Rightarrow a=\pm 6\\$

Three vertices of a tetrahedron are

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

$(-4, 1, 3)$ . If the centroid of the tetrahedron be

$(1, -2, 5)$ then the fourth vertex is

0%
$(2, -4, 18)$
0%
$(1, -4, 18)$
0%

$\left (\dfrac {3}{2}, \dfrac {-3}{2}, \dfrac {7}{4}\right )$
0%
none of these

Explanation

if the fourth vertex is

$(x_1,y_1,z_1)$

Then

$0+6-4+x_1=0, \therefore x_1=2$

similarily

$0-5+1+y_1=0 \therefore y_1=-4$

$0-1+3+z_1=20$

$\therefore z_1=18$

$\therefore$ the fourth vertex is

$(2,-4,18)$

$= (1, 1, 4)$ and B

$= (5,-3, 4)$ 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%
$4$
0%
$\sqrt{\displaystyle \frac{32}{9}}$
0%
$\sqrt{2}$

Explanation

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

$AB = \sqrt {(-4)^2+4^2}$

$AB = \sqrt {32}$

Given that

$AP=PQ=QB$

But,

$AB=AP+PQ+QB$

Hence,

$AB=3 \times PQ$

So,

$PQ = \cfrac {\sqrt {32}}{3} = \sqrt {\cfrac {32}{9}}$

If P

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

$(5, 4, -6)$ and R

$(9, 8, -10)$ are collinear, then R divides PQ in the ratio

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

Explanation

Let R divides PQ in ratio m:n.

By Section Formula

coordinates of R are:

$R=(\dfrac{5m+3n}{m+n},\dfrac{4m+2n}{m+n},\dfrac{-6m-4n}{m+n})$

we have coordinates of R=(9,8,-10)

Therefore,

$9=\dfrac{5m+3n}{m+n}$

$9m+9n=5m+3n$

$-6n=4m$

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

Therefore R divides PQ in 3:2 ratio and negative sign indicates,it cuts Externally.

Therefore option (B) is Correct.

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

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

0%
$\displaystyle \frac{\sqrt{\mathrm{3}+\mathrm{4}+\mathrm{5}}}{4}$
0%
$\displaystyle \frac{\sqrt{\mathrm{3}+\mathrm{4}+\mathrm{5}}}{3}$
0%
$\displaystyle \frac{\sqrt{\mathrm{3}^{2}+\mathrm{4}^{2}+\mathrm{5}^{2}}}{16}$
0%
$\displaystyle \frac{\sqrt{\mathrm{3}^{2}+\mathrm{4}^{2}+\mathrm{5}^{2}}}{4}$

Explanation

$G=(\dfrac{a}{4},\cfrac{b}{4},\cfrac{c}{4})$

$\therefore OG=\sqrt{(\cfrac{a^2}{16}+\cfrac{b^2}{16}+\cfrac{c^2}{16})}$

Thus, here

$a=3,b=4,c=5$

substituting in above equation we get

$OG=\sqrt{\cfrac{3^2+4^2+5^2}{4}}$

The position vectors of the four angular point of a tetrahedron

$OABC$ are

$(0, 0, 0)$ ;

$(0, 0, 2)$ ;

$(0, 4, 0)$ and

$(6, 0, 0)$ respectively. Find the coordinates of cenroid

0%
$\left (2, \displaystyle \frac {4}{3}, \displaystyle \frac {2}{3}\right )$
0%
$\left (\displaystyle \frac {6}{4}, 1, \displaystyle \frac {2}{4}\right )$
0%
$(0, 0, 0)$
0%
none of these

Explanation

Angular points of tetrahedron OABC are

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

To find the coordinates of the centroid of the tetrahedron whose vertices are

$(x_1, y_1, z_1 ), (x_2, y_2, z_2 ), (x_3, y_3, z_3 )$ and

$(x_4 , y_4 , z_4 ).$

Hence, the centroid is

$(\cfrac{x_1+x_2+x_3+x_4}{4}),(\cfrac{y_1+y_2+y_3+y_4}{4}),(\cfrac{z_1+z_2+z_3+z_4}{4})$

Now, substituting the values we get

$(\cfrac{0+0+0+6}{4}),(\cfrac{0+0+4+0}{4}),(\cfrac{0+2+0+0}{4})$

$\therefore$ the coordinates of the centroid are :

$(\cfrac{6}{4},1,\cfrac{2}{4})$

Let

$A= \left ( 1,2,3 \right )B= \left ( -1,-2,-1 \right )C= \left ( 2,3,2 \right )$ and

$D= \left ( 4,7,6 \right )$ . Then

$ABCD$ is a

0%
rectangle
0%
square
0%
parallelogram
0%
none of these

Explanation

${=}$

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

${=}$

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

${=}$

$\sqrt{36}$
AB

${=}$ 6
Similarly you find that BC

${=}$

$\sqrt{43}$ CD

${=}$ 6 and DA

${=}$

$\sqrt{43}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC

${=}$

$\sqrt{{(2-1)}^{2}+{(3-2)}^{2}+{(2-3)}^{2}}$
AC

${=}$

$\sqrt{3}$
similarly BD

${=}$

$\sqrt{155}$
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
hence it is parallelogram

$A= \left ( 5,-1,1 \right ),B= \left ( 7,-4,7 \right ),C= \left ( 1,-6,10 \right ),D= \left ( -1,-3,4 \right )$ . Then

$ABCD$ is a

0%
square
0%
rectangle
0%
rhombus
0%
none of these

Explanation

${=}$

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

${=}$

$\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$
AB

${=}$

$\sqrt{49}$
AB

${=}$ 7
Similarly you find that BC

${=}$

$\sqrt{49}$ CD

${=}$ 7 and DA

${=}$ 7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC

${=}$

$\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$
AC

${=}$

$\sqrt{122}$
similarly BD

${=}$

$\sqrt{74}$
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus

The equation of the plane which is parallel to the

$xy-$ plane is

0%
$x=y$
0%
$z=c$
0%
$y=c$
0%
$z=xy$

Explanation

The equation of a plane which is parallel to the

$xy-$ plane is

$z=c$

$C_1:{x^2+y^2}-20x+64=0$ and

$C_2:{x^2+y^2}+30x+144=0$ . Then the length of the shortest line segment

$PQ$ which touches

$C_1$ at

$P$ and to

$C_2$ at

$Q$ is

0%
$10$
0%
$15$
0%
$22$
0%
$27$

Explanation

Given

$C_1 : x^2+y^2-20x+64=0 \Rightarrow (x-10)^2+y^2=36$
and

$C_2 : x^2+y^2+30x+144=0 \Rightarrow (x+15)^2+y^2=81$
So centre and radius of

$C_1$ and

$C_2$ are

$(10,0)$ ,

$(-15,0)$ and

$6,9$ respectively.
Then, distance between

$C_1$ and

$C_2$ is

$\sqrt{(15+10)^{2}+(0-0)^{2}}=25$ .

$PQ$ touches

$C_{1}$ at

$P$ and

$C_{2}$ at

$Q$ .

Then, shortest length of

$PQ$ is

$=25-(9+6)=25-15=10$

The plane

$XOZ$ divides the join of

$(1,-1,5)$ and

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

$\lambda :1$ , then

$\lambda$ is

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

Explanation

${\textbf{Step - 1: Using section formula.}}$

${\text{ Let given points (1,-1,5) and (2,3,4) be}}$

$(x_1,y_1,z_1)$

${\text{and}}$

$(x_2,y_2,z_2)$

${\text{The given ratios are}}$

$\lambda : 1$

${\text{be}}$

$m:n$

${\text{Section formula are:}}$

$\mathbf{\left( {\dfrac{{mx_2 + nx_1}}{{m + n}},\dfrac{{my_2 + ny_1}}{{m + n}},\dfrac{{mz_2 + nz_1}}{{m + n}}} \right)}$

$\text{putting values we get, } \left( {\dfrac{{2\lambda + 1}}{{\lambda + 1}},\dfrac{{3\lambda - 1}}{{\lambda + 1}},\dfrac{{4\lambda + 5}}{{\lambda + 1}}} \right)$

${\textbf{Step - 2: Taking y co-ordinate equals to zero}}$

${\text{Since, this points lies in XOZ plane then its y co-ordinate should be zero}}$

$\Rightarrow$ $\dfrac{{3\lambda - 1}}{{\lambda + 1}} = 0$

$\Rightarrow$ ${{3\lambda - 1}}= 0$

$\Rightarrow$ $3\lambda= 1$

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

${\textbf{Hence, the correct option is (D)}}$ $\mathbf{\lambda = \dfrac{1}{3}}$

The distance of the point

$(1,-2,3)$ from the plane

$x-y+z=5$ measured parallel to the line

$\displaystyle \frac{x}{2}=\displaystyle \frac{y}{3}=\displaystyle \frac{z}{-6}$ is

0%
$1$
0%
$19$
0%
$155$
0%
$\sqrt{271}$

Explanation

Equation of the line through

$(1,-2,3)$ parallel to the line

$\displaystyle \frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z}{ -6 }$ is

$\displaystyle \frac { x-1 }{ 2 } =\frac { y+2 }{ 3 } =\frac { z-3 }{ -6 } =r$ (say) ...(1)
Then any point on (1) is

$(2r+1,3r-2,-6r+3)$
If this point lies on the plane

$x-y+z=5$ then

$(2r+1)-(3r-2)+(-6r+3)=5$

$\displaystyle \Rightarrow -7r+6=5\Rightarrow r=\frac { 1 }{ 7 }$
Hence, the point is

$\displaystyle \left( \frac { 9 }{ 7 } ,\frac { -11 }{ 7 } ,\frac { 15 }{ 7 } \right)$
Distance between

$(1,-2,3)$ and

$\displaystyle \left( \frac { 9 }{ 7 } ,\frac { -11 }{ 7 } ,\frac { 15 }{ 7 } \right)$

$\displaystyle =\sqrt { \frac { 4 }{ 49 } +\frac { 9 }{ 49 } +\frac { 36 }{ 49 } } =\sqrt { \frac { 49 }{ 49 } } =1$

The points

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

$D(3, 1, -2)$ are

0%
collinear
0%
vertices of a rectangle
0%
vertices of a square
0%
vertices of a rhombus

Explanation

Given,

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

$D(3,1,-2)$

$AB$

${=}$

$\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$

${=}$

$\sqrt{11}$

$BC$

${=}$

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

${=}$

$\sqrt{6}$

$CD$

${=}$

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

${=}$

$\sqrt{11}$

$DA$

${=}$

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

${=}$

$\sqrt{6}$

$AB{=}CD$ and

$BC{=}DA$

i.e. opposite sides are equal.

Now Direction ratio of

$AB =(1,3,-1)$ and Direction ratio of

$BC= (2,-1,-1)$ ,

$\vec {AB}=\hat i+\hat j+\hat k$ &

$\vec {BC}=2\hat i-\hat j-\hat k$

$\vec {AB}\cdot \vec {BC}=2-1-1=0$

As the dot product between

$AB$ and

$BC$ is

$0$ , that means angle between them is

$90^o$ .

Similarly check for

$BC$ and

$CD$ , then you find that all angles are equal to

$90^o$ and opposite sides are equal.

Hence, it is a rectangle.

A line passes through two point

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

$B (8, -1, 2)$ . The coordinates of a point on this line at a distance of

$14$ units from

$A$ are

0%
$(14, 1, 5)$
0%
$(-10, -7, 7)$
0%
$(86, 25, 41)$
0%
None of these

Explanation

Given points are

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

$B(8,-1,2)$
Therefore direction ratio of

$AB$ are

$l = \dfrac{6}{7}, m=\dfrac{2}{7},n= \dfrac{3}{7}$ or

$l=\dfrac{-6}{7},m= \dfrac{-2}{7},n= \dfrac{-3}{7}$
Hence, coordinates of a point

$14$ unit from point

$A$ is given as,

$(2+14l,-3+14m,-1+14n)$

$\Rightarrow (14,1,5)$ or

$(-10,-7,-7)$
Hence, option 'A' is correct.

The chord of contact of tangents from a point

$P$ to a circle passes through

$q$ . If

$l_1$ and

$l_2$ are the lengths of the tangents from

$P$ and

$Q$ to the circle, then

$PQ$ is equal to

0%
$\dfrac{l_1+l_2}{2}$
0%
$\dfrac{l_1-l_2}{2}$
0%
$\sqrt{|{l^2_1}-{l^2_2}|}$
0%
$\sqrt[2]{{l^2_1}+{l^2_2}}$

Explanation

Let

$P\equiv \left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$Q\equiv \left( { x }_{ 2 },{ y }_{ 2 } \right)$
Let the equation of given circle be

${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$
The equation of chord of contact of tangent drawn from the point

$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$ to the given circle is

${ xx }_{ 1 }+{ yy }_{ 1 }={ a }^{ 2 }$ .
Since it passes through

$Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$

$\therefore { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 }={ a }^{ 2 }$ ...(1)
Now

${ l }_{ 1 }=\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-{ a }^{ 2 } } ,{ l }_{ 2 }=\sqrt { { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }-{ a }^{ 2 } }$
and

$\\ PQ=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \\ =\sqrt { \left( { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 } \right) +\left( { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 } \right) -2\left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \right) } \\ =\sqrt { \left( { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 } \right) +\left( { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }-{ a }^{ 2 } \right) } \\ =\sqrt { \left( { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-{ a }^{ 2 } \right) \left( { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }-{ a }^{ 2 } \right) } \\ =\sqrt { { l }_{ 1 }^{ 2 }+{ l }_{ 2 }^{ 2 } }$

Find the coordinates of the point on the

$x$ -axis that is equidistant from

$P(4,3,1)$ and

$Q(-2,-6,-2)$ .

0%
$\displaystyle \left( \frac { 3 }{ 2 } ,0,0 \right)$
0%
$\displaystyle \left( - \frac { 3 }{ 2 } ,0,0 \right)$
0%
$\displaystyle \left( 0,-\frac { 3 }{ 2 } ,0 \right)$
0%
$\displaystyle \left( 0,\frac { 3 }{ 2 } ,0 \right)$

Explanation

Let

$R(x,0,0)$ be the point on

$x$ -axis which is equidistant from

$P(4,3,1)$ and

$Q(-2,-6,-2)$

$\Rightarrow { \left( x-4 \right) }^{ 2 }+{ \left( -3 \right) }^{ 2 }+{ \left( -1 \right) }^{ 2 }={ \left( x+2 \right) }^{ 2 }+{ 6 }^{ 2 }+{ 2 }^{ 2 }$ gives

$-12x=18$

$x=-1.5$

Hence,

$\displaystyle R\equiv \left( -\frac { 3 }{ 2 } ,0,0 \right)$

$(1,-1,0),(-2,1,8)$ and

$(-1,2,7)$ are three consecutive vertices of a parallelogram then the fourth vertex is

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

Explanation

Let the fourth vertex be

$(h,k,l)$

The midpoints of diagonal will be same

The midpoints are

$(0,\frac{1}{2},\frac{7}{2})$ and

$(\frac{h-2}{2} , \frac{k+1}{2},\frac{l+8}{2})$

If we equate both we get

$h-2=0 , k+1=1 , l+8=7$

$\Rightarrow h=2 , k=0 , l=-1$

So the fourth vertex is

$(2,0,-1)$

$P(x,y,z)$ is a point on the line segment joining

$Q(2,2,4)$ and

$R(3,5,6)$ such that the projection of

$\overrightarrow { OP }$ on the axes are

$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 } ,\frac { 26 }{ 5 }$ respectively, then

$P$ divides

$QR$ in ratio

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

Explanation

Since $\overrightarrow { OP }$ has projection $\displaystyle \dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 }$ and $\displaystyle \dfrac { 26 }{ 5 }$ on the coordinate axes, therefore $\displaystyle \overrightarrow { OP } =\dfrac { 13 }{ 5 } i+\dfrac { 19 }{ 5 } j+\dfrac { 26 }{ 5 } k$

Let $P$ divides the join of $Q\left( 2,2,4 \right)$ and $R\left( 3,5,6 \right)$ in the ratio $\lambda :1$ .

The position vector of $P$ is $\displaystyle \left( \dfrac { 3\lambda +2 }{ \lambda +1 } \right) i+\left( \dfrac { 5\lambda +2 }{ \lambda +1 } \right) j+\left( \dfrac { 6\lambda +4 }{ \lambda +1 } \right) k$

$\displaystyle \therefore \dfrac { 13 }{ 5 } i+\dfrac { 19 }{ 5 } j+\dfrac { 26 }{ 5 } k=\left( \dfrac { 3\lambda +2 }{ \lambda +1 } \right) i+\left( \dfrac { 5\lambda +2 }{ \lambda +1 } \right) j+\left( \dfrac { 6\lambda +4 }{ \lambda +1 } \right) k$

$\displaystyle \Rightarrow \dfrac { 3\lambda +2 }{ \lambda +1 } =\dfrac { 13 }{ 5 } ,\dfrac { 5\lambda +2 }{ \lambda +1 } =\dfrac { 19 }{ 5 } ,\dfrac { 6\lambda +4 }{ \lambda +1 } =\dfrac { 26 }{ 5 }$

$\Rightarrow \lambda =\dfrac { 3 }{ 2 }$

The distance of the point

$\left( 1,-2,3 \right)$ from the plane

$x-y+z=5$ measured parallel to the line

$\displaystyle \frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z-1 }{ -6 }$ is

0%
$1$
0%
$2$
0%
$4$
0%
None of these

Explanation

Equation of the line through $\left( 1,-2,3 \right)$ parallel to the line $\displaystyle \dfrac { x }{ 2 } =\dfrac { y }{ 3 } =\dfrac { z-1 }{ -6 }$ is

$\displaystyle \dfrac { x-1 }{ 2 } =\dfrac { y+2 }{ 3 } =\dfrac { z-1 }{ -6 } =r$ (say) ... $(1)$

Then any point on $(1)$ is $\left( 2r+1,3r-2,-6r+3 \right)$ .

If this point lies on the plane $x-y+z=5$ , then

$\displaystyle \left( 2r+1 \right) -\left( 3r-2 \right) +\left( -6r+3 \right) =5\Rightarrow -7r+6=5\Rightarrow r=\dfrac { 1 }{ 7 }$

Hence, the point is $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 } \right)$

Distance between $\left( 1,-2,3 \right)$ and $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 } \right)$

$\displaystyle =\sqrt { \left( \dfrac { 4 }{ 49 } +\dfrac { 9 }{ 49 } +\dfrac { 36 }{ 49 } \right) } =\sqrt { \dfrac { 49 }{ 49 } } =1$

The ratio in which the plane

$\displaystyle \bar {r} .(\bar {i} - 2 \bar {j} + 3 \bar {k}) = 17$ divides the line joining the points

$\displaystyle -2 \bar {i} + 4 \bar {j} + 7 \bar {k}$ and

$\displaystyle 3 \bar {i} - 5 \bar {j} + 8 \bar {k}$ is

0%
$1 : 10$
0%
$3 : 10$
0%
$3 : 5$
0%
$1 : 5$

Explanation

Let

$\vec {r}$ be a point on a plane

$\vec {r} .(\vec i-2\vec j+3\vec k)=17$ which divides line joining the points

$-2\vec i+4\vec j+7\vec k$ &

$3\vec i-5\vec j+8\vec k$ in the ratio

$n:m$
Therefore,

$\vec {r}=\vec { r } =\dfrac { m\left( -2\vec i+4\vec j+7\vec k \right) +n\left( 3\vec i-5\vec j+8\vec k \right) }{ m+n } =\dfrac { \left( -2m+3n \right) i+\left( 4m-5n \right) j+\left( 7m+8n \right) k }{ m+n }$

Substituting

$\vec {r}$ in equation of plane, we get

$\left( \dfrac { \left( -2m+3n \right) \vec i+\left( 4m-5n \right) \vec j+\left( 7m+8n \right) \vec k }{ m+n } \right) .\left( \vec i-2\vec j+3\vec k \right) =17$

$\Rightarrow -2m+3n-8m+10n+21m+24n=17m+17n$

$\Rightarrow -6m+20n=0$

$\Rightarrow n:m=3:10$

Ans: B

A point on the line

$\displaystyle \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+1}{3}$ at a distance

$\sqrt{6}$ from the origin is

0%
$\left (\displaystyle \frac{-5}{7},\frac{-10}{7},\frac{13}{7}\right)$
0%
$(1,2,-1)$
0%
$\left (\displaystyle \frac{5}{7},\frac{10}{7},\frac{-13}{7}\right)$
0%
$(-1,-2,1)$

Explanation

Given line is,

$\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z+1}{3} = k$ (assume)
So any point on this line is

$P(k+1, 2k+2, 3k -1)$
Distance of any point on this line from the origin is,

$\sqrt{(k+1)^2 + (2k+2)^2 + (3k-1)^2} = \sqrt{6}$ (given)

$\Rightarrow 14k^2 + 4k = 0$

$\Rightarrow k = 0, \dfrac{-2}{7}$

Therefore, point on the line is,

$(1, 2, -1)$ or

$\left (\dfrac{5}{7}, \dfrac{10}{7}, -\dfrac{13}{7}\right)$

A point

$Q$ at a distance

$3$ from the point

$P(1,1,1)$ lying on the line joining the points

$A(0,-1,3)$ and

$P$ , has the coordinates

0%
$(2,3,-1)$
0%
$(4,7,-5)$
0%
$(0,-1,3)$
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$(-2,-5,7)$

Explanation

Let the coordinates of point

$Q$ be

$(a,b,c)$

The distance of

$Q$ from

$P(1,1,1)$ is

$3$

The equation of line

$AP$ is

$\displaystyle \frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{-2}$

Therefore the point on the line

$AP$ will look like

$Q(t+1,2t+1,1-2t)$

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

$|QP| = \pm3t=3$

$\Rightarrow t=\pm1$

So the possible coordinates of

$Q$ are

$(2,3,-1)$ and

$(0,-1,3)$

Therefore the correct options are

$A$ and

$C$

If two vertices of a triangle

$ABC$ are

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

$B(2,-3,0)$ ,and the centroid is

$(2,0,2)$ then the vertex

$C$ has the coordinates

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$(5,1,2)$
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$\left ( 1,-\displaystyle \frac{1}{3},\frac{7}{3} \right )$
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$\left ( 3,-\displaystyle \frac{2}{3},\frac{5}{3} \right )$
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none of these

Explanation

Let

$C$ be

$(a,b,c)$
By using the formula of centroid of triangle, we get

$G(2,0,2) = \left(\dfrac{-1+2+a}{3},\dfrac{2-3+b}{3},\dfrac{4+0+c}{3}\right)$

$\implies (2,0,2) = \left(\dfrac{1+a}{3},\dfrac{b-1}{3},\dfrac{4+c}{3}\right)$

Comparing the coordinates we get

$\dfrac{a+1}{3}=2, \dfrac{b-1}{3}=0, \dfrac{4+c}{3}=2$

$\implies a=5, b=1, c=2$
Hence,

$C=(5,1,2)$

Three vertices of a tetrahedron are

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

$(-4,1,3)$ . If the centroid of the tetrahedron be

$(1,-2,5)$ then the fourth vertex is

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$(2,-4,18)$
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$(2,-4,-18)$
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$\left (\displaystyle \frac{3}{4},\frac{-3}{2},\frac{7}{4} \right )$
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none of these

Explanation

The three vertices of tetrahedron is

$(0,0,0)$ ,

$(6,-5,-1)$ and

$(-4,1,3)$

Let the fourth vertex be

$(x,y,z)$

Given that the centroid is

$(1,-2,5)$

So we have

$\frac{x+0+6-4}{4}=1$ ,

$\frac{y+0-5+1}{4}=-2$ and

$\frac{z+0+3-1}{4}=5$

$\Rightarrow x=2$ ,

$y=-4$ and

$z=18$

Therefore the correct option is

$A$

The points

$(4, -5, 1)$ ,

$(3, -4, 0)$ ,

$(6, -7, 3)$ ,

$(7, -8, 4)$ are vertices of a

0%
square
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parallelogram
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rectangle
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rhombus

Explanation

${\textbf{Step -1: Finding distance between points .}}$

${\text{Let A(4, -5, 1)}}$

${\text{B=(3, -4, 0)}}$

${\text{C=(6, -7, 3)}}$

${\text{D=(7, -8, 4)}}$

${\text{ Now let the quadrilateral be ABCD}}$

${\text{Then,}}$

$AB=-i+j-k |AB|$

$=\sqrt3$

$BC= 3i-3j+3k |BC|$

$=3\sqrt 3$

$CD=i-j+k|CD|$

$=\sqrt3$

$AD=3i-3j+3k|AD|$

$=3\sqrt3$

$\textbf{Step -2: Checking distances between opposite sides . }$

${\text{Hence opposite sides are equal and parallel. Therefore the above}}$

${\text{points form a parallelogram. Now all the sides are not equal. Hence}}$

${\text{it cannot qualify as a rhombus or square.}}$

${\text{Now dot product of AB and BC is not zero. Hence adjacent sides are not perpendicular to each other.}}$

${\text{Therefore it is not a rectangle also.}}$

${\textbf{ Hence , option B is correct.}}$

Four vertices of a tetrahedron are

$(0,0,0),(4,0,0),(0,-8,0)$ and

$(0,0,12)$ ,Its centroid has the coordinates

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$\left ( \displaystyle \frac{4}{3},-\frac{8}{3},4 \right )$
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$(2,-4,6)$
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$(1,-2,3)$
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none of these

Explanation

the centroid of the coordinates is

$\left(\dfrac {x_1+x_2+x_3+x_4}{4},\dfrac{y_1+y_2+y_3+y_4}{4},\dfrac{z_1+z_2+z_3+z_4}{4}\right)$

Thus by substituting the vertices we get

$=\left(\dfrac{0+4+0+0}{4},\dfrac{0+0-8+0}{4},\dfrac{0+0+0+12}{4}\right)$

$=\left(\dfrac{4}{4},\dfrac{-8}{4},\dfrac{12}{4}\right)$

$\therefore$ the centroid of the coordinates is

$(1,-2,3)$

In geometry, we take a point, a line and a plane as undefined terms.

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True
0%
False
0%
Ambiguous
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Data Insufficient

A rectangular parallelopiped is formed by drawing planes through the points

$(-1,2,5)$ and

$(1,-1,-1)$ and parallel to the coordinate planes. the length of the diagonal of the parallelopiped is

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$2$
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$3$
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$6$
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$7$

Explanation

The plane forming the parallelipiped are

$x=-1,x=1;y=2,y=-1$ and

$z=5,z=-1$

Hence, the lengths of the edges of the parallelopiped are

$1-\left( -1 \right) =2,\left| -1-2 \right| =3$ and

$\left| -1-5 \right| =6$

$($ length of an edge of the parallelopiped is the distance between the parallel plane sperpendicular to the edge

$)$

$\therefore$ Length of diagonal of the parallelopiped

$=\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 49 } =7$

The ratio of

$yz$ -plane divide the line joining the points

$A(3, 1,- 5), B(1, 4, -6)$ is

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

Explanation

Let

$yz$ -plane divide the line segment joining the points

$A(3,1,-5)$ &

$B(1,4,-6)$ in the ratio

$m:n$
Then,

$(0,y,z)=\left( \dfrac { 3m+n }{ m+n } ,\dfrac { m+4n }{ m+n } ,\dfrac { -5m-6n }{ m+n } \right)$

$\Rightarrow \dfrac { 3m+n }{ m+n } =0$

$\Rightarrow m:n=-1:3$

Ans: B

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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