MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Maths
Introduction To Three Dimensional Geometry
Quiz 10
The distance of P(2,-3) from the x-axis is......
Report Question
0%
2
0%
-3
0%
3
0%
$$\sqrt{13}$$
If the distance between the points (2,-3) and (5,b) is 5 , then b=.........
Report Question
0%
1
0%
2
0%
7
0%
5
Find the centroid of a triangle, mid-points of whose sides are $$(!,2,-3),(3,0,1)$$ and $$(-1,1,-4)$$
Report Question
0%
$$(-1,-1,-2)$$
0%
$$(-1,2,-2)$$
0%
$$(-1,1,-2)$$
0%
$$(1,1,-2)$$
The point in the $$xy -$$ plane which is equidistant from $$(2, 0, 3), (0, 3, 2)$$ and $$(0,0, 1)$$ is
Report Question
0%
$$(1, 2,3)$$
0%
$$(-3, 2, 0)$$
0%
$$(3, -2. 0)$$
0%
$$(3, 2, 0)$$
0%
$$(3, 2, 1)$$
Explanation
Let the points are $$A(2,0,3),B(0,3,2)$$ and $$D(0,0,1)$$
We know that Z-coordinate of every point an xy-plane is zero so let $$p(x,y,0) $$ be a point on xy-plane such that $$PA=PB=PC$$
Now, $$PA=PB$$
$$\Rightarrow PA^2=PB^2$$
$$\Rightarrow (x-2)^2+(y-0)^2+(0-3)^2=(x-0)^2+(y-3)^2+(0-2)$$
$$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0$$......(i)
and , $$PB=PC$$
$$\Rightarrow PB^2=PC^2$$
$$\Rightarrow (x-0)^2+(y-3)^2(0-2)^2=(x-0)^2+(y-0)^2+(0-1)^2$$
$$\Rightarrow -6y+12=0$$
$$\Rightarrow y=2$$......(ii)
Putting $$y=2$$ in equation (i), we get $$x=3$$
Hence, the required point is $$(3,2,0)$$
The foot of the perpendicular from the point $$A(7, 14, 5)$$ to the plane $$2x+4y-z=2$$ is?
Report Question
0%
$$(3, 1, 8)$$
0%
$$(1, 2, 8)$$
0%
$$(3, -3, 5)$$
0%
$$(5, -3, -4)$$
Explanation
Let N be the foot of the perpendicular drawn from the point $$A(7, 14, 5)$$ and perpendicular to the plane $$2x+4y-z=2$$
Then, the equation of the line PN is $$\dfrac{x-7}{2}=\dfrac{y-14}{4}=\dfrac{z-5}{-1}=\lambda$$ (say)
Let the coordinates of N be $$N(2\lambda +7, 4\lambda +14, -\lambda +5)$$
Since N lies on the plane $$2x+4y-z=2$$, so
$$2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2$$
$$\Rightarrow 21\lambda =-63$$
$$\Rightarrow \lambda =-3$$
$$\therefore$$ required foot of the perpendicular is
$$N(-6+7, -12+14, 3+5)$$, i.e., $$N(1, 2, 8)$$.
The coordinates of the point where the line through the points $$A(5,1,6)$$ and $$B(3,4,1)$$ crosses the yz-plane is
Report Question
0%
$$(0,17,-13)$$
0%
$$\left( 0,\cfrac { -17 }{ 2 } ,\cfrac { 13 }{ 2 } \right) $$
0%
$$\left( 0,\cfrac { 17 }{ 2 } ,\cfrac { -13 }{ 2 } \right) $$
0%
none of these
Explanation
Equation of the line passing through the points $$A(5, 1, 6)$$ and $$B(3,4, 1)$$ is
$$\dfrac{x-5}{2}=\dfrac{4-1}{-3} = \dfrac{7-6}{5}$$ ....(i)
Let the line (i) cross the yz plane at point $$(o, h, k)$$
Then $$(o, h, k)$$ point lie on the line (I)
So, $$\dfrac{0-5}{2} = \dfrac{h-1}{-3} = \dfrac{k-6}{5}$$
Then, $$\dfrac{k-6}{5} = - \dfrac{5}{2}$$
or, $$k-6 = -\dfrac{25}{2}$$
or, $$k = 6- \dfrac{-25}{2} =\dfrac{12-25}{2} = -\dfrac{13}{2}$$
Again, $$\dfrac{h-1}{-3} = \dfrac{-5}{2}$$
or, $$h-1 = \dfrac{15}{2}$$
or, $$h=\dfrac{15}{2} + 1 = \dfrac{17}{2}$$
Therefore the crossing points is $$\left(0, \dfrac{17}{2}, -\dfrac{13}{2}\right)$$
The point on the line $$\dfrac {x - 2}{1} = \dfrac {y + 3}{-2} = \dfrac {z + 5}{-2}$$ at a distance of 6 from the point $$\left ( 2,-3,-5 \right )$$ is
Report Question
0%
$$\left ( 3,-5,-3 \right )$$
0%
$$\left ( 4,-7,-9 \right )$$
0%
$$\left ( 0,2,-1 \right )$$
0%
$$\left ( -3,5,3 \right )$$
The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are $$d_1,d_2,d_3$$. Then $$d_1^2 +d_2^2 + d_3^2 = kd^2$$ where K is
Report Question
0%
$$1$$
0%
$$5$$
0%
$$3$$
0%
$$2$$
The shortest distance between the lines $$\dfrac {x - 3}{2} = \dfrac {y + 15}{-7} = \dfrac {z - 9}{5}$$ and $$\dfrac {x + 1}{2} = \dfrac {y - 1}{1} = \dfrac {z - 9}{-3}$$ is
Report Question
0%
$$2 \sqrt 3$$
0%
$$4 \sqrt 3$$
0%
$$3 \sqrt 6$$
0%
$$5 \sqrt 6$$
The ratio in which the plane $$\overrightarrow{r} \cdot \left (\overrightarrow{i} - 2 \overrightarrow{j} + 3 \overrightarrow{k} \right ) = 17$$ divides the line joining the points $$ -2 \overrightarrow{i} + 4 \overrightarrow{j} + 7 \overrightarrow{k} $$ and $$ 3 \overrightarrow{i} - 5 \overrightarrow{j} + 8 \overrightarrow{k} $$
Report Question
0%
1 : 5
0%
1 : 10
0%
3 : 5
0%
3 : 10
Explanation
Let the plane $$\overrightarrow{r} \cdot \left ( \overrightarrow{i} - 2 \overrightarrow{j} + 3 \overrightarrow{k} \right ) = 17 $$ divided the line joining the points $$ -2 \overrightarrow{i} + 4 \overrightarrow{j} + 7 \overrightarrow{k} $$ and $$ 3 \overrightarrow{i} - 5 \overrightarrow{j} + 8 \overrightarrow{k} $$ in the ratio $$t : 1$$ at point P.
Therefore, point P is
$$\dfrac{3t - 2} {t + 1} \overrightarrow{i} + \dfrac{-5t + 4} {t + 1} \overrightarrow{j} + \dfrac{8t + 7} {t + 1} \overrightarrow{k}$$
This lies on the given plane
$$\therefore \dfrac{3t - 2} {t + 1} (1) + \dfrac{-5t + 4} {t + 1} (- 2) + \dfrac{8t + 7} {t + 1} (3) = 17$$
Solving, we get
$$t = \dfrac{3} {10}$$
The point on the line $$\frac{x - 2} {1} = \frac{y + 3} {-2} = \frac{z + 5} {-2} $$ at a distance of 6 from the point $$\left ( 2, -3, -5 \right )$$ is
Report Question
0%
$$\left ( 3, -5, -3 \right )$$
0%
$$\left ( 4, -7, -9 \right )$$
0%
$$\left ( 0,2, -1 \right )$$
0%
$$\left ( -3, 5, 3 \right )$$
Explanation
b. Direction cosines of the given line are $$\frac{1} {3}, - \frac{2} {3}, -\frac{2} {3}$$
Hence, the equation of line can be point in the form $$\frac{x - 2} {1/3} = \frac{y + 3} {-2/3} = -\frac{z + 5} {-2/3} = r $$
Therefore, any point on the line is $$\left (2 + \frac{r} {3}, - 3 - \frac{2r} {3}, - 5 - \frac{2r} {3} \right )$$, where $$r = \pm 6$$.
Points are$$\left (4, -7, -9 \right )$$ and $$\left (0, 1, -1 \right )$$
The points $$A(1,2,-1),B(2,5,-2),C(4,4,-3)$$ and $$D(3,1,-2)$$ are
Report Question
0%
collinear
0%
vertices of a rectangle
0%
vertices of a square
0%
vertices of a rhombus
Explanation
AB$${=}$$ $$\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$$
AB$${=}$$ $$\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$$
AB$${=}$$ $$\sqrt{11}$$
Similarly you find that BC$${=}$$ $$\sqrt{6}$$ CD$${=}$$ $$\sqrt{11}$$ and DA$${=}$$$$\sqrt{6}$$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC$${=}$$ $$\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$$
AC$${=}$$ $$\sqrt{17}$$
similarly BD$${=}$$ $$\sqrt{17}$$
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle
Let $$A \left ( 2\hat{i}+3\hat{j}+5\hat{k} \right )B\left ( -\hat{i}+3\hat{j}+2\hat{k} \right ) $$and $$ C \left ( \lambda \hat{i}+5\hat{j}+\mu\hat{k} \right )$$ are vertices of a triangle and its median through $$A$$ is equally inclined to the positive directions of the axes. The value of $$\lambda+\mu$$ is equal to
Report Question
0%
$$-7$$
0%
$$2$$
0%
$$7$$
0%
$$17$$
Explanation
We have,
$$ \vec{AD} = \dfrac 12 \left( \vec b + \vec c - 2 \vec a \right) $$
$$ \vec{AD} = \dfrac{ \lambda -5 }{2} \hat i + \hat j + \dfrac{\mu-8}{2} \hat{k} $$
The line is equally inclined to the axes. Hence, its direction ratios must be of the form $$1:1:1$$.
Hence, $$\lambda= 7 $$ and $$\mu = 10 $$
If $$(0, b, 0)$$ is the centroid of the triangle formed by the points $$(4, 2, -3)$$ , $$({a}, -5, 1)$$ and $$(2, -6, 2)$$ . If $$a ,b$$ are the roots of the quadratic equation $$ x^2+px+q = 0 $$, then $$p,q$$ are
Report Question
0%
$$9,18$$
0%
$$-9,-18$$
0%
$$3,-18$$
0%
$$-3,18$$
Explanation
Since $$a,b$$ are the roots of the equation
$$x^{2}+px+q=0$$
$$\Rightarrow a+b=-p$$ and
$$ab=q$$
Centroid of triangle is $$\displaystyle \left(\dfrac{a+6}{3},-3,0\right)$$
Given, centroid $$(0,b,0)$$
Comparing, we get $$ b=-3$$
and
$$\dfrac{a+6}{3}=0$$
$$\Rightarrow a=-6$$
Hence,
$$p=9, q=18$$
Hence, option A.
lf $$OABC$$ is a tetrahedron such that the $$OA^{2}+BC^{2}=OB^{2}+CA^{2}=OC^{2}+AB^{2}$$, then which of the following is/are correct
Report Question
0%
$$AB\perp OC$$
0%
$$OB\neq CA$$
0%
$$OC=AB$$
0%
$$AB\perp BC$$
Explanation
$$OABC$$ is a tetrahedron such that $$OA^{2}+BC^{2}=OB^{2}+CA^{2}=OC^{2}+AB^{2}$$
Let $$O(0,0,0),A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2}),C(x_{3},y_{3},z_{3})$$ are coordinates of vertices.
$$OA^{2}+BC^{2}$$ = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+((x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2})$$
= $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{2}x_{3}+y_{2}y_{3}+z_{2}z_{3})$$
Similarly,
$$OB^{2}+CA^{2}$$ = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{1}x_{3}+y_{1}y_{3}+z_{1}z_{3})$$
$$OC^{2}+AB^{2}$$ = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2})$$
$$\therefore$$ $$OA^{2}+BC^{2}=OB^{2}+CA^{2}\Rightarrow x_{2}x_{3}+y_{2}y_{3}+z_{2}z_{3} = x_{1}x_{3}+y_{1}y_{3}+z_{1}z_{3}$$
$$\Rightarrow (x_{1}-x_{2})x_{3}+(y_{1}-y_{2})y_{3}+(z_{1}-z_{2})z_{3} = 0$$
$$\Rightarrow (x_1-x_2)(x_3-0) + (y_1-y_2)(y_3-0)+(z_1-z_2)(z_3-0) = 0$$
$$\Rightarrow \vec{AB}\cdot\vec{OC} = 0$$
Hence, $$OC$$ is perpendicular to $$AB.$$
The plane $$\displaystyle ax + by + cz + (-3) = 0$$ meet the co-ordinate axes in $$A, B, C$$. The centroid of the triangle is
Report Question
0%
$$\displaystyle (3a, 3b, 3c)$$
0%
$$\displaystyle \left(\frac {3}{a}. \frac {3}{b}, \frac {3}{c}\right)$$
0%
$$\displaystyle \left(\frac {a}{3}. \frac {b}{3}, \frac {c}{3}\right)$$
0%
$$\displaystyle \left(\frac {1}{a}. \frac {1}{b}, \frac {1}{c}\right)$$
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 $$\left(\dfrac{3}{a},0,0\right)$$
similarily, the point on y axis becomes $$\left(0,\dfrac{3}{b},0\right)$$
And that on z axis becomes $$\left(0,0,\dfrac{3}{c}\right)$$
The centroid of the triangle formed by these points would be
$$\left(\dfrac {\dfrac{3}{a}+0+0}{3} , \dfrac {{0}+\dfrac{3}{a}+0}{3} ,\dfrac{0+0+\dfrac{3}{c}}{3}\right)$$
$$\therefore \left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)$$
$$ABCD$$ is a parallelogram. $$L$$ is a point on $$BC$$ which divides $$BC$$ in the ratio $$1 : 2$$. $$AL$$ intersects $$BD$$ at $$P$$. $$M$$ is a point on $$DC$$ which divides $$DC$$ in the ratio $$1 : 2$$ and AM intersects $$BD$$ in $$Q$$.
Point $$P$$ divides $$AL$$ in the ratio
Report Question
0%
$$1 : 2$$
0%
$$1 : 3$$
0%
$$3 : 1$$
0%
$$2 : 1$$
Explanation
Given, $$BL:LC=1:2$$ and $$DM:MC=1:2$$
Now, $$BL:BC=1:3$$
Consider $$\triangle APD$$ and $$\triangle LPB$$
$$\angle ADP=\angle LBP$$
$$\angle APD=\angle BPL$$
By AA corollary,
$$\triangle APD \sim \triangle LPB$$
$$\Rightarrow $$ Corresponding parts are proportional.
$$\Rightarrow \dfrac{AP}{LP}=\dfrac{AD}{BL}$$
$$\Rightarrow \dfrac{AP}{LP}=\dfrac{BC}{BL}$$ ($$\because AD=BC$$)
$$\Rightarrow \dfrac{AP}{LP}=3:1$$
If the vertices of a triangle are $$(-1,6,-4),(2,1,1)$$ and $$(5,-1,0)$$ then the centroid of the triangle is
Report Question
0%
$$(6,6,-3)$$
0%
$$(2,2,-1)$$
0%
$$\left ( 3,3,-\displaystyle \frac{3}{2} \right )$$
0%
none of these
Explanation
the centroid of the triangle is
$$(\dfrac {x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3})$$
Thus by substituting the vertices we get
$$=(\dfrac{-1+2+5}{3},\dfrac{6+1-1}{3},\dfrac{-4+1+0}{3})$$
$$=(\dfrac{6}{3},\dfrac{6}{3},\dfrac{-3}{3})$$
$$\therefore$$ the centroid of the triangle is $$(2,2,-1)$$
The coordinates of a point which is equidistant from the point $$(0,0,0),(a,0,0),(0,b,0)$$ and $$(0,0,c)$$ are given by
Report Question
0%
$$\displaystyle \left( \frac { a }{ 2 } ,\frac { b }{ 2 } ,\frac { c }{ 2 } \right) $$
0%
$$\displaystyle \left( \frac { -a }{ 2 } ,\frac { -b }{ 2 } ,\frac { c }{ 2 } \right) $$
0%
$$\displaystyle \left( \frac { a }{ 2 } ,\frac { -b }{ 2 } ,\frac { -c }{ 2 } \right) $$
0%
$$\displaystyle \left( \frac { -a }{ 2 } ,\frac { b }{ 2 } ,\frac { -c }{ 2 } \right) $$
Explanation
Let $$P(x,y,z)$$ be the required point.
Then $$OP=PA=PB=PC$$
Now $$OP=PA \Longrightarrow OP^2+PA^2 \Longrightarrow x^2+y^2+z^2$$
= $$(x-a)^2+(y-0)^2+(z-0)^2 \Longrightarrow x=\dfrac{a}{2}$$
Similarily, $$OP=PB=y=\dfrac{b}{2}$$ and $$OP=PC=z=\dfrac{c}{2}$$
hence the coordinate of the required point are $$\left(\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}\right)$$
If $$A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C= \left ( \sqrt{2},\sqrt{2},0 \right )$$ and $$D= \left ( \displaystyle \frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17} \right )$$, then $$ABCD$$ is a
Report Question
0%
rhombus
0%
square
0%
parallelogram
0%
none of these
Explanation
Given points are, $$A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C=
\left ( \sqrt{2},\sqrt{2},0 \right )$$ and $$D= \left ( \displaystyle
\frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17}
\right )$$
Length $$AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$$
Length $$BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$$
Length $$CD = \sqrt{\left(
\dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$$
Length $$AD = \sqrt{\left(
\dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$$
Angle between two vectors $$\overline{AB} = p_{1}\hat{i}+q_{1}\hat{j}+r_{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$$ and $$\overline{BC} = p_{2}\hat{i}+q_{2}\hat{j}+r_{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$$ is $$cos\theta = 0 \Rightarrow \theta = 90^o$$
All sides are equal and angle between $$AB$$ and $$BC$$ is $$90^o$$, Hence, its a square.
The equation of motion of a rocket are: $$x=2t,y=-4t,z=4t,$$ where the time $$t$$ is given in seconds and the coordinate of a moving point in kilometers. At what distance will the rocket be from the starting point $$O(0,0,0)$$ in $$10$$ seconds ?
Report Question
0%
$$60$$ km
0%
$$30$$ km
0%
$$45$$ km
0%
None of these
Explanation
Eliminating t from the given equation, we get the equation of the path $$\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t $$
Thus the path of the Rocket represents a straight line passing through the origin for $$t=10sec.$$
we have $$x=20,y=-40,z=40$$
Let $$\vec r=x\vec i+y\vec j+z\vec k$$
$$\Longrightarrow |\vec r|=\sqrt{{x^2}+{y^2}+{z^2}}=\sqrt{400+1600+1600}=60km$$
What is the distance in space between $$(1,0,5)$$ and $$(-3,6,3)$$?
Report Question
0%
$$4$$
0%
$$6$$
0%
$$2\sqrt { 11 } $$
0%
$$2\sqrt { 14 } $$
0%
$$12$$
Explanation
The distance between the points $$(1,0,5)$$ and $$(-3,6,3)$$ is given below:
$$D=\sqrt { \left( 3-5 \right) ^{ 2 }+\left( 6-0 \right) ^{ 2 }+\left( -3-1 \right) ^{ 2 } } $$
$$=\sqrt { \left( -2 \right) ^{ 2 }+\left( 6 \right) ^{ 2 }+\left( -4 \right) ^{ 2 } } $$
$$=\sqrt { 4+36+16 } =\sqrt { 56 } $$
$$=2\sqrt { 14 }$$
A triangle $$ABC$$ is placed so that the midpoints of its sides are on the $$x, y$$ and $$z$$ axes respectively. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively $$\displaystyle \alpha ,\beta ,\gamma$$, then the coordinates of the centroid of the triangle $$ABC$$ are
Report Question
0%
$$\displaystyle \left (- \dfrac {\alpha }{3}, \dfrac {\beta }{3}, \dfrac {\gamma }{3} \right )$$
0%
$$\displaystyle \left (\dfrac {\alpha }{3},- \dfrac {\beta}{3}, \dfrac {\gamma}{3}\right)$$
0%
$$\displaystyle \left ( \dfrac {\alpha }{3},\dfrac { \beta }{3},-\dfrac {\gamma }{3} \right )$$
0%
$$\displaystyle \left ( \dfrac {\alpha }{3}, \dfrac {\beta }{3},\dfrac {\gamma }{3} \right )$$
Explanation
Equation of the plane containing the triangle
$$ABC$$ is $$\displaystyle \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$$
which meets the axes in $$\displaystyle \left ( \alpha ,0,0 \right ),\left ( 0,\beta ,0 \right )$$ and $$\displaystyle \left ( 0,0,y \right )$$.
Let the coordinates of $$A$$ be $$\displaystyle \left (x_1,y_1,z_1 \right )$$
Since the middle point of $$AB$$ lies on the $$z$$-axis it is $$\displaystyle \left (0,0,\gamma \right )$$ and thus the coordinates of $$B$$ are $$\displaystyle \left (-x_1,-y_1,2\gamma -z_1 \right )$$
Similarly the coordinates of $$C$$ are $$\displaystyle \left (-x_1,2\beta -y_1,-z_1 \right )$$
So that the middle point of $$\displaystyle BC=\left ( -x_1,\beta -y_1,\gamma -z_1 \right ) \displaystyle =\left ( \alpha ,0,0 \right )$$
$$\displaystyle \Rightarrow x_1=-\alpha ,y_1=\beta ,z_1=\gamma$$.
And thus the coordinates of $$A$$ are $$\displaystyle \left ( -\alpha ,\beta ,\gamma \right )$$
Similarly the coordinates of $$B$$ are $$\displaystyle \left ( \alpha ,-\beta ,\gamma \right )$$ and those of $$C$$ are $$\displaystyle \left ( \alpha ,\beta ,-\gamma \right )$$.
Hence, the coordinates of the centroid of the triangle $$ABC$$ are $$\displaystyle \left ( \dfrac {\alpha }{3},\dfrac {\beta }{3}, \dfrac {\gamma }{3} \right )$$
Point A is $$\displaystyle a+2b,$$ and a divides AB in the ratio 2 :The position vector of B is
Report Question
0%
$$\displaystyle 2a-b$$
0%
$$\displaystyle b-2a$$
0%
$$\displaystyle a-3b$$
0%
b
Explanation
Let us consider x be the position vector of B,
then a divides AB in the ratio $$2 : 3$$
$$a$$ =$$\dfrac {2x +3(a +2b)}{2+3}$$
$$\Rightarrow x= a - 3b$$
$$D(2, 1, 0), E(2, 0, 0), F(0, 1, 0)$$ are mid point of the sides $$BC, CA, AB$$ of $$\Delta$$ $$ABC$$ respectively, The the centroid of $$\Delta$$ABC is
Report Question
0%
$$\left ( \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$
0%
$$\left ( \displaystyle \frac{4}{3},\, \displaystyle \frac{2}{3},\, 0 \right )$$
0%
$$\left (- \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$
0%
$$\left ( \displaystyle \frac{2}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$$
Explanation
Centroid of triangle coincide with the centroid of triangle formed by joing the mid-point of sides of triangle
So, centroid of $$\triangle ABC =$$ centroid of $$\triangle DEF$$
$$=\left(\dfrac{2+2+0}{3},\dfrac{1+0+1}{3},\dfrac{0+0+0}{3}\right)$$
$$=\left(\dfrac{4}{3},\dfrac{2}{3},0\right)$$
If $$P(x,y,x)$$ is a point on the line segment joining $$Q(2,2,4)$$ and $$R(3,5,6)$$ such that the projection of $$\overline { OP } $$ on the axis are $$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 } ,\frac { 26 }{ 5 } ,$$ respectively, then $$P$$ divides $$QR$$ in the ratio
Report Question
0%
$$1:2$$
0%
$$3:2$$
0%
$$2:3$$
0%
$$1:3$$
Explanation
Since $$OP$$ has projection $$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 } $$ and $$\displaystyle \frac { 26 }{ 5 } $$ on the coordinate axis,
therefore $$\displaystyle OP=\frac { 13 }{ 5 }\vec i+\frac { 19 }{ 5 } \vec j+\frac { 26 }{ 5 } \vec k$$.
Suppose $$P$$ divides the join of $$Q(2,2,4)$$ and $$R(3,5,6)$$ in the ratio $$\lambda :1$$.
Then, the position vector of $$P$$ is
$$\displaystyle \left( \frac { 3\lambda +2 }{ \lambda +1 } \right)\vec i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \vec j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \vec k$$
$$\displaystyle \therefore \frac { 13 }{ 5 } \vec i+\frac { 19 }{ 5 }\vec j+\frac { 26 }{ 5 }\vec k=\left( \frac { 3\lambda +2 }{ \lambda +1 } \right) \vec i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \vec j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \vec k$$
$$\displaystyle \Rightarrow \frac { 3\lambda +2 }{ \lambda +1 } =\frac { 13 }{ 5 } ;\frac { 5\lambda +2 }{ \lambda +1 } =\frac { 19 }{ 5 } ;\frac { 6\lambda +4 }{ \lambda +1 } =\frac { 26 }{ 5 } $$
$$\displaystyle \Rightarrow 2\lambda =3$$
$$\Rightarrow \lambda =\dfrac { 3 }{ 2 } $$
There are three points with position vectors $$ -2a+3b+5c, a+2b+3c $$ and$$ 7a-c$$. What is the relation between the three points?
Report Question
0%
Collinear
0%
Forms a triangle
0%
In different plane
0%
None of the above
Explanation
The relation between the three points are collinear
Thus option A is correct answer
The plane $$ax+by+cz+d=0$$ divides the line joining the points $$\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 } \right) $$ in the ratio
Report Question
0%
$$\displaystyle \frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) } $$
0%
$$\displaystyle \frac { \left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) } $$
0%
$$\displaystyle \frac { a{ x }_{ 1 }{ x }_{ 2 }+b{ y }_{ 1 }{ y }_{ 2 }+c{ z }_{ 1 }{ z }_{ 2 } }{ { d }^{ 2 } } $$
0%
None of these
Explanation
Let the given plane meet the line joining the given points in $$\left( { x }_{ 3 },{ y }_{ 3 },{ z }_{ 3 } \right) $$.
Then $$a{ x }_{ 3 }+b{ y }_{ 3 }+c{ z }_{ 3 }+d=0$$ ...(1)
Also let the point $$\left( { x }_{ 3 },{ y }_{ 3 },{ z }_{ 3 } \right) $$ divide the line joining the given points in the ratio $$m:n$$
Then
$$\displaystyle { x }_{ 3 }=\frac { m{ x }_{ 1 }+n{ x }_{ 2 } }{ m+n } ,{ y }_{ 3 }=\frac { m{ y }_{ 1 }+n{ y }_{ 2 } }{ m+n } $$ and $$\displaystyle { z }_{ 3 }=\frac { m{ z }_{ 1 }+n{ z }_{ 2 } }{ m+n } $$
Substituting these values in (1), we get
$$\displaystyle a\left( \frac { m{ x }_{ 1 }+n{ x }_{ 2 } }{ m+n } \right) +b\left( \frac { m{ y }_{ 1 }+n{ y }_{ 2 } }{ m+n } \right) +c\left( \frac { m{ z }_{ 1 }+n{ z }_{ 2 } }{ m+n } \right) +d=0$$
$$\Rightarrow a\left( m{ x }_{ 1 }+n{ x }_{ 2 } \right) +b\left( m{ y }_{ 1 }+n{ y }_{ 2 } \right) +c\left( m{ z }_{ 1 }+n{ z }_{ 2 } \right) +d\left( m+n \right) =0$$
$$\Rightarrow m\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) +n\left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) =0$$
$$\displaystyle\Rightarrow \frac { n }{ m } =\frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) }$$
Hence, option A.
The set of points in space 4 inches from a given line and 4 inches from a given point on this line is ______ , if given point lies on the given line.
Report Question
0%
a set consisting of two points
0%
a set consisting of four points
0%
a set consisting of two circles
0%
the empty set
Let $$\displaystyle a,b,c \epsilon R$$ such that $$abc = p$$ and $$qa-b = 0$$, where $$ p$$ and $$q$$ are fixed positive number, then minimum distance of the point $$(a, b, c)$$ from origin in the three dimensional coordinate system is:
Report Question
0%
$$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{2q} \right )^{1/3}$$
0%
$$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{q} \right )^{1/3}$$
0%
$$\displaystyle \sqrt{3}(p)^{1/3}$$
0%
$$\displaystyle \sqrt{2}\left ( \frac{p}{q} \right )^{1/2}$$
Explanation
We know that the distance between point $$(a,b,c)$$ and origin $$(0,0,0)$$ is:
$$d= \sqrt({a^2+b^2+c^2)}$$.
Using the inequality: $$AM\geq GM$$, for terms $$a^2, b^2$$ and $$c^2$$:
$$\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge \sqrt [ 3 ]{ { (abc) }^{ 2 } } $$.
$$abc = p$$ (Given)
$$\implies \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge { (p) }^{ \frac{2}{3} } $$.
Taking square root on both sides of the inequality:
$$ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } \ge \sqrt { 3 } { p }^{ 1/3 }$$.
Hence, Option C is correct.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Engineering Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
