Three Dimensional Geometry Test

Result

Three Dimensional Geometry Test - 47

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

ABC is a triangle where $$A = ( 2,3,5 ) , B = ( - 1,2,2 )$$ and $$C (\lambda,5 , \mu )$$ if the median through A is equally inclined to the positive axis then $$\lambda + \mu$$ is

A
7

B
6

C
15

D
9

Solution

$$AD$$ is median $$D\left( {\frac{{\lambda - 1}}{2},\frac{7}{2},\frac{{\mu + 2}}{2}} \right)$$
$$\overrightarrow {AD} = \left( {\frac{{\lambda - 1}}{2} - 2,\frac{7}{2} - 3,\frac{{\mu + 2}}{2} - 5} \right)$$
$$\overrightarrow {AD} $$ is equally in lined. $$ \therefore$$paralle to $$(1,1,1)$$
$$\begin{array}{l} \frac { { \lambda -5 } }{ 2 } =\frac { 1 }{ 2 } =\frac { { \mu +8 } }{ 2 } \\ \lambda =6 \\ \mu =9 \\ \lambda +\mu =15 \end{array}$$
Then,
Option $$C$$ is correct answer.
Question 2
1 / -0

A line makes angles $$\alpha, \beta, \gamma$$ with the positive direction of the axes of reference. The value of $$\cos{2\alpha}+\cos{2\beta}+\cos{2\gamma}$$ is

A
$$1$$

B
$$3$$

C
$$-1$$

D
$$0$$

Solution

$$\cos ^{2}\alpha+\cos^{2}\beta+\cos^{2}r=1$$
$$\cos 2\alpha+\cos ^{2}\beta+\cos 2r=2\cos^{2}\alpha-1+2\cos^{2}\beta-1+2\cos^{2}r-1$$
$$=2(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}r)-3$$
$$=2(1)-3$$
$$=-1$$
Question 3
1 / -0

Projection of a vector on $$3$$ coordinate axes are $$6, - 3, 2$$ respectively. Then DC's of vector are-

A
$$6, - 3, 2$$

B
$$\dfrac{6}{5},\dfrac{{ - 3}}{5},\dfrac{2}{5}$$

C
$$\dfrac{-6}{7},\dfrac{{ - 3}}{2},\dfrac{2}{7}$$

D
$$\dfrac{ 6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$$

Solution

Projection of a vector on coordinate axis are $$X_2-X_1, Y_2-Y_1, Z_2-Z_1$$
$$X_2-X_1=6, Y_2-Y_1=-3, Z_2-Z_1=2$$
$$\sqrt {{{\left( {{X_2} - {X_1}} \right)}^2} + {{\left( {{Y_2} - {Y_1}} \right)}^2} + {{\left( {{Z_2} - {Z_1}} \right)}^2}} = \sqrt {36 + 9 + 4} = 7$$
The Direction cosines of the vector are $$\dfrac{6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$$
Question 4
1 / -0

If the foot of the perpendicular from $$(0,0,0)$$ to a plane is $$P(1,2,2)$$. Then, the equation of the plane is

A
$$-x+2y+8z-9=0$$

B
$$x+2y+2z-9=0$$

C
$$x+y+z-5=0$$

D
$$x+2y-3z+3=0$$

Solution

Equation of normal $$\Rightarrow \ \hat i+2\hat j+2\hat k$$
equation of plane $$\Rightarrow $$
$$(x-1)1+(xy-2)2+(z-2)2=0$$
$$x-1+2y-4+2z-4=0$$
$$x+2y+2z-9=0$$
Question 5
1 / -0

$$P\left(1,1,1\right)$$ and $$Q\left(\lambda,\lambda,\lambda\right)$$ are two points in the space such that $$PQ=\sqrt{27},$$ the value of $$\lambda$$ can be

A
$$-4$$

B
$$-2$$

C
$$2$$

D
$$0$$

Solution

$${PQ}^{2}={\left(\lambda-1\right)}^{2}+{\left(\lambda-1\right)}^{2}+{\left(\lambda-1\right)}^{2}$$
$$\Rightarrow\,3{\left(\lambda-1\right)}^{2}=27$$
$$\Rightarrow\,{\left(\lambda-1\right)}^{2}=9$$
$$\Rightarrow\,\lambda-1=\pm\,3$$
$$\Rightarrow\,\lambda=1\pm\,3$$
$$\Rightarrow\,\lambda=1+3,$$or $$\,1-3$$
$$\therefore\,\lambda=4,$$ or $$-2$$
Question 6
1 / -0

If the points $$\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$$ are collinear, then

A
$$k$$ has only one real value

B
$$k$$ has two real value

C
$$k$$ has no real values

D
$$k$$ has infinite number of real values

Solution

As the $$3$$ point should be collinear area of triangle termed by then should be zero

considering $$2$$ direction to be $$(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$$
& $$(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$$

$$(2\bar{b})\times (1-k)\bar{b}=0$$

this will be for any value of $$k$$ as cross produced of $$2$$ linear vector $$=0$$
Question 7
1 / -0

If the points $$(\alpha, - 1), (2, 1)$$ and $$(4, 5)$$ are collinear, then find $$\alpha $$ by vector method.

A
$$4$$

B
$$1$$

C
$$8$$

D
None of these

Solution

If there points are collinear then vectors from one to another will have scalar triple produced $$0$$.Point $$\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$$
$$\left( 2-\alpha \right) \hat { i } +2\hat { j } -\bar { A }$$
$$2\hat { i } +4\hat { j } -\bar { B }$$
$$\left( 4-\alpha \right) \hat { i } +6\hat { j } -\bar { C }$$
$$ \bar { A } .\left( \bar { B } \times \bar { C } \right) =0$$
$$\left( \left( 2-\alpha \right) \hat { i } +2\hat { j } \right) \left( 2\hat { i } +4\hat { j } \right) \times \left( \left( 4-\alpha \right) \hat { i } +6\hat { j } \right) \\ \left( \left( 2-\alpha \right) \hat { i } +2\hat { j } \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k } \right] =0$$
$$4\alpha =4$$
$$\alpha =1$$
Also the direction vector will be proportion
$$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$$
$$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$$
$$\lambda=\dfrac{1}{2}$$ as $$2=4\lambda$$
$$2-\alpha=1$$
$$\therefore \alpha=1$$
Question 8
1 / -0

$$A(2,3,7),B(-1,3,2)$$ and $$C(q,5,r)$$ are the vertices of $$\Delta ABC$$. If the median through A is equally inclined to the coordinate axes then the coordinates of the vertex C is

A
$$(7,5,14)$$

B
$$(7,5,12)$$

C
$$(7,5,10)$$

D
$$(7,5,16)$$

Solution
Question 9
1 / -0

A line $$AB$$ in three-dimensional space males angles $${45}^{o}$$ and $${120}^{o}$$ with the positive x-axis and the positive y-axis respectively. If $$AB$$ makes an acute angle $$\theta$$ with the positive z-axis, then $$\theta$$ equal

A
$${45}^{o}$$

B
$${60}^{o}$$

C
$${75}^{o}$$

D
$${30}^{o}$$

Solution

'l', 'm', 'n' be the Direction cosine of the line
$$ l = cos \alpha = cos45^{\circ} = \dfrac{1}{\sqrt{2}}$$
$$ m = cos\beta = cos120^{\circ} = \dfrac{-1}{2}$$
$$ n = cos\theta $$
we know
$$ l^{2}+m^{2}+n^{2} = 1 $$
$$ \Rightarrow d\dfrac{1}{2}+\dfrac{1}{4}+cos^{2}\theta = 1 $$
$$ \Rightarrow cos^{2}\theta = 1-\dfrac{1}{2}- \dfrac{1}{4}$$
$$ \Rightarrow cos^{2}\theta = \dfrac{1}{4}$$
$$ \Rightarrow cos = \dfrac{1}{2}$$
$$ \therefore \theta = 60^{\circ}$$
Question 10
1 / -0

The Cartesian equation of line $$6x - 2 = 3y + 1 = 2z - 2$$ is given by

A
$$\dfrac{{3x - 1}}{3} = \dfrac{{3y + 1}}{6} = \dfrac{{z - 1}}{3}$$

B
$$\dfrac{{3x + 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$$

C
$$\dfrac{{3x - 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$$

D
$$\dfrac{{3x - 1}}{6} = \dfrac{{3y - 1}}{3} = \dfrac{{z - 1}}{3}$$

Solution

Solution -

$$ 6x-2 = 3y+1 = 2z-2 $$

$$ \dfrac{6x-2}{6} = \dfrac{3y+1}{6} = \dfrac{2z-2}{6} $$

$$ \dfrac{x-\dfrac{1}{3}}{1} = \dfrac{y+\dfrac{1}{3}}{2} = \dfrac{z-1}{3} $$
DR $$ (1,2,3) $$

Point passing through line $$(1/3,-1/3,1) $$

$$ \dfrac{3x-1}{3} = \dfrac{3y+1}{6} = \dfrac{z-1}{3} $$

A is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 47

Result

Three Dimensional Geometry Test - 47

Score

-

Rank

-

SHARING IS CARING

Question 1

7

6

15

9

Solution

Question 2

$$1$$

$$3$$

$$-1$$

$$0$$

Solution

Question 3

$$6, - 3, 2$$

$$\dfrac{6}{5},\dfrac{{ - 3}}{5},\dfrac{2}{5}$$

$$\dfrac{-6}{7},\dfrac{{ - 3}}{2},\dfrac{2}{7}$$

$$\dfrac{ 6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$$

Solution

Question 4

$$-x+2y+8z-9=0$$

$$x+2y+2z-9=0$$

$$x+y+z-5=0$$

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

Solution

Question 5

$$-4$$

$$-2$$

$$2$$

$$0$$

Solution

Question 6

$$k$$ has only one real value

$$k$$ has two real value

$$k$$ has no real values

$$k$$ has infinite number of real values

Solution

Question 7

$$4$$

$$1$$

$$8$$

None of these

Solution

Question 8

$$(7,5,14)$$

$$(7,5,12)$$

$$(7,5,10)$$

$$(7,5,16)$$

Solution

Question 9

$${45}^{o}$$

$${60}^{o}$$

$${75}^{o}$$

$${30}^{o}$$

Solution

Question 10

$$\dfrac{{3x - 1}}{3} = \dfrac{{3y + 1}}{6} = \dfrac{{z - 1}}{3}$$

$$\dfrac{{3x + 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$$

$$\dfrac{{3x - 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$$

$$\dfrac{{3x - 1}}{6} = \dfrac{{3y - 1}}{3} = \dfrac{{z - 1}}{3}$$

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.