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CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 11 - MCQExams.com
CBSE
Class 11 Engineering Maths
Introduction To Three Dimensional Geometry
Quiz 11
If $$P\left( x,y,z \right) $$ is a point on the line segment joining $$Q\left( 2,2,4 \right) $$ and $$R\left( 3,5,6 \right) $$ such that the projections of $$OP$$ on the axis are $$\cfrac { 13 }{ 5 } ,\cfrac { 19 }{ 5 } ,\cfrac { 26 }{ 5 } $$ respectively, then $$P$$ divides $$QR$$ in the ratio
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$$1:2$$
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$$3:2$$
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$$2:3$$
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$$1:3$$
Explanation
Since, $$\overline { OP } $$ has projections $$\cfrac { 13 }{ 5 } ,\cfrac { 19 }{ 5 } ,\cfrac { 26 }{ 5 } $$ on the coordinate axes
$$\therefore \overline { OP } =\cfrac { 13 }{ 5 } \hat { i } +\cfrac { 19 }{ 5 } \hat { j } +\cfrac { 26 }{ 5 } \hat { k } $$
Suppose $$P$$ divides the line segment joining of $$Q(2,2,4)$$ and $$R(3,5,6)$$ in the ratio $$\lambda:1$$
then the position vector of $$P$$ is
$$\left( \cfrac { 3\lambda +2 }{ \lambda +1 } \right) \hat { i } +\left( \cfrac { 5\lambda +2 }{ \lambda +1 } \right) \hat { j } +\left( \cfrac { 6\lambda +4 }{ \lambda +1 } \right) \hat { k } $$
$$\therefore \cfrac { 13 }{ 5 } \hat { i } +\cfrac { 19 }{ 5 } \hat { j } +\cfrac { 26 }{ 5 } \hat { k } =\left( \cfrac { 3\lambda +2 }{ \lambda +1 } \right) \hat { i } +\left( \cfrac { 5\lambda +2 }{ \lambda +1 } \right) \hat { j } +\left( \cfrac { 6\lambda +4 }{ \lambda +1 } \right) \hat { k } $$
$$\Rightarrow \lambda =3/2$$
$$\therefore$$ Required ratio in which $$P$$ divides $$QR$$ is $$3:2$$
The points $$A(1,2,3); B-(-1,-2,-1); C(2,3,2)$$ and $$D(4,7,6)$$ form
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Square
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Rectangle
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Parallelogram
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Rhombus
Find the coordinates of the points which trisect the line segment AB, given that
A =( 2,1 , - 3 ) and B =( 5 , - 8,3 ).
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$$( 3,2 , - 1 ) ; ( 4,5 , - 1 )$$
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$$( 3 , - 2 , - 1 ) ; ( 4 , - 5,1 )$$
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$$( 3 , - 2,1 ) ; ( - 4,5 , - 1 )$$
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$$( 3 , - 2 , - 1 ) ; ( 4,5 , - 1 )$$
30 consider at three dimensional figure represented by $$xy{z^2} = 2$$, then its minimum distance from origin is
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0%
2
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4
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3
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1
The distances of the point $$P(1,2,3)$$ from the coordinates axes are:
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$$\sqrt {13} ,\sqrt {10} ,\sqrt 5 $$
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$$\sqrt {11} ,\sqrt {10} ,\sqrt 5 $$
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$$\sqrt {13} ,\sqrt {20} ,\sqrt {15} $$
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$$\sqrt {23} ,\sqrt {10} ,\sqrt 5 $$
If O and O' are circumcenter and orthocenter of a $$\Delta ABC$$ where $$\overline{OA} + \overline{OB} + \overline{OC}$$ is $$\lambda \overline{OO'}$$ then the value of $$\lambda$$ is
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$$1$$
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$$2$$
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$$3$$
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$$4$$
The distance between the orthocentre and circumcentre of the triangle formed by the points $$(1, 2, 3), (3, -1, 5), (4, 0, -3)$$ is
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$$\dfrac {\sqrt {66}}{2}$$
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$$\dfrac {\sqrt {17}}{2}$$
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$$\dfrac {\sqrt {55}}{2}$$
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$$\dfrac {\sqrt {37}}{2}$$
In the $$xy-plane$$, the length of the shortest from $$(0, 0)$$ to $$(12, 16)$$ that does not go inside the circle $$(x - 6)^{2} + (y + 8)^{2} = 25$$ is
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$$10\sqrt {3}$$
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$$10\sqrt {5}$$
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$$10\sqrt {3} + \dfrac {5\pi}{3}$$
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$$10 + 5\pi$$
If $$R$$ divides the line segment joining $$P(2,3,4)$$ and $$Q(4,5,6)$$ in the ratio $$-3:2$$, then the parameter which represent $$R$$ is
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$$3$$
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$$2$$
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$$1$$
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$$-1$$
Explanation
$$x=\cfrac { -3\times 4+2\times 2 }{ -1 } \quad \quad y=\cfrac { -3\times 5+2\times 3 }{ -1 }$$
$$x=\cfrac { -12+4 }{ -1 } \quad \quad$$
$$y =\cfrac { -15+6 }{ -1 } $$
$$x=8\quad \quad y=9$$
$$z=\cfrac { -3\times 6+2\times 4 }{ -1 } =\cfrac { -10 }{ -1 } $$
$$z=10$$
$$R=\left( 8,9,10 \right) $$
Consider at three dimensional figure represented by $$xy{ z }^{ 2 }=2$$, then its minimum distance from origin is
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2
0%
4
0%
6
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8
The point which is equidistant from the points $$(-1,1,3),(2,1,2),(0,5,6)$$ and $$(3,2,2)$$ is
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$$(-1,3,4)$$
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$$(3,1,4)$$
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$$(1,3,4)$$
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$$(4,1,3)$4
Minimum distance between the curves
$$y^{2}=4x$$ & $$x^{2}+y^{2} -12x+31=0$$ is -
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$$\sqrt {21}$$
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$$\sqrt {26}-\sqrt {5}$$
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$$\sqrt {20}-\sqrt {5}$$
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$$\sqrt {21}-\sqrt {5}$$
Let $$A(2,3,5),B(-1,3,2)$$ and $$C(\lambda ,5,\mu )$$ are the vertices of a triangle and its median through A meets side BC at D. AD is equally inclined with the axes. If E is the point on BC such that $$BE:EC=1:2.$$
Project of
BA
on
BC
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$$\dfrac { 23 }{ \sqrt { 33 } } $$
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$$\dfrac { \sqrt { 33 } }{ 24 } $$
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$$\dfrac { 24 }{ \sqrt { 33 } } $$
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$$None$$ $$of$$ $$these$$
If A = (2,-3,1), B = (3,-4,6) and C is a point of trisection of AB, then $${ C }_{ y }$$
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$$\frac { 11 }{ 3 } $$
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-11
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$$\frac { 10 }{ 3 } $$
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$$\frac { -11 }{ 3 } $$
The coordinates of the orthocentre of the triangle that has the coordinates of mid points of its sides as (0 , 0) (1 , 2) and ( -6 , 3) is :
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(0 , 0)
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(-4 , 5)
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(-5 , 5)
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(-4 , 4)
Let $$A(2,3,5),B(-1,3,2)$$ and $$C(\lambda ,5,\mu )$$ are the vertices of a triangle and its median through A meets side BC at D. AD is equally inclined with the axes. If E is the point on BC such that $$BE:EC=1:2.$$
Equation of plane containing triangle ABC
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$$x+y+3=0$$
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$$x-z-3=0$$
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$$x-z+3=0$$
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$$x-y+3=0$$
Perpendicular distance from the origin to the line joining the points $$(a\cos{\theta},a\sin{\theta})(a\cos{\theta},a\sin{\theta})$$ is
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$$2a\cos{(\theta-\phi)}$$
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$$a\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } $$
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$$4a\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } $$
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$$a\cos { \left( \cfrac { \theta +\phi }{ 2 } \right) } $$
Consider a variable plane $$lx+my+nz=k(k>0)and\quad l,m,n$$ are direction cosines of normal of the plane. Let the given plane intersects the co-ordinate axes at A,B and C, then the minimum area of $$\triangle ABC$$ is _______.
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$$\dfrac { 3\sqrt { 3k } ^{ 2 } }{ 2 } $$
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$$\dfrac { 3\sqrt { 3k } ^{ 2 } }{ 4 } $$
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$$3\sqrt { 3 } { k }^{ 2 }$$
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$$12\sqrt { 3 } { k }^{ 2 }$$
The shortest distance between the point $$\left( \dfrac { 3 }{ 2 } ,0 \right) $$ and the curve $$y=\sqrt { x } $$, $$(x>0)$$, is:
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$$\dfrac { 3 }{ 2 } $$
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$$\dfrac { \sqrt { 3 } }{ 2 } $$
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$$\dfrac { \sqrt { 5 } }{ 2 } $$
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$$\dfrac { 5 }{ 4 } $$
Q, R, S are the points $$(-2, -1), (0, 3) (4, 0)$$ respectively. Then the coordinates of P such that PQRS is a parallelogram is ________________.
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$$(2, -6)$$
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$$(-6, 2)$$
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$$(2, -4)$$
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$$(-3, 2)$$
If $$A=(1, 2, 3)$$ and $$B(3, 5, 7)$$ and P, Q are the points on AB such that AP$$=$$PQ$$\neq$$QB, then the mid point of PQ is?
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$$(2, 3, 5)$$
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$$\left(2, \dfrac{7}{2}, 5\right)$$
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$$(2, 4, 5)$$
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$$(4, 7, 0)$$
If $$A=(1, -2, -1), B=(4, 0, -3); C=(1, 2, -1)$$ and $$D=(2, -4, -5)$$, then the distance between AB and CD is?
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$$\dfrac{2}{3}$$
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$$\dfrac{4}{3}$$
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$$\dfrac{3}{2}$$
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$$\dfrac{5}{3}$$
The equation of plane which is passing through the point $$(1,2,3)$$ and which is at maximum distance from the point $$(-1,0,2)$$ is
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$$2x+2y+z=9$$
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$$2x+z=5$$
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$$3x+y-z=2$$
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none of these
The distance of the point $$(2,1,-1)$$ from the line $$\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-3}{-3}$$ measured parallel to the plane $$x+2y+z=4$$ is
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$$\sqrt{10}$$
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$$\sqrt{20}$$
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$$\sqrt{5}$$
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$$\sqrt{30}$$
A line passes through two points A(2, -3, -1) and B(8, -1, 2) the coordinates of a point on this line nearer to the origin at a distance of 14 units from A are
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(14 , 1, 5)
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(-10, -7, -7)
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(10 , 7, 7)
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(-14, -1, -5)
If $$\lambda$$ is the length of any edge of a regular tetrahedron, then the distance of any vertex form the opposite face is-
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$$\dfrac{2}{3}\lambda^{2}$$
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$$\sqrt{\dfrac{2}{3}}\lambda$$
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$$\dfrac{\sqrt{2}}{3}\lambda$$
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$$None of these$$
Explanation
$$\begin{array}{l} In\, \, { 30^{ 0 } }-{ 60^{ 0 } }-{ 90^{ 0 } }-side\, \, is\, \, in\, \, ratio,\, \\ 1:\sqrt { 3 } :2 \end{array}$$
In regular tetraheron, height (h)
$$\begin{array}{l} 5\frac { { \sqrt { 2 } } }{ 3 } \\ and\, \, side\, \, ,s=\lambda \\ \therefore \lambda =\sqrt { \frac { 2 }{ 3 } } \lambda \end{array}$$
The distance of the point $$(2,3)$$ form the line $$x-2y+5=0$$ measured in a direction parallel to the line $$x-3y=0$$ is
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$$2\sqrt{10}$$
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$$\sqrt{10}$$
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$$2\sqrt{5}$$
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$$None\ of\ these$$
If $$ A=(1,-2,-1), B=(4,0,-3), C=(1,2,-1), D=(2,-4,-5) $$, then distance between A B and CD is
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$$
\frac{1}{3}
$$
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$$
\frac{2}{3}
$$
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1
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$$
\frac{4}{3}
$$
$$A(1, -1, -3)$$, $$B(2,1,-2)$$ & $$(-5, 2, -6)$$ are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is:
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$$\sqrt { 10 } /4$$
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$$3\sqrt { 10 } /4$$
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$$\sqrt { 10 } $$
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None
If x-coordinates of a point P on the joining the points $$Q(2, 2,1)$$ and $$R(5, 1, -2)$$ is 4, then the z-coordinates of P is
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-2
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-1
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1
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2
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