MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 4

The value(s) of

$\lambda$ , for which the triangle with vertices

$(6,10,10),(1,0,-5)$ and

$(6,-10,\lambda)$ will be a right angled triangle is/ are

0%
$1$
0%
$\displaystyle \dfrac{70}{3},0$
0%
$35$
0%
$\displaystyle 0,-\dfrac{70}{3}$

Explanation

Let the given points be

$A,B$ and

$C$ respectively.
Then,

${ AB }^{ 2 }=350,{ AC }^{ 2 }=500\lambda -20\lambda +{ \lambda }^{ 2 }{ BC }^{ 2 }=150+10\lambda +{ \lambda }^{ 2 }$
Now,

$\displaystyle{ AB }^{ 2 }+{ AC }^{ 2 }={ BC }^{ 2 }\Rightarrow 350+500-20\lambda +{ \lambda }^{ 2 }=150+10\lambda +{ \lambda }^{ 2 }\Rightarrow \lambda =\dfrac { 70 }{ 3 }$
Next,

${ AB }^{ 2 }+BC^{ 2 }={ AC }^{ 2 }\Rightarrow 250+150+10\lambda +{ \lambda }^{ 2 }=500-20\lambda +{ \lambda }^{ 2 }\Rightarrow \lambda =0$
Further,

${ BC }^{ 2 }+AC^{ 2 }={ AB }^{ 2 }\Rightarrow 150+10\lambda +12+500-20\lambda +{ \lambda }^{ 2 }=350\Rightarrow { \lambda }^{ 2 }-5\lambda +150=0$ ,
which has no real solution.

$\therefore$ the triangle is right angle for

$\displaystyle\lambda=0,\dfrac { 70 }{ 3 }$

$A(\cos\alpha,\sin\alpha, 0),B(\cos\beta,\sin\beta, 0)$ ,

$C(\cos\gamma,\sin\gamma,0)$ are vertices of

$\Delta ABC$ and let

$\cos \alpha+\cos\beta+\cos\gamma=3{a}$ ,

$\sin\alpha+\sin\beta+\sin\gamma =3b$ , then correct matching of the following is:

List : I	List : II
A. Circumcentre	$(3a,3b,0)$
B. Centroid	$(0,0,0)$
C. Ortho centre	$(a,b,0)$

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

Explanation

We know,

${ \sin }^{ 2 }\alpha +{ \cos }^{ 2 }\alpha =1$
Similarly it is true for other three coordinates.
These are points on a circle of unit radius.
By observation

$(0,0,0)$ would be the circumcentre of the circle.
Centroid

$=\cfrac { (\cos{ \alpha }_{ 1 }+\cos{ \alpha }_{ 2 }+\cos{ \alpha }_{ 3 }) }{ 3 } ,\cfrac { (\sin{ \alpha }_{ 1 }+\sin{ \alpha }_{ 2 }+\sin{ \alpha }_{ 3 }) }{ 3 } ,0$
Centroid

$=(\cfrac { 3a }{ 3 } ,\cfrac { 3b }{ 3 } ,0)$ =

$(a,b,0)$
Centroid divides line segment joining orthocentre and circumcentre in the ratio 2:1.
Orthocentre

$=(3a,3b,0)$

Arrange the points:

$\mathrm{A}(1,2-3), \mathrm{B}(-1,2,-3), \mathrm{C}(-1,-2-3)$ and

$\mathrm{D}(1,-2, -3)$ in the increasing order of their octant numbers:

0%
$A,B,C,D$
0%
$B,C,D,A$
0%
$C,D,A,B$
0%
$D,C,B,A$

Explanation

Octant	$I$	$II$	$III$	$IV$	$V$	$VI$	$VII$	$VIII$
Signs:	$+,+,+$	$-,+,+$	$-,-,+$	$+,-,+$	$+,+,-$	$-,+,-$	$-,-,-$	$+,-,-$

Based on this, increasing order is

$A,B ,C,D$

$P(0,5,6),Q(1,4,7),R(2,3,7)$ and

$S(3,5,16)$ are four points in the space. The point nearest to the origin

$O(0,0,0)$ is

0%
$P$
0%
$Q$
0%
$R$
0%
$S$

Explanation

The

$4$ points are as given.
We calculate their individual distance from the origin.

$OP =$

${({5}^{2} + {6}^{2})}^{0.5}$ =

${(61)}^{0.5}$

$OQ =$

${({1}^{2} + {4}^{2} + {7}^{2} )}^{0.5}$ =

${(66)}^{0.5}$

$OR =$

${({2}^{2} + {3}^{2} + {7}^{2} )}^{0.5}$ =

${(62)}^{0.5}$

$OS =$

${({3}^{2} + {5}^{2} + {16}^{2} )}^{0.5}$ =

${(290)}^{0.5}$
Hence,

$P$ is the nearest to the origin.

In the

$\Delta ABC$ , if

$AB=\sqrt{2}; AC=\sqrt{20}, B=(3,2,0)$ and

$C=(0,1,4)$ , then the length of the median passing through

$A$ is

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

Explanation

Using Apollonius theorem: In any triangle

$ABC$ , if

$AD$ is a median,

then

${ AB }^{ 2 }+{ AC }^{ 2 }=2({ BD }^{ 2 }+{ AD }^{ 2 })$

$D$ is the midpoint of

$BC$
Coordinates of

$D$ are

$\left(\dfrac{3}{2},\dfrac{3}{2},2\right)$ and

$(BD)^2=\dfrac{26}{4}$

$\Rightarrow { AB }^{ 2 }+{ AC }^{ 2 }=2({ BD }^{ 2 }+{ AD }^{ 2 })$

$\Rightarrow AD = \dfrac { 3 }{ \sqrt { 2 } }$

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%
$13$
0%
$\sqrt{13}$
0%
$2\sqrt{13}$
0%
$169$

Explanation

By looking

$(-2, 4, 6)$ and

$(3, 16, 6)$ extremities of a diagonal both lie in same plane (

$xy$ - plane).

Base diagonal is diagonal joining these two extremities.

So, length of diagonal

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

$=13$

Assertion (A): The points

$A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus
Reason (R):

$AB = BC = CD = DA$ and

$AC = BD$

0%
Both A and R are individually true and R is the correct explanation of A
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Both A and R individually true but R is not the correct explanation of A
0%
A is true but R is false
0%
Both A and R false

Explanation

Given: The points

$A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus.

So using distance formula Reason is not true.

Thus both A and R false.

If the plane a

$2x-3y+5_{Z}-2=0$ divides the line segment joining

$(1, 2, 3)$ and

$(2, 1, k)$ in the ratio

$9 : 11$ , then

$k$ is

0%
$1$
0%
$-2$
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$-10$
0%
$\displaystyle -\dfrac{1}{2}$

Explanation

Coordinate of the point which divides the line segment joining the points

$(1,2,3)$ and

$(2,1,k)$ in the ratio

$9:11$ are

$\left(\dfrac{29}{20}, \dfrac{31}{20}, \dfrac{9k+33}{20}\right)$

Also, this point will lie on the given plane

$\Rightarrow 2\times\dfrac{29}{20}-3\times\dfrac{31}{20}+5\times\dfrac{9k+33}{20}-2 =0$

$\Rightarrow k = -2$

Hence, option B is correct.

The point equidistant from the points

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

$(0,0,3)$ is

0%
$(1,2,3)$
0%
$\left (\displaystyle \dfrac{1}{2},1,\dfrac{3}{2}\right)$
0%
$\left (-\displaystyle \dfrac{1}{2}, -1,-\displaystyle \dfrac{3}{2}\right)$
0%
$(1,-2,3)$

Explanation

Equation of sphere passes through origin is given by

$x^2+y^2+z^2+ax+by+cz=0 ...(1)$

Given, it also passes through

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

$1+a=0\Rightarrow a = -1$

$4+2b=0\Rightarrow b = -2$

and

$9+3c=0\Rightarrow c = -3$

$\therefore$ the equation of the sphere becomes

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

Comparing with

$x^2+y^2+z^2+2gx+2fy+2kz+C=0$

Therefore, the point equidistant from all the given four point will be the centre of the sphere

$(1)$ passing through all these points which is

$\left(\dfrac{1}{2},1, \dfrac{3}{2}\right)i.e. the \ centre$ .

A point on the line

$\displaystyle \frac{{x + 2}}{1} = \frac{{y - 3}}{{ - 4}} = \frac{{z - 1}}{{2\sqrt 2 }}$ at a distance 6 from the point (2, 3, 1) is

0%
$(4-21, 1+12\sqrt{2})$
0%
$\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$
0%
$\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$
0%
$\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$

Explanation

$(\lambda -2, -4\lambda+3, 2\sqrt 2\lambda+1)\leftrightarrow (-2, 3, 1)$

$(\lambda-2+2)^2+(-4\lambda+3-3)^2+2\sqrt 2\lambda+1-1)^2=36$

$\lambda^2+16\lambda^2+8\lambda^2=36$

$\lambda=\pm \frac {6}{5}$

$\therefore point=(\frac {-4}{5}, \frac {-9}{5}, 1)$

$A (1, 2, 3), B ( 2, 3, 1), C (3, 1, 2)$ then the length of the altitude through

$C$ is

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

Explanation

$\bigtriangleup \ ABC$ is equilateral
So length of median

$=$ length of attitude

$D$ is mid point of

$AB$
P.V of

$D=(\dfrac{3}{2},\dfrac{5}{2},2)$

$CD=\sqrt{\dfrac{9}{4}+\dfrac{9}{4}}$

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

A plane intersects the co ordinate axes at

$A, B, C$ . If

$O= (0, 0, 0)$ and

$(1, 1, 1)$ is the centroid of the tetrahedron

$O ABC$ , then the sum of the reciprocals of the intercepts of the plane

0%
$12$
0%
$\displaystyle \dfrac{4}{3}$
0%
$1$
0%
$\displaystyle \dfrac{3}{4}$

Explanation

Let the point of intersections be,

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

$C(0,0,c)$
The coordinates of the centroid are

$\left ( \dfrac{a}{4} , \dfrac{b}{4} , \dfrac{c}{4} \right )$
Comparing it with the coordinates given, we get

$a = 4 , b = 4 , c = 4$
Hence,

$\left ( \dfrac{1}{a} , \dfrac{1}{b} , \dfrac{1}{c} \right) = \dfrac{3}{4}$

Assertion(A): If centroid and circumcentre of a triangle are known its orthocentre can be found.
Reason (R) : Centriod, orthocentre and circumcentre of a triangle are collinear

0%
Both A and R are individually true and R is the correct explanation of A
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Both A and R individually true but R is not the correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

Centroid, orthocentre and circumcentre are collinear and centroid divides

the line joining orthocentre and circumcentre in the ratio

$2:1$ so if any two points are given then this one can be found.

Hence option 'A' is the correct choice

The plane

$ax+by+cz+(-3)=0$ meet the co-ordinate axes in A,B,C. Then centroid
of the triangle is

0%
$(3a,3b,3c)$
0%
$(\displaystyle \frac{3}{a}\frac{3}{b},\frac{3}{c})$
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$\left ( \dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3} \right )$
0%
$(\displaystyle \dfrac{1}{a}, \dfrac{1}{b},\dfrac{1}{c})$

Explanation

For finding the coordinates of the point where the plane

$ax + by + cz - 3 = 0$ cuts the x axis, we equate y and z to zero.

The point becomes

$\left (\cfrac{3}{a}, 0, 0 \right )$

Similarly, the point on y axis becomes

$\left (0, \cfrac{3}{b}, 0 \right )$

And that on z axis becomes

$\left (0, 0, \cfrac{3}{c} \right )$

The centroid of the triangle formed by these points would be

$\left ( \cfrac{\cfrac{3}{a} + 0 + 0}{3}, \cfrac{0 + \cfrac{3}{b} + 0}{3}, \cfrac{0 + 0 + \cfrac{3}{c}}{3} \right )$

$= \left ( \cfrac{1}{a}, \cfrac{1}{b}, \cfrac{1}{c} \right )$

The name of the figure formed by the points

$(-1, -3, 4), (5, -1,1), (7, -4, 7)$ and

$(1, -6, 10)$ is a

0%
square
0%
rhombus
0%
parallelogram
0%
rectangle

Explanation

Keeping the above points as vertices and using the distance formula,

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$
We get that the sides of the parallelogram formed by the above lines are equal and all the sides are of

$7$ units.
Hence the parallelogram will be either a rhombus or a square. For the parallelogram to be square, all the adjacent sides of the parallelogram should make an angle of

$\dfrac{\pi}{2}$ .
Consider the vector equation of

$AB$ as

$(5-(-1))i'+(-1-(-3))j'+(1-4)k'$

$6i'+2j'-3k'$
Consider the vector equation of

$BC$ as

$(7-5)i'+(-4-(-1))j'+(7-1)k'$

$2i'-3j'+6k'$
Taking dot product, we get

$6(2)+2(-3)-3(6)$

$=12-6-18$

$=-12$
Thus the parallelogram is a rhombus.

The end points of a body diagonal of a rectangular parallelepiped whose faces are parallel to the coordinate planes are

$(2, 3, 5)\ and\ (5, 7, 10)$ . The lengths of its sides are

0%
$5, 7, 3$
0%
$6, 5, 3$
0%
$3, 6, 2$
0%
$3, 4, 5$

Explanation

$\bar { AB } =3\hat { i } +4\hat { j } +5\hat { k }$

$\overrightarrow { AB }$ is the resultant of

$3\hat { i }$ (side parallel to x axis),

$4\hat { j }$ (side parallel to y axis),

$5\hat { k }$ (side parallel to x axis)

$(3,4,5)$ are lengths of sides.

A tetrahedron is a three dimensional figure bounded by non coplanar triangular planes. So, a tetrahedron has four non-coplanar points as its vertices. Suppose a tetrehedron has points A,B,C,D as its vertices which have coordinates

$(x1, y1, z_{1}) (x_{2}, y_{2}, z_{2})$ ,

$(x_{3}, y_{3}, z_{3})$ and

$(x_{4}, y_{4}, z_{4})$ , respectively in a rectangular three dimensional space. Then, the coordinates of its centroid are

$[\dfrac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\dfrac{y_{1}+y_{2}+y_{3}+y_{4}}{4},\dfrac{z_{1}+z_{2}+z_{3}+z_{4}}{4}]$ .
Let a tetrahedron have three of its vertices represented by the points

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

$(4,1,3)$ and its centroid lies at the point

$(1,2,5)$ . Now, answer the following question. The coordinate of the fourth vertex of the tetrahedron is:

0%
$(-6, 2,16)$
0%
$(1, -2,13)$
0%
$(-2,4,-2)$
0%
$(1, -1,1)$

Explanation

Given three vertices of tetrahedron are

$(0,0,0), (6,5,1),(4,1,3)$

Now equation of plane passing through

$(0,0,0)$ is given by,

$ax+by+cz=0$

Also this line passing through other two points

$\Longrightarrow 6a+5b+c=0, 4a+b+3c=0$

Solving these, we get

$a=-c,b=c$

Hence required plane will be

$-x+y+z=0$

From the above equationn the coordinate of the fourth vertex of the tetrahedron is

$(-6,2,16)$

The equation of median through C to side AB is

0%

$r = - \hat i + \hat j + \hat k + p (3\hat i - 2\hat k)$
0%

$r = - \hat i + \hat j + \hat k + p (3\hat i + 2\hat k)$
0%

$r = - \hat i + \hat j + \hat k + p (-3\hat i + 2\hat k)$
0%

$r = -\hat i + \hat j + \hat k + p (3\hat i + 2 \hat j )$

Explanation

Given, coordinates of A = (1 , 2 , 3 )

coordinates of B = (0 , 0 , 1 )

coordinates of C = (-1 , 1 , 1 )

Now CD is the median through C on AB.

Now we have to find the coordinate of D.

Hence, D =

$\left( \dfrac{1+0}{2} , \dfrac{2+0}{2} , \dfrac{3+1}{2} \right)$ [Since D is the midpoint of AB ]

$\therefore D = \left( \dfrac{1}{2} , 1 , 2 \right) \, \, [\therefore \, \textrm{midpoint} \, = \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} , \dfrac{z_1 + z_2}{2} ]$

$\therefore$ The direction radios of CD

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

And this line passes through C

$\therefore$ Eqn of line CD ,

$\dfrac{x+1}{3/2} = \dfrac{y-1}{0} = \dfrac{z-1}{1}$

$\therefore \vec{r} = (-\hat{i} + \hat{j} + \hat{k}) + P \, (3 \hat{i} + 2 \hat{k})$

1911423_121132_ans_f3ffbc15406d4a35955848a99a04aeef.png

If the extremities of a diagonal of a square are

$(1, -2, 3)$ and

$(4, 2, 3)$ then the area of the square is

0%
$25$
0%
$50$
0%
$\displaystyle \frac{25}{2}$
0%
$\sqrt{50}$

Explanation

$a$ is the length of a side of square then the length of diagonal is given by

$\sqrt{2}a$ . Distance between two given points is

$\sqrt{(1-4)^2+(-2-2)^2+(3-3)^2}=5=\sqrt{2}a$ . Hence the area is given by

$a^2=\dfrac{25}{2}$ .

$\mathrm{P} (-1,-1,-1 )$ and

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

$PQ=\sqrt{48}$ , then value(s) of

$\lambda$ can be

0%
$3$
0%
$-5$
0%
$4$
0%
$2$

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

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

$(\lambda ,\lambda ,\lambda )$ is

$\sqrt { { ({ x }_{ 1 }-{ x }_{ 2 }) }^{ 2 }+{ ({ y }_{ 1 }-{ y }_{ 2 }) }^{ 2 } + (z_1-z_2)^2 }$

Distance between

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

$(\lambda ,\lambda ,\lambda )$ is

$\sqrt { { (\lambda +1) }^{ 2 }+{ (\lambda +1) }^{ 2 }+{ (\lambda +1) }^{ 2 } }$

$\sqrt { 3\times { (\lambda +1) }^{ 2 } } =\sqrt { 48 } =4\sqrt { 3 } \\ \Rightarrow (\lambda +1)=\pm 4\\ \Rightarrow \lambda =3,-5$

$L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$

$L_2:\dfrac{x-2}{3}=\dfrac{y-4}{2}=\dfrac{z-5}{5}$ be two given lines, point P lies on

$L_1$ and Q lies on

$L_2$ then distance between P and Q can be

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{9}$
0%
$15$
0%
$30$

Explanation

$\dfrac{|\begin{vmatrix}2-1&4-2 &5-3 \\2 & 3 &4 \\ 3& 2 & 5\end{vmatrix}|}{|\begin{vmatrix}i&j &k \\2 & 3 & 4\\3 & 2 & 5\end{vmatrix}|}=\dfrac{1}{\sqrt{78}}$

From which of the following the distance of the point

$(1, 2, 3)$ is

$\sqrt{10}$ ?

0%
Origin
0%
$x-$ axis
0%
$y-$ axis
0%
$z-$ axis

Explanation

The point is

$P(1,2,3)$ so distance of point from

$y$ axis is

${=}$

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

$=\sqrt{10}$

Hence, option C is correct.

The plane

$ax+by + cz + d = 0$ divides the line joining

$(x_1, y_1, z_1)$ and

$(x_2, y_2, z_2)$ in the ratio

0%
$\displaystyle - \frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 +d}$
0%
$\displaystyle - \frac{ax_1\times ax_2 + by_1\times by_2 + cz_1\times cz_2 + d^2}{ax_2 + by_2 + cz_2 +d}$
0%
$\displaystyle - \frac{a + b + c + d}{x_1+x_2 + y_1+ y_2 + z_1 + z_2 }$
0%
$\displaystyle \frac{ax_1^2 + by_1^2 + cz_1^2 + d}{ax_2^2 + by_2^2 + cz_2^2 +d}$

Explanation

Let the plane $ax+by + cz + d =0$ divides the lines joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $k:1$ as shown in figure.

$\therefore$ Coordinates of $\displaystyle P \left ( \dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k+1}, \dfrac{kz_2 + z_1}{k+1} \right )$ must satisfy $ax + by + ca + d = 0$

i.e., $\displaystyle \left ( a \dfrac{kx_2 + x_1}{k + 1} \right ) + b\left ( \dfrac{ky_2 + y_1}{k + 1} \right) + c \left ( \dfrac{kz_2 + z_1}{k + 1} \right) + d = 0$

$\Rightarrow a(kx_2 + x_1) + b (ky_2 + y_1) + c (kz_2 + z_1) + c(kz_2 + z_1) + d (k + 1)= 0$

$\Rightarrow k (ax_2 + by_2 + cz_2 + d) + (ax_1 + by_1 + cz_1 + d)= 0$

$\Rightarrow \displaystyle k = - \dfrac{(ax_1 + by_1 + cz_1 + d)}{(ax_2 + by_2 + cz_2 + d)}$

Hence, option A is correct.

If the centroid of triangle whose vertices are

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

$(4, 7, c)$ be the origin, then the values of

$a, b$ and

$c$ are

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

Explanation

Let

$P\left( { x }_{ 1 },{ y }_{ 1 } \right) ,Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and

$R\left( { x }_{ 3 },{ y }_{ 3 } \right)$ be coordinates of triangle

Now the centroid of triangle is,

$\left( \cfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right)$

By using the formula of centroid of triangle,

$(0,0,0)\ {=}\ [(a-2+4)/3,(1+b+7)/3,(3-5+c)/3]$

$(0,0,0)\ {=}\ [(a+2)/3,(b+8)/3,(c-2)/3]$
on solving, we get

$a\ {=}\ -2$ ;

$b\ {=}\ -8$ ;

$c\ {=}\ 2$

Find the ratio in which

$2x + 3y + 5z = 1$ divides the line joining the points

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

$(1,\ -5,\ 7)$ .

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

Explanation

$2x + 3y + 5z = 1$ divides $(1, 0, -3)$ and $(1, -5, 7)$ in the ratio of $k : 1$ at point $P$ .

Then, $\displaystyle P \left ( \displaystyle \dfrac{k + 1}{k + 1} , \displaystyle \dfrac{-5k}{k + 1}, \displaystyle \dfrac{7k - 3}{k + 1} \right )$ which must satisfy $2x + 3y + 5z = 1$

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

$\Rightarrow 2k + 2 - 15 k + 35 k - 15 = k + 1$

$\Rightarrow 21 k = 14$

$\Rightarrow k = \displaystyle \dfrac{2}{3}$

$\therefore 2x + 3y + 5z = 1$ divides $(1, 0, -3)$ and $(1, -5, 7)$ in the ratio of $2 : 3$ .

If the sum of the squares of the distance of a point from the three coordinate axes be

$36$ , then its distance from the origin is

0%
$6$ units
0%
$3$ $\sqrt{2}$ units
0%
$2$ $\sqrt{3}$ units
0%
none of these

Explanation

Let

$(x, y, z)$ be the point.

Given sum of the squares of distance from point to the axes is

$36$ .

$\Rightarrow (x^2+y^2) +(y^2+z^2) + (z^2+x^2) = 36$

$\Rightarrow 2(x^2+y^2+z^2) = 36 \Rightarrow x^2 + y^2 + z^2 = 18$

So the distance of the point from the origin is

$= \sqrt{x^2 + y^2 + z^2} = 3\sqrt{2}$

Hence, option B; is correct.

The circum radius of the triangle formed by the points

$(0, 0, 0)$ ,

$(0, 0, 12)$ and

$(3, 4, 0)$ is

0%
$\sqrt{156}$
0%
$13$
0%
$\displaystyle \frac { 13 }{ 2 }$
0%
$8$

Explanation

Center of the circum circle of a right angle triangle is on the mid of the hypotenuse.

Hence the radius is the half of the length of the hypotenuse.

$\dfrac{\sqrt{3^2+4^2+12^2}}{2}=\dfrac{13}{2}$

The coordinates of the point where the line segment joining

$A(5,1,6)$ and

$B (3,4,1)$ crosses the yz plane are

0%
$(0,\dfrac{17}{2}, \dfrac{13}{2})$
0%
$(0,-\dfrac{17}{2},\dfrac{13}{2})$
0%
$(0,\dfrac{17}{2},-\dfrac{13}{2})$
0%
$(0,-\dfrac{17}{2},-\dfrac{13}{2})$

Explanation

The direction ratios of line joining the points

$(5,1,6)$ and

$(3,4,1)$ are

$2,-3,5$

Therefore the equation of line is

$\frac { x-5 }{ 2 } =\frac { y-1 }{ -3 } =\frac { z-6 }{ 5 }$

Given that the line crosses

$yz$ plane . In

$yz$ plane,

$x=0$

$\Rightarrow \frac { 0-5 }{ 2 } =\frac { y-1 }{ -3 } =\frac { z-6 }{ 5 }$

$\Rightarrow y= \frac{17}{2} , z=-\frac{13}{2}$

Therefore the correct option is

$C$

If the plane

$7x + 11y+ 13z= 3003$ meets the axes in

$A, B, C$ , then the centroid of

$\Delta ABC$ is

0%
$(143, 91, 77)$
0%
$(143,77, 91)$
0%
$(91, 143, 77)$
0%
( $143, 66, 91)$

Explanation

intercept form

$\dfrac{7x}{3003}+\dfrac{11y}{3003}+\dfrac{13z}{3003}=1$

$\therefore A[429,0,0], B[0,273,0], C[0,0,231]$ are the vertices of triangle ABC.

then centroid is

$(\dfrac{429+0+0}{3}+\dfrac{273}{3}+\dfrac{237}{3})$

$\Longrightarrow (143,91,77)$

Four vertices of a tetrahedron are

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

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

0%
$\left (\dfrac {4}{3}, -\dfrac {8}{3}, 4\right )$
0%
$(2, -4, 6)$
0%
$(1, -2, 3)$
0%
none of these

Explanation

If the vertices of tetrahedron are,

$(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3),(x_4,y_4,z_4)$ , then the centroid of tetrahedron 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 from the vertices given in the question, the centroid is

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

$=(1,-2,3)$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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