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

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
  • $$1$$
  • $$\displaystyle \dfrac{70}{3},0$$
  • $$35$$
  • $$\displaystyle 0,-\dfrac{70}{3}$$
If $$A(\cos\alpha,\sin\alpha, 0),B(\cos\beta,\sin\beta, 0)$$, $$C(\cos\gamma,\sin\gamma,0)$$ 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
$$(a,b,0)$$
  • $$4\ 3\ 2$$
  • $$2\ 3\ 1$$
  • $$1\ 2\ 3$$
  • $$2\ 3\ 4$$
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:
  • $$A,B,C,D$$
  • $$B,C,D,A$$
  • $$C,D,A,B$$
  • $$D,C,B,A$$
$$P(0,5,6),Q(1,4,7),R(2,3,7)$$ and $$S(3,5,16)$$ are four points in the space. The point nearest to the origin $$O(0,0,0)$$ is
  • $$P$$
  • $$Q$$
  • $$R$$
  • $$S$$
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
  • $$\displaystyle \dfrac{3}{2}$$
  • $$\displaystyle \dfrac{9}{2}$$
  • $$\displaystyle \dfrac{3}{\sqrt{2}}$$
  • $$\displaystyle \dfrac{\sqrt{3}}{2}$$
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
  • $$13$$
  • $$\sqrt{13}$$
  • $$2\sqrt{13}$$
  • $$169$$
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 $$AC = BD$$
  • Both A and R are individually true and R is the correct explanation of A
  • Both A and R individually true but R is not the correct explanation of A
  • A is true but R is false
  • 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
  • $$1$$
  • $$-2$$
  • $$-10$$
  • $$\displaystyle -\dfrac{1}{2}$$
The point equidistant from the points $$(0,0,0), (1,0,0), (0,2,0)$$ and $$(0,0,3)$$ is
  • $$(1,2,3)$$
  • $$\left (\displaystyle \dfrac{1}{2},1,\dfrac{3}{2}\right)$$
  • $$\left (-\displaystyle \dfrac{1}{2}, -1,-\displaystyle \dfrac{3}{2}\right)$$
  • $$(1,-2,3)$$
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
  • $$(4-21, 1+12\sqrt{2})$$
  • $$\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$$
  • $$\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$$
  • $$\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$$
If $$A (1, 2, 3),  B ( 2, 3, 1),  C (3, 1, 2)$$ then the length  of the altitude through $$C$$ is 
  • $$3$$
  • $$3\sqrt{3}$$
  • $$3\sqrt{2}$$
  • $$\dfrac{3}{\sqrt{2}}$$
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
  • $$12$$
  • $$\displaystyle \dfrac{4}{3}$$
  • $$1$$
  • $$\displaystyle \dfrac{3}{4}$$
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
  • Both A and R are individually true and R is the correct explanation of A
  • Both A and R individually true but R is not the correct explanation of A
  • A is true but R is false
  • A is false but R is true
The plane $$ax+by+cz+(-3)=0$$ meet the co-ordinate axes in A,B,C. Then centroid
of the triangle is
  • $$(3a,3b,3c)$$
  • $$(\displaystyle \frac{3}{a}\frac{3}{b},\frac{3}{c})$$
  • $$\left ( \dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3} \right )$$
  • $$(\displaystyle \dfrac{1}{a}, \dfrac{1}{b},\dfrac{1}{c})$$
The name of the figure formed by the points $$(-1, -3, 4), (5, -1,1), (7, -4, 7)$$ and $$(1, -6, 10)$$ is a
  • square
  • rhombus
  • parallelogram
  • rectangle
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 
  • $$5, 7, 3$$
  • $$6, 5, 3$$
  • $$3, 6, 2$$
  • $$3, 4, 5$$
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:
  • $$(-6, 2,16)$$
  • $$(1, -2,13)$$
  • $$(-2,4,-2)$$
  • $$(1, -1,1)$$
The equation of median through C to side AB is

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

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

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

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

If the extremities of a diagonal of a square are $$(1, -2, 3)$$ and $$(4, 2, 3)$$ then the area of the square is
  • $$25$$
  • $$50$$
  • $$\displaystyle \frac{25}{2}$$
  • $$\sqrt{50}$$
$$\mathrm{P} (-1,-1,-1 )$$ and $$ Q(\lambda, \lambda, \lambda)$$ are two points in the space such that $$PQ=\sqrt{48}$$, then value(s) of $$\lambda $$ can be
  • $$3$$
  • $$-5$$
  • $$4$$
  • $$2$$
$$L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$$
$$L_2:\dfrac{x-2}{3}=\dfrac{y-4}{2}=\dfrac{z-5}{5}$$ be two given lines, point P lies on $$L_1$$ and Q lies on $$L_2$$ then distance between P and Q can be
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{9}$$
  • $$15$$
  • $$30$$
From which of the following the distance of the point $$(1, 2, 3)$$ is $$\sqrt{10}$$?
  • Origin
  • $$x-$$axis
  • $$y-$$axis
  • $$z-$$axis
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
  • $$\displaystyle - \frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 +d}$$
  • $$\displaystyle - \frac{ax_1\times ax_2 + by_1\times by_2 + cz_1\times cz_2 + d^2}{ax_2 + by_2 + cz_2 +d}$$
  • $$\displaystyle - \frac{a + b + c + d}{x_1+x_2 + y_1+ y_2 + z_1 + z_2 }$$
  • $$\displaystyle \frac{ax_1^2 + by_1^2 + cz_1^2 + d}{ax_2^2 + by_2^2 + cz_2^2 +d}$$
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
  • $$-2, -8, -2$$
  • $$2, 8, -2$$
  • $$-2, -8, 2$$
  • $$7, -1, 0$$
Find the ratio in which $$2x + 3y + 5z = 1$$ divides the line joining the points $$(1,\ 0,\ -3)$$ and $$(1,\ -5,\ 7)$$.
  • $$1 : 2$$
  • $$2 : 1$$
  • $$3 : 2$$
  • $$2 : 3$$
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
  • $$6$$ units
  • $$3$$ $$\sqrt{2}$$ units
  • $$2$$ $$\sqrt{3}$$ units
  • none of these
The circum radius of the triangle formed by the points $$(0, 0, 0)$$, $$(0, 0, 12)$$ and $$(3, 4, 0)$$ is
  • $$\sqrt{156}$$
  • $$13$$
  • $$\displaystyle \frac { 13 }{ 2 } $$
  • $$8$$
The coordinates of the point where the line segment joining $$A(5,1,6)$$ and $$B (3,4,1)$$ crosses the yz plane are
  • $$(0,\dfrac{17}{2}, \dfrac{13}{2})$$
  • $$(0,-\dfrac{17}{2},\dfrac{13}{2})$$
  • $$(0,\dfrac{17}{2},-\dfrac{13}{2})$$
  • $$(0,-\dfrac{17}{2},-\dfrac{13}{2})$$
If the plane $$7x + 11y+ 13z= 3003$$ meets the axes in $$A, B, C$$, then the centroid of $$\Delta ABC$$ is
  • $$(143, 91, 77)$$
  • $$(143,77, 91)$$
  • $$(91, 143, 77)$$
  • ($$143, 66, 91)$$
Four vertices of a tetrahedron are $$(0, 0, 0), (4, 0, 0), (0, -8, 0)$$ and $$(0, 0, 12)$$. Its centroid has the coordinates
  • $$\left (\dfrac {4}{3}, -\dfrac {8}{3}, 4\right )$$
  • $$(2, -4, 6)$$
  • $$(1, -2, 3)$$
  • none of these
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