Three Dimensional Geometry Test

Result

Three Dimensional Geometry Test - 31

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If the image of the point $$\displaystyle \left ( 1, \: 1, \: 1 \right )$$ by a plane $$\displaystyle \left ( 3, \: -1, \: 5 \right )$$ then the equation of the plane is

A
$$\displaystyle x - y + 2z = 8$$

B
$$\displaystyle x - y + 2z = 16$$

C
$$\displaystyle x - y + 2z = 14$$

D
None of these

Solution

Let $$A = (1,1,1)$$ and $$A'=(3,-1,5)$$
So mid point of $$AA'$$ is $$(2,0,3)$$
and direction ratio of $$AA'$$ are $$2,-2$$ and $$4$$.
Since $$A'$$ is the image of $$A$$ so equation of plane is given by,
$$2(x-2)-2(y-0)+4(z-3)=0$$
$$\Rightarrow x-y+2z=8$$
Question 2
1 / -0

Which of the triplet can not represent direction cosine of a line

A
$$\displaystyle \left( \frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } \right) $$

B
$$\displaystyle \left( \frac { 3 }{ \sqrt { 50 } } ,\frac { 4 }{ \sqrt { 50 } } ,\frac { 5 }{ \sqrt { 50 } } \right) $$

C
$$\displaystyle \left( \frac { 4 }{ \sqrt { 77 } } ,\frac { 5 }{ \sqrt { 77 } } ,\frac { 6 }{ \sqrt { 77 } } \right) $$

D
$$\displaystyle \left( \frac { 2 }{ \sqrt { 25 } } ,\frac { 3 }{ \sqrt { 25 } } ,\frac { 4 }{ \sqrt { 25 } } \right) $$

Solution

In general, the direction cosine of a line is defined as the cosine of the angles between the positive directed lines and the coordinate axes.
So the direction cosine of a line $$(\dfrac{2}{\sqrt 25},\dfrac{3}{\sqrt 25},\dfrac{4}{\sqrt 25})$$ cannot be represented in triplet form
Option D is right answer
Question 3
1 / -0

The direction cosines of the normal to the plane $$\displaystyle 5\left ( x - 2 \right ) = 3\left ( y - z \right )$$ are

A
$$\displaystyle 5, \: -3, \: 3$$

B
$$\displaystyle \frac{5}{\sqrt{43}}, \: \frac{-3}{\sqrt{43}}, \: \frac{3}{\sqrt{43}}$$

C
$$\displaystyle \frac{1}{2}, \: \frac{-3}{10}, \: \frac{3}{10}$$

D
$$\displaystyle 1, \: \frac{-3}{5}, \: \frac{3}{5}$$

Solution

Given plane is, $$5(x-2) = 3(y-z)$$
$$\Rightarrow 5x-3y+3z-10 = 0$$
So direction ratios of normal to the plane are $$5, -3$$ and $$3$$.
Thus direction cosines are, $$\dfrac{5}{\sqrt{5^2+3^2+3^2}}, -\dfrac{3}{\sqrt{5^2+3^2+3^2}}, \dfrac{3}{\sqrt{5^2+3^2+3^2}}$$
or $$ \dfrac{5}{\sqrt{43}}, \dfrac{-3}{\sqrt{43}}, \dfrac{3}{\sqrt{43}}$$
Question 4
1 / -0

The position vectors of three points are $$2\vec{a}-\vec{b}+3\vec{c}$$, $$\vec{a}-2\vec{b}+\lambda \vec{c}$$ and $$\mu \vec{a}-5\vec{b}$$ where $$\vec{a}, \vec{b}, \vec{c}$$ are non coplanar vectors, then the points are collinear when

A
$$\displaystyle \lambda =-2, \mu =\dfrac{9}{4}$$

B
$$\displaystyle \lambda =-\dfrac{9}{4}, \mu =2$$

C
$$\displaystyle \lambda =\dfrac{9}{4}, \mu =-2$$

D
None of these

Solution

When points $$x, y, z$$ are collinear, we have $$\alpha x + \beta y = (\alpha + \beta)z$$
Similarly, $$x(2\vec{a} - \vec{b} + 3\vec{c}) + y(\vec{a} - 2\vec{b} + \lambda \vec{c}) = (x + y)(\mu \vec{a} - 5\vec{b})$$
$$\Rightarrow$$ comparing the coefficients of $$\vec{a} \rightarrow 2x + y = x\mu + y\mu $$
$$\vec{b} \rightarrow - x - 2y = - 5x - 5y$$
$$\vec{c} \rightarrow 3x + \lambda y = 0$$
$$\Rightarrow 4x = -3y$$ and so $$\lambda = \dfrac{9}{4}$$
Also, $$\mu = -2$$
Question 5
1 / -0

The angle between the straight lines $$\displaystyle \frac { x+1 }{ 2 } =\frac { y-2 }{ 5 } =\frac { z+3 }{ 4 } $$ and $$\displaystyle \frac { x-1 }{ 1 } =\frac { y+2 }{ 2 } =\frac { z-3 }{ -3 } $$ is

A
$$45^0$$

B
$$30^0$$

C
$$60^0$$

D
$$90^0$$

Solution

The direction ratios of the lines are $${ a }_{ 1 }=2,{ b }_{ 1 }=5,{ c }_{ 1 }=4,{ a }_{ 2 }=1,{ b }_{ 2 }=2,{ c }_{ 2 }=-3$$
Thus $$\cos { \theta } =\displaystyle\frac { { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 } }{ \sqrt { { \left( { a }_{ 1 } \right) }^{ 2 }+{ \left( { b }_{ 1 } \right) }^{ 2 }+{ \left( { c }_{ 1 } \right) }^{ 2 } } \sqrt { { \left( { a }_{ 2 } \right) }^{ 2 }+{ \left( { b }_{ 2 } \right) }^{ 2 }+{ \left( { c }_{ 2 } \right) }^{ 2 } } } $$
$$\Rightarrow \cos { \theta } =\displaystyle\frac { 2\times 1+5\times 2+4\times -3 }{ \sqrt { { \left( 2 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } \sqrt { { \left( 1 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 }+{ \left( -3 \right) }^{ 2 } } } =0$$
$$\Rightarrow \theta =\cos ^{ -1 }{ 0 } =\dfrac { \pi }{ 2 } $$
Question 6
1 / -0

Let $$A= \left ( 1,2,2 \right )$$, $$B=\left ( 2,3,6 \right )$$and $$C= \left ( 3,4,12 \right )$$. The direction cosines of a line equally inclined with $$OA,OB$$ and $$OC$$ , where $$O$$ is the origin, are

A
$$\displaystyle \frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0$$

B
$$\displaystyle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$$

C
$$\displaystyle \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$

D
$$\displaystyle \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}$$

Solution

Let the direction cosines of the line be $$(a,b,c)$$
Since it's equally inclined to $$OA,\; OB,\; OC$$
$$ \dfrac{a+2b+2c}{3}=\dfrac{2a+3b+6c}{7}=\dfrac{3a+4b+12c}{13}=k$$
$$ \Rightarrow a+2b+2c=3k...(1) \\ 2a+3b+6c=7k ....(2)\\ 3a+4b+12c=13k....(3)$$
Eliminating $$a$$ from $$(1)$$ and $$(2)$$, we get
$$b-2c=-k....(3)$$
Similarly, eliminating $$a$$ from $$(1)$$ and $$(3)$$, we get
$$2b-6c=-4k...(4)$$
From $$(3)$$ and $$(4)$$, we get $$b=c=k...(5)$$
Using the above value in $$(1)$$ we get $$a=-k....(6)$$
From $$(5)$$ and $$(6)$$, $$a:b:c=-1:1:1$$
Also, $$a^2+b^2+c^2=1 \Rightarrow k= \pm \dfrac{1}{\sqrt{3}}$$
Hence, direction cosines are $$ \left(-\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} \right)$$ or $$ \left(\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}} \right)$$
Question 7
1 / -0

A st. line which makes angle of $$\displaystyle 60^{\circ}$$ with each of y - and z - axes, is inclined with x - axis at an angle

A
$$\displaystyle 45^{\circ}$$

B
$$\displaystyle 30^{\circ}$$

C
$$\displaystyle 75^{\circ}$$

D
$$\displaystyle 60^{\circ}$$

Solution

As we know in $$3D$$ the relation between angles made by a vector with $$x$$ axis $$(x),y$$ axis $$(y), z$$ axis$$(z)$$ is:-

$${\cos}^{2}{x}+{\cos}^{2}{y}+{\cos}^{2}{z}=1$$

$$\Rightarrow\,{\cos}^{2}{{60}^{\circ}}+{\cos}^{2}{{60}^{\circ}}+{\cos}^{2}{z}=1$$

$$\Rightarrow\,{\left(\dfrac{1}{2}\right)}^{2}+{\left(\dfrac{1}{2}\right)}^{2}+{\cos}^{2}{z}=1$$

$$\Rightarrow\,\dfrac{1}{4}+\dfrac{1}{4}+{\cos}^{2}{z}=1$$

$$\Rightarrow\,\dfrac{2}{4}+{\cos}^{2}{z}=1$$

$$\Rightarrow\,\dfrac{1}{2}+{\cos}^{2}{z}=1$$

$$\Rightarrow\,{\cos}^{2}{z}=1-\dfrac{1}{2}=\dfrac{1}{2}$$

$$\Rightarrow\,\cos{z}=\dfrac{1}{\sqrt{2}}$$

$$\therefore\,z={45}^{\circ}$$
Question 8
1 / -0

Find the equations of the plane through the point $$\displaystyle \left ( x_{1},y_{1},z_{1} \right )$$ and perpendicular to the straight line $$\displaystyle \frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$$

A
$$\displaystyle l\left ( x-x_{1} \right )+m\left ( y-y_{1} \right )+n\left ( z-z_{1} \right )=0.$$

B
$$\displaystyle l\left ( x-x_{1} \right )+m\left ( y-y_{1} \right )-n\left ( z-z_{1} \right )=0.$$

C
$$\displaystyle l\left ( x-x_{1} \right )-m\left ( y-y_{1} \right )+n\left ( z-z_{1} \right )=0.$$

D
none of these

Solution

Since the line is perpendicular to the plane, hence the normal vector of the plane is parallel to the direction vector of the line.
Hence the equation of the plane will be
$$l(x)+m(y)+n(z)=d$$
It is given that the plane passes through $$x_{1},x_{2},x_{3}$$
Hence
$$lx_{1}+my_{1}+nz_{1}=d$$
Substituting in the equation of the plane, we get
$$lx+my+nz=lx_{1}+my_{1}+nz_{1}$$
Or
$$l(x-x_{1})+m(y-y_{1})+n(z-z_{1})=0$$
Question 9
1 / -0

Which of the following triplets give the direction cosines of a line ?

A
$$\displaystyle 1, 1, 1$$

B
$$\displaystyle 1, -1, 1$$

C
$$\displaystyle 1, 1, -1$$

D
$$\displaystyle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$

Solution

If $$l,m,n$$ are the directions cosine of a line then $$l^2+m^2+n^2=1$$
Thus we get $$\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}$$
Option D is correct answer
Question 10
1 / -0

If a line makes angles $$\displaystyle \alpha ,\beta ,\gamma $$ with axes of co-ordinates, then $$\displaystyle \cos 2\alpha +\cos 2\beta +\cos 2\gamma$$ is equla to

A
$$\displaystyle -2$$

B
$$\displaystyle -1$$

C
$$1$$

D
$$2$$

Solution

We know that direction cosines are $$\cos\alpha, \cos\beta, \cos\gamma$$
Then sum of the squares of the direction cosine is 1
$$i.e. \cos^2\alpha+\cos^2\beta+\cos^2\gamma =1$$
we know Trignometric identity that $$ \sin^2A+\cos^2A=1$$
From the above identity we can write as $$\cos^2A=1-\sin^2A$$
Now replacing we get $$ (1-\sin^2\alpha) + (1-\sin^2\beta) + (1-\sin^2\gamma)$$
By adding all we get $$ 3-(\sin^2\alpha+\sin^2\beta+\sin^2\gamma)=1$$
$$\therefore \cos2\alpha+\cos2\beta+\cos2\gamma = 2-3 = -1$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 31

Result

Three Dimensional Geometry Test - 31

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle x - y + 2z = 8$$

$$\displaystyle x - y + 2z = 16$$

$$\displaystyle x - y + 2z = 14$$

None of these

Solution

Question 2

$$\displaystyle \left( \frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } \right) $$

$$\displaystyle \left( \frac { 3 }{ \sqrt { 50 } } ,\frac { 4 }{ \sqrt { 50 } } ,\frac { 5 }{ \sqrt { 50 } } \right) $$

$$\displaystyle \left( \frac { 4 }{ \sqrt { 77 } } ,\frac { 5 }{ \sqrt { 77 } } ,\frac { 6 }{ \sqrt { 77 } } \right) $$

$$\displaystyle \left( \frac { 2 }{ \sqrt { 25 } } ,\frac { 3 }{ \sqrt { 25 } } ,\frac { 4 }{ \sqrt { 25 } } \right) $$

Solution

Question 3

$$\displaystyle 5, \: -3, \: 3$$

$$\displaystyle \frac{5}{\sqrt{43}}, \: \frac{-3}{\sqrt{43}}, \: \frac{3}{\sqrt{43}}$$

$$\displaystyle \frac{1}{2}, \: \frac{-3}{10}, \: \frac{3}{10}$$

$$\displaystyle 1, \: \frac{-3}{5}, \: \frac{3}{5}$$

Solution

Question 4

$$\displaystyle \lambda =-2, \mu =\dfrac{9}{4}$$

$$\displaystyle \lambda =-\dfrac{9}{4}, \mu =2$$

$$\displaystyle \lambda =\dfrac{9}{4}, \mu =-2$$

None of these

Solution

Question 5

$$45^0$$

$$30^0$$

$$60^0$$

$$90^0$$

Solution

Question 6

$$\displaystyle \frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0$$

$$\displaystyle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$$

$$\displaystyle \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$

$$\displaystyle \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}$$

Solution

Question 7

$$\displaystyle 45^{\circ}$$

$$\displaystyle 30^{\circ}$$

$$\displaystyle 75^{\circ}$$

$$\displaystyle 60^{\circ}$$

Solution

Question 8

$$\displaystyle l\left ( x-x_{1} \right )+m\left ( y-y_{1} \right )+n\left ( z-z_{1} \right )=0.$$

$$\displaystyle l\left ( x-x_{1} \right )+m\left ( y-y_{1} \right )-n\left ( z-z_{1} \right )=0.$$

$$\displaystyle l\left ( x-x_{1} \right )-m\left ( y-y_{1} \right )+n\left ( z-z_{1} \right )=0.$$

none of these

Solution

Question 9

$$\displaystyle 1, 1, 1$$

$$\displaystyle 1, -1, 1$$

$$\displaystyle 1, 1, -1$$

$$\displaystyle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$

Solution

Question 10

$$\displaystyle -2$$

$$\displaystyle -1$$

$$1$$

$$2$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.