If $$O=(0,0,0),OP=5$$ and the d.rs of OP are $$1,2,2$$ then $$P_x+P_y+P_z=$$

$$\left(\dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3}\right)$$

MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 7

The planes

$2x - y + 4z = 5$ and

$5x - 2.5y + 10z = 6$ are

0%
Parallel
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Perpendicular
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Intersect
0%
intersect $x$ axis

Explanation

Planes are

$2x-y+4z=5$

and

$5x-2.5y+10z=6$

Multiply both sides by 2 to the second equation

$\Rightarrow 10x-5y+20=12$

Now divide both sides by

$2$

$\Rightarrow 2x-y+4z=\dfrac{12}{5}$

Clearly both planes are parallel

The equation of plane passing through

$(-1,0,-1)$ parallel to

$xz$ plane is

0%
$y=-2$
0%
$y=0$
0%
$-x-z=0$
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None of the above

Explanation

Given that the plane is parallel to

$xz$ plane and the plane passes through

$(-1,0,-1)$

Since the plane is parallel to

$xz$ plane , the

$y$ coordinate should be constant

Given that it passes through point

$(-1,0,-1)$ , therefore the plane lies on

$xz$ plane

Therefore the equation of plane is

$y=0$

The correct options are

$B$

The distance of the point

$P(a, b, c)$ from the

$x$ -axis is.

0%
$\sqrt{b^2+c^2}$
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$\sqrt{a^2+c^2}$
0%
$\sqrt{a^2+b^2}$
0%
$a$

Explanation

$\textbf{Step 1: Find Co-ordinate of Point on X axis and Find its Distance from Point P}$

$\text{Given:Co-ordinate of Point $P=(x_1,y_1,z_1)$=(a,b,c)}$

$\text{ On X-axis, y and z coordinate will be zero.}$

$\therefore \text{Coordinate on the X-axis where perpendicular from point will meet is $(x_2,y_2,z_3)$= (a,0,0)}$

$\text{By Distance Formula,}$

$\text{Distance Between 2 Points=}$

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

$\Rightarrow \text{Distance Between P and X axis}$

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

$=\sqrt{b^2+c^2}$

$\textbf{Therefore,Distance Between Point P and X axis is }$

$\sqrt{b^2+c^2}$

If a line

$OP$ of length

$r$ (Where '

$O$ ' is the origin) makes an angle

$\alpha$ with x-axis and lies on the xz-plane, then what are the coordinates of

$P$ ?

0%
$\left( r\cos { \alpha } ,0,r\sin { \alpha } \right)$
0%
$\left( 0,0,r\sin { \alpha } \right)$
0%
$\left( r\cos { \alpha } ,0,0 \right)$
0%
$\left( 0,0,r\cos { \alpha } \right)$

Explanation

As OP lies on XZ plane so y co-ordinate is 0.
QP=

$r\cos { \alpha }$
PR=

$r\sin { \alpha }$
Co-ordinate of P=

$(r\cos { \alpha },0,r\sin{ \alpha })$
Option A is correct

Which of the following is true for a plane?

0%
A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus.
0%
Value of $y$ in a $zx$ plane is non-zero.
0%
Value of $z$ in a $xy$ plane is zero.
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None of the above

Points

$A(3,2,4),B\left( \cfrac { 33 }{ 5 } ,\cfrac { 28 }{ 5 } ,\cfrac { 38 }{ 5 } \right)$ , and

$C(9,8,10)$ are given. The ratio in which

$B$ divides

$\overline { AC }$ is

0%
$5:3$
0%
$2:1$
0%
$1:3$
0%
$3:2$

Explanation

$B$ divides

$AC$ in the ratio is

${x}_{1}-{x}_{2}:{x}_{2}-{x}_{3}$

$3-\cfrac { 33 }{ 5 } :\cfrac { 33 }{ 5 } -9$

$3:2$

The locus of a point, which is equidistant from the points

$(1, 1)$ and

$(3, 3)$ , is

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$y = x + 4$
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$x + y = 4$
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$x = 2$
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$y = 2$
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$y = -x$

Explanation

Let

$P(h, k)$ be a point, which is equidistant from the points

$A(1, 1)$ and

$B(3, 3)$ .
i.e.,

$PA = PB \Rightarrow (PA)^{2} = (PB)^{2}$ (by distance formula)

$\Rightarrow 1 - 2h + 1 - 2k = 9 - 6h + 9 - 6k$

$\Rightarrow 4h + 4k = 16$

$\Rightarrow h + k = 4$
So, required locus is

$x + y = 4$ .

Let

$O$ be the origin and

$A$ be the point

$(64, 0)$ . If

$P$ and

$Q$ divide

$OA$ in the ratio

$1 : 2 : 3$ , then the point

$P$ is

0%
$\left (\dfrac {32}{3}, 0\right )$
0%
$(32, 0)$
0%
$\left (\dfrac {64}{3}, 0\right )$
0%
$(16, 0)$
0%
$\left (\dfrac {16}{3}, 0\right )$

Explanation

Now, by internal section formula,

$P\equiv \left (\dfrac {1\times 64 + 5\times 10}{1 + 5} \cdot \dfrac {1\times 0 + 5\times 0}{1 + 5}\right )$

$\equiv \left (\dfrac {64}{6}, 0\right ) = \left (\dfrac {32}{3}, 0\right )$ .

The shortest distance between z-axis and the line

$x+y+2z-3=0=2x+3y+4z-4$ , is _____________

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

Explanation

$x+y+2z-3=0=2x+3y+4z-4$ at

$z-axis, x=y=0$

$x+y=3-2z$ and

$2x+3y=4(1-z)$

solving

$x$ and

$y$ in function

$z$ ,

$2x+3(3-2z-x)=4(1-z)$

$\implies 2x+9-6z-3x=4-4z$

$\implies x=9-6z-4+4z$

$\implies x=5-2z\quad equation (2)$

$\implies (5-2z)+y=3-2z$

$\implies y=3-2z-5+2z$

$y=-2\quad equation (2)$

So,

$\sqrt {x^2+y^2}=\sqrt {(-2)^2+(5-2z)^2}$

at,

$z=\cfrac {5}{2}$ , this would be minimum.

$z_{1}$ and

$z_{2}$ are

$z$ co-ordinates of the points of trisection of the segment joining the points

$A(2, 1, 4), B(-1, 3, 6)$ then

$z_{1} + z_{2} =$

0%
$1$
0%
$4$
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$5$
0%
$10$

Explanation

For

$z_{1}$ , the ratio of line segment is

$1:2$

For

$z_{2}$ , the ratio of line segment is

$2:1$

By internal division formula,

$Z=z_{1}+z_{1}$

$=\dfrac{1(6)+2(4)}{1+2}+\dfrac{2(6)+1(4)}{2+1}$

$=\dfrac{6+8}{3}+\dfrac{12+4}{3}$

$=\dfrac{14+16}{3}=\dfrac{30}{3}=10$

Hence,

$z_{1}+z_{2}=10$

In the triangle with vertices

$A(1, -1, 2), B(5, -6, 2)$ and

$C(1,3,-1)$ find the altitude

$n=|BD|$ .

0%
$5$
0%
$10$
0%
$5\sqrt{2}$
0%
$\displaystyle\frac{10}{\sqrt{2}}$

Explanation

Given

$A(1,-1,2) , B(5,-6,2) , C(1,3,-1)$

Let the coordinates of

$D$ be

$D(h,k,l)$

the equation of line

$AC$ is

$\dfrac{x-1}{0}=\dfrac{y+1}{4}=\dfrac{z-2}{-3}$

So the point of line

$AC$ will be in the form of

$( 1,4t-1,2-3t)$

Since

$D$ lies on

$AC$ , we have

$h=1,k=4t-1,l=2-3t$

The drs of

$BD$ are

$h-5,k+6,l-2$ and the drs of

$AC$ are

$0,4,-3$

Since

$BD$ and

$AC$ are perpendicular, we have

$4(k+6)-3(l-2)=0$

$\Rightarrow 4(4t+5)-3(-3t)=0$

$\Rightarrow t=-\dfrac{4}{5}$

So, coordinates of

$D$ are

$D\left(1,-\dfrac{21}{5},\dfrac{22}{5}\right)$

So, the value of

$n=\sqrt { 16+\dfrac { 81 }{ 25 } +\dfrac { 144 }{ 25 } } =\sqrt { 25 } =5$

The distance between the X-axis and the point

$(3, 12, 5)$ is

0%
3
0%
13
0%
14
0%
12
0%
5

Explanation

The distance between the

$x$ -axis and point

$(3, 12, 5) =$ Distance between

$PQ$

$= \sqrt{(3 - 3)^2 + (12 - 0)^2 + (5 - 0)^2}$

$= \sqrt{0^2 + 12^2 + 5^2} = \sqrt{144 + 25}$

$= \sqrt{169} = 13$

The point which divides the line joining the points

$(1, 3, 4)$ and

$(4,3,1)$ internally in the ratio

$2:1 ,$ is

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$(2, -3, 3 )$
0%
$(2, 3, 3)$
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$\left( \dfrac {5}{2} , 3 , \dfrac {5}{2} \right)$
0%
$( -3, 3, 2 )$
0%
$( 3,3,2)$

Explanation

Therefore,

$R (x,y,z) = \left( \dfrac {2 \times 4 + 1 \times 1 }{2 +1} , \dfrac {2 \times 3 + 1 \times 3 }{2+1} , \dfrac { 2 \times 1 + 1 \times 4 }{2+1} \right)$

$= \left( \dfrac{9}{3}, \dfrac{9}{3}, \dfrac{6}{3} \right) = (3,3,2)$

The coordinates of the foot of the perpendicular drawn from the point

$A(1,0,3)$ to the join of the points

$B(4,7,1)$ and

$C(3,5,3)$ are

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$(5,7,17)$
0%
$\left( \cfrac { -5 }{ 7 } ,\cfrac { 7 }{ 3 } ,\cfrac { -17 }{ 3 } \right)$
0%
$\left( \cfrac { 5 }{ 7 } ,\cfrac { -7 }{ 3 } ,\cfrac { 17 }{ 3 } \right)$
0%
$\left( \cfrac { 5 }{ 7 } ,\cfrac { 7 }{ 3 } ,\cfrac { 17 }{ 3 } \right)$

Explanation

The cartesian equation of the line passing through given 2-point form

$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$

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

let P be the coordinates of foot

$P\equiv (\lambda +3,2\lambda+5,-2\lambda +3 )$

DR of

$PA\equiv (\lambda+2,2\lambda+5,-2\lambda)$

So,

$(\lambda+2)1+(2\lambda+5)2+(-2\lambda)(-2)=0$

$9\lambda +12=0$

$\lambda =-\dfrac{4}{3}$

$P\equiv \left ( \dfrac{5}{3},\dfrac{7}{3},\dfrac{17}{3} \right )$

If xy -plane and yz-plane divides the line segment joining A(2,4,5) and B(3,5,-4) in the ratio a:b and p:q respectively then value of

$\left( {{a \over b},{p \over q}} \right)$ may be

0%
${\dfrac{23}{12}}$
0%
${\dfrac{7}{5}}$
0%
${\dfrac{7}{12}}$
0%
${\dfrac{21}{10}}$

Explanation

Let the co-ordinate of xy plane be

$(x, y, 0)$

According to the section formula

$0 = \dfrac{-4a + 5b}{a + b}$ {for z co-cordinates}

$\Rightarrow 5b = 4a$

$\dfrac{a}{b} = \dfrac{5}{4}$

Let the co-ordinate of yz-plane be (0, y, z)

According to the section formula,

$0 = \dfrac{2q + 3p}{p + q}$

$\therefore \dfrac{p}{q} = \dfrac{-2}{3}$

$(\dfrac{a}{b} + \dfrac{p}{q}) = (\dfrac{5}{4} - \dfrac{2}{3}) = (\dfrac{15 - 8}{12}) = \dfrac{7}{12}$

Option C should be correct

A swimmer can swim

$2$ km in

$15$ minutes in a lake and in a river he can swim a distance of

$4$ km in

$20$ minutes along the stream. If a paper boat is put in the river, then the distance covered by it in

$\displaystyle$ 2

$\, \frac{1}{2}$ 2 hours will be

0%
$18$ km
0%
$12$ km
0%
$8$ km
0%
$10$ km

Explanation

Speed of the man in still water

$\cfrac{2}{15/60}$ =

$8$ km/hr
Speed of the man in downstream =

$\cfrac{4}{20/60}$ =

$12$ km/hr
Speed of stream =

$4$ km/hr
Distance covered by paper boat in

$2$

$\frac{1}{2}$ hours =

$\cfrac{5}{2} \, \times$

$4$ =

$10$ km

In a

$\triangle {ABC}$ , side

$AB$ has the equation

$2x+3y=29$ and the side

$AC$ has the equation

$x+2y=16$ . If the mid point of

$BC$ is

$(5,6)$ , then the equation of

$BC$ is

0%
$2x+y=7$
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$x+y=1$
0%
$2x-y=17$
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None of these

Explanation

Let co-ordinates of

$B$ be

$(x_1, y_1)$ &

$C$ be

$(x_2, y_2)$

$\therefore$

$(5,6)$ is the mid point,

so,

$\dfrac{x_1 + x_2}{2} = 5, \dfrac{y_1 + y_2}{2} = 6$

$\Rightarrow x_1 + x_2 = 10, y_1 + y_2 = 12$

$B(x_1, y_1)$ lies on the line

$2x + 3y = 29$

$\therefore 2x_1 + 3y_1 = 29$ ----(1)

$C(x_2, y_2)$ lies on the line

$x + 2y = 16,$

$\therefore x_2 + 2y_2 = 16$ ----(2)

$\therefore$ putting

$x_1, y_1$ in the form of

$x_2, y_2$ in (1)

$2(10 - x_2) + 3(12 - y_2) = 29$ {

$x_1 = 10 - x_2, y_1 = 12 - y_2$ }

$\Rightarrow 20 - 2x_2 + 36 - 3y_2 = 29$

$\Rightarrow 2x_2 + 3y_2 = 27$ ----(3)

on subtracting

$(3)$ and

$(2)$

$\times$

$2$

$-y_2 = -5$

$y_2 = 5$

Putting

$y_2 \, in (2)$

$x_2 + 2(5) = 16$

$x_2 = 6$

$x_1 = 10 - x_2$

$= 4$

$y_1 = 12 - 5 = 7$

Equation :

$\dfrac{x - x_1}{x_2 - x_1} = \dfrac{y - y_1}{y_2 - y_1}$

$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$

$\Rightarrow -x + 4 = y - 7$

$\Rightarrow x + y = 11$

Plane

$ax + by + cz = 1$ intersect axes in

$A, B, C$ respectively. If

$G\left (\dfrac {1}{6}, -\dfrac {1}{3}, 1\right )$ is a centroid of

$\triangle ABC$ then

$a + b + 3c =$ _________.

0%
$\dfrac {4}{3}$
0%
$4$
0%
$2$
0%
$\dfrac {5}{6}$

Explanation

let

$A\left (\dfrac {1}{a}, 0, 0\right ) B\left (0, \dfrac {1}{b}, 0\right ) C\left (0, 0, \dfrac {1}{c}\right )$

$centroid \Rightarrow \left (\dfrac {1}{3a}, \dfrac {1}{3b}, \dfrac {1}{3c}\right ) = \left (\dfrac {1}{6}, \dfrac {-1}{3}, 1\right )$

On comparing we get,

$3a=6 \Rightarrow a=2\\3b=-3\Rightarrow b=-1\\3c=1\Rightarrow c=\dfrac{1}{3}$

$\therefore a = 2, b = -1, c = \dfrac{1}{3}$

$\therefore a + b + 3c = 2$
Only (C) is correct.

$xy-$ plane and

$yz-$ plane divides the line segment joining

$A(2,4,5)$ and

$B(3,5,-4)$ in the ratio

$a:b$ and

$p:q$ respectively then value of

$\left(\dfrac {a}{b}+\dfrac {p}{q}\right)$ may be

0%
$\dfrac {23}{12}$
0%
$\dfrac {7}{5}$
0%
$\dfrac {7}{12}$
0%
$\dfrac {21}{10}$

Explanation

Let the point on

$xy$ -plane which divide line AB in the ratio a:b be

$\left( {{x_1},{y_1},0} \right)$ and the point on

$yz$ -plane which divide line AB in the ratio p:q be

$\left( {0,{y_2},{z_2}} \right)$ .

Then

$\left( {{x_1},{y_1},0} \right) = \left( {\dfrac{{3a + 2b}}{{a + b}},\dfrac{{5a + 4b}}{{a + b}},\dfrac{{ - 4a + 5b}}{{a + b}}} \right)$

$\Rightarrow \dfrac{{ - 4a + 5b}}{{a + b}} = 0$

$\Rightarrow 4a = 5b$

$\Rightarrow \dfrac{a}{b} = \dfrac{5}{4}$

and

$\left( {0,{y_2},{z_2}} \right) = \left( {\dfrac{{3p + 2q}}{{p + q}},\dfrac{{5p + 4q}}{{p + q}},\dfrac{{ - 4p + 5q}}{{p + q}}} \right)$

$\Rightarrow \dfrac{{3p + 2q}}{{p + q}} = 0$

$\Rightarrow 3p = - 2q$

$\Rightarrow \dfrac{p}{q} = \dfrac{{ - 2}}{3}$

Now,

$\dfrac{a}{b} + \dfrac{p}{q} = \dfrac{5}{4} - \dfrac{2}{3} = \dfrac{{15 - 8}}{{12}} = \dfrac{7}{{12}}$

Locus of a point

$P$ which such that

$PA = PB$ where

$A = (0, 3, 2)$ and

$B = (2, 4, 1)$ is

0%
$2x + y - z = 4$
0%
$x - 2y + z + 1 = 0$
0%
$9x - 2y + 4z - 5 = 0$
0%
None of these

Explanation

Let the co-ordinate of

$P$ be

$(x,y,z)$ .

Given the points

$A(0,3.2)$ and

$B(2,4,1)$ .

Given,

$PA=PB$ .

or,

$\sqrt{x^2+(y-3)^2+(z-2)^2}=\sqrt{(x-2)^2+(y-4)^2+(z-1)^2}$

or,

$-6y+9-4z+4=-4x+4-8y+16-2z+1$ [ Squaring and then eliminating the higher power of

$x,y,z$ ]

or,

$4x+2y-2z=8$

or,

$2x+y-z=4$ .

So locus of

$P$ is

$2x+y-z=4$ .

Points

$(5, 0, 2), (2, -6, 0), (4, -9, 6)$ &

$(7, -3, 8)$ are vertices of a

0%
Square
0%
Rhombus
0%
Rectangle
0%
Parallelogram

Explanation

$\textbf{Step -1: Find the distance between the given points.}$

$\text{Let }A(5,0,2),B(2,-6,0),C(4,-9,6)\text{ and }D(7,-3,8).$

$AB=\sqrt{(2-5)^2+(-6-0)^2+(0-2)^2}$

$=\sqrt{9+36+4}$

$=\sqrt{49}$

$=7$

$BC=\sqrt{(4-2)^2+(-9-(-6))^2+(6-0)^2}$

$=\sqrt{4+9+36}$

$=\sqrt{49}$

$=7$

$CD=\sqrt{(7-4)^2+(-3-(-9))^2+(8-6)^2}$

$=\sqrt{9+36+4}$

$=\sqrt{49}$

$=7$

$DA=\sqrt{(5-7)^2+(0-(-3))^2+(2-8)^2}$

$=\sqrt{4+9+36}$

$=\sqrt{49}$

$=7$

$AC=\sqrt{(4-5)^2+(-9-0)^2+(6-2)^2}$

$=\sqrt{1+81+16}$

$=\sqrt{98}$

$BD=\sqrt{(7-2)^2+(-3-(-6))^2+(8-0)^2}$

$=\sqrt{25+9+64}$

$=\sqrt{98}$

$\textbf{Step -2: Find the correct option.}$

$\because AB=BC=CD=DA$

$\therefore\text{All sides are equal.}$

$\text{and }AC=BD$

$\text{They must be its diagonal and they are also equal.}$

$\text{So, the figure formed from the given vertices is a square.}$

$\textbf{Hence , the correct option is A.}$

The values of a for which

$(8,-7,a),(5,2,4)$ and

$(6,-1,2)$ are collinear , is given by

0%
$2$
0%
$-2$
0%
$-1$
0%
$1$

Given

$A(3, 2, -4), B(5, 4, -6)$ &

$C(9, 8, -10)$ are collinear. Ratio in which

$B$ divides

$AC$

0%
$1:2$
0%
$2:1$
0%
$-1:2$
0%
None of these

Explanation

Given A

$(3,2,-4)$

$(5,4,-6)$

$(9,8,-10)$

Let the ratio be

$\lambda :1$

$\therefore 5= \cfrac { (3\times 1)+(9\times \lambda ) }{ \lambda +1 } \\$

$\Rightarrow 5\lambda +5=3+9\lambda$

$\Rightarrow 4\lambda =2$

$\therefore \lambda =\cfrac { 1 }{ 2 } \\ \text{Hence the } Ratio=\lambda :1$

$\quad \quad \quad =1:2$

A cube of side 5 has one vertex at the point (1,0,-1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.

0%
(1,0,1),
0%
(0,-1,0),
0%
(0,0,-1),
0%
(1,0,0)

Explanation

Consider the problem

Below, are four complete cube face on

$XZ-plane,\,(y=0)$

Given point

$(1,0,-1)$

End of the edge parallel to negative

$x-axis$

$(0,0-1)$

Origin

$(0,0,0)$

End of the edge parallel to positive

$z-axis$

$(1,0,0)$

And, below

Four point complete the opposite face of cube.

consider

$P$ , end of edge parallel to negative

$y-axis$

$(1,-1,-1)$

Edge from

$P$ parallel to positive

$z-axis$

$(0,-1,0)$

Edge from

$P$ parallel to negative

$x-axis$

$(0,-1,-1)$

And

$(0,-1,0)$

$O=(0,0,0),OP=5$ and the d.rs of OP are

$1,2,2$ then

$P_x+P_y+P_z=$

0%
$25$
0%
$\dfrac{25}{9}$
0%
$\dfrac{25}{3}$
0%
$\left(\dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3}\right)$

Find the co-ordinates of a point lying on the line

$\dfrac{x -2}{3} = \dfrac{y + 3}{4} = \dfrac{z - 1}{7}$ which is at a distance

$10$ units from

$(2, -3, 1)$ .

0%
$(32,37,71)$
0%
$(-28,-43,-69)$
0%
$(-32,-37,-71)$
0%
None of these

Explanation

Given that the required point lies on the line

$\dfrac{x-2}{3}=\dfrac{y+3}{4}=\dfrac{z-1}{7}$ .

The required point is at a distance of

$10$ units from

$(2,-3,1)$

By option verification,

(A) Substituting the point

$(32,37,71)$

$\implies \dfrac{32-2}{3}=\dfrac{37+3}{4}=\dfrac{71-1}{7}$

$\implies 10=10=10$

Therefore, distance is

$\sqrt{(32-2)^2+(37+3)^2+(71-1)^2} =86.02$ units

Hence,

$(32,37,71)$ is not the required point.

(B) Substituting the point

$(-28,-43,-69)$

$\implies \dfrac{-28-2}{3}=\dfrac{-43+3}{4}=\dfrac{-69-1}{7}$

$\implies -10=-10=-10$

Therefore, distance is

$\sqrt{(-28-2)^2+(-43+3)^2+(-69-1)^2} =86.02$ units

Hence,

$(-28,-43,-69)$ is not the required point.

$(-32,-37,-71)$

$\implies \dfrac{-32-2}{3}=\dfrac{-37+3}{4}=\dfrac{-71-1}{7}$

$\implies 11.3=11.3=10.28$

Therefore, distance is

$\sqrt{(-32-2)^2+(-37+3)^2+(-71-1)^2} =86.57$ units

Hence,

$(-32,-37,-71)$ is not the required point.

The ratio in which the join of

$(1, -2, 4)$ and

$(4, 2, -1)$ divided by the

$XY$ plane is

0%
$1:3$
0%
$3:1$
0%
$4:1$
0%
$1:4$

Explanation

Let

$P$ be the point where the line joining the given two points

$(1,-2,4)$ and

$(4,2,-1)$ intersects the

$X-Y$ plane in

$m:n$ ratio. We are to find

$m:n$ .

Now the co-ordinate of the point

$P$ be

$\left(\dfrac{4m+n}{m+n},\dfrac{2m-2n}{m+n},\dfrac{-m+4n}{m+n}\right)$ .

As the point

$P$ lies on the

$X-Y$ plane,

$\dfrac{-m+4n}{m+n}=0$

or,

$-m+4n=0$

or,

$\dfrac{m}{n}=\dfrac{4}{1}$

or,

$m:n=4:1$

A plane meets the co-ordinate axes in A,B,C such that the centroid of the triangle ABC is the point

$(p,q,r)$ . The equation of the plane is

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$\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 0$
0%
$\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 1$
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$\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 2$
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none of these

Explanation

The equation of plane is

$\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$ ……. (1)

Using centroid formula in XYZ plane is

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

Then,

$\left( \dfrac{A}{3},\dfrac{B}{3},\dfrac{C}{3} \right)=\left( p,q,r \right)$

Therefore,

$A=3p,\,\,\,B=3q,\,\,C=3r$

Put the value of $A,B$ and $C$ in equation of plane (1) and we get

$\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$

$\dfrac{x}{3p}+\dfrac{y}{3q}+\dfrac{z}{3r}=1$

Hence,

$\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=3$

option (D) is correct answer.

If a point

$P$ from where line drawn cuts coordinates axes at

$A$ and

$B$ (with

$A$ on

$x-$ axis and

$B$ on

$y-$ axis ) satisfies

$\alpha \dfrac{x^{2}}{PB^{2}}+\beta \dfrac{y^{2}}{PA^{2}}=1$ , then

$\alpha+\beta$ is

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

Explanation

$\alpha\dfrac{x^2}{PB^2}+\beta\dfrac{y^2}{PA^2}=1$

$PA=\sqrt{4+4}=2\sqrt 2$

$PB=2\sqrt 2$

put value of

$PA$ &

$PB$ in given eqn

$\alpha\dfrac{4}{8}+\beta\dfrac{4}{8}=1$

$\alpha+\beta=2$ Option B

You can take any value of point

$A$ &

$B$

Answer will remain same.

1425935_1057037_ans_d99a81b187c54f099b92efd492e5bc35.png

The distance between the points

$(\cos \theta, \sin \theta)$ and

$(\sin \theta - \cos \theta)$ is

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$\sqrt {3}$
0%
$\sqrt {2}$
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$2$
0%
$1$

Explanation

Distance between the point

$(\cos\theta, \sin\theta)$ and

$(\sin\theta, -\cos\theta)$ is :

$= \sqrt{(\sin\theta - \cos\theta)^2 + (-\cos\theta - \sin\theta)^2}$

$= \sqrt{1 - 2 \sin\theta \cos\theta + 1 + 2 \sin\theta \cos\theta}$

$(\cos^2\theta + \sin^2\theta = 1)$

$= \sqrt{2}$

Option B is the correct answer

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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