Three Dimensional Geometry Test

Result

Three Dimensional Geometry Test - 39

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

A straight line passes through (1, -2, 3) and perpendicular to the plane $$2x + 3y - z = 7$$.Find the direction ratios of normal to plane.

A
$$< 2, 3, -1 >$$

B
$$< 2, 3, 1 >$$

C
$$< -1, 2, 3 >$$

D
None of the above

Solution

$$\text{concept: for any plane}\,ax+by+cz+d=0,\,\text{normal vector to this plane is}\,a\hat{i}+b\hat j+c\hat k$$

the normal vector of the plane $$2x+3y-z=7$$ is
$$2\hat{i}+3\hat{j}-\hat{k}$$
so the direction ratios of normal to plane are
$$<2,3,-1>$$
Question 2
1 / -0

What are the direction ratios of the line if it passes through the intersection of the planes $$x=3z+4$$ and $$y=2z-3$$?

A
$$(1,2,3)$$

B
$$(2,1,3)$$

C
$$(3,2,1)$$

D
$$(1,3,2)$$

Solution

Equations of the planes are $$ x=3z+4$$ and $$y=2z−3$$
$$\therefore$$ The equation of the plane passing through the line of intersection of these planes is
$$ x=3z+4$$ and $$y=2z−3$$
Thus The direction Ratios of the equation passes through intersection of the planes is $$(3,2,1)$$
Question 3
1 / -0

What is the sum of the squares of direction cosines of the line joining the points (1, 2, -3) and (-2, 3, 1) ?

A
0

B
1

C
3

D
$$\frac{2}{\sqrt{26}}$$

Solution

the sum of the squares of direction cosines of the line is always 1
Question 4
1 / -0

Direction ratios of the line which is perpendicular to the lines with direction ratios $$-1,2,2$$ and $$0,2,1$$ are

A
$$1,1,2$$

B
$$2,-1,2$$

C
$$-2,1,2$$

D
$$2,1,-2$$

Solution

Let $$l,m,n$$ be the direction ratios of the required line

Given : $$l,m,n$$ are perpendicular to $$-1,2,2$$

So, their dot product will be zero

$$\implies l(-1)+m(2)+n(2)=0$$

$$\implies -l+2m+2n=0$$ .... $$(i)$$

Also, $$l,m.n$$ are perpendicular to $$0,2,1$$

$$\implies 2m+n=0$$

$$\implies m=-\dfrac{n}{2}$$

Substituting $$m$$ in $$(i)$$ we get

$$-l-n+2n=0 \implies l=n$$

So, the direction ratios are in proportion to $$n:-\dfrac{n}{2}:n$$

Simplifying we get $$2,-1,2$$

Thus, the direction ratios are $$2,-1,2$$.
Question 5
1 / -0

The direction cosines of the straight line given by the planes $$x=0$$ and $$z=0$$ are

A
$$1,0,0$$

B
$$0,0,1$$

C
$$1,1,0$$

D
$$0,1,0$$

E
$$0,1,1$$

Solution

Given, $$x=z=0$$
It represents Z-axis
$$\therefore$$ Direction cosines$$=(0,1,0)$$
Question 6
1 / -0

If $$(1, -2, -2)$$ and $$(0, 2, 1)$$ are direction ratios of two lines, then the direction cosines of a perpendicular to both the lines are

A
$$\left( \dfrac{1}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$$

B
$$\left( \dfrac{2}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$$

C
$$\left( -\dfrac{2}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$$

D
$$\left( \dfrac{2}{\sqrt{14}}, - \dfrac{1}{\sqrt{14}} , \dfrac{3}{\sqrt{14}}\right )$$

Solution

Let a vector $$ \vec { A } =\hat { i } -2\hat { j } -2\hat { k } $$
$$\vec { B } =2\hat { j } +\hat { k } $$
So, vector perpendicular to $$ \vec { A } \& \vec { B }$$ will be
$$=\vec { A } \times \vec { B } $$$$=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & -2 \\ 0 & 2 & 1 \end{matrix} \right| $$
$$=2\hat { i } -1\hat { j } +2\hat { k } $$
So, the direction ration of the perpendicular vector or line are (2,-1,2)
Direction cosines are $$\cfrac { 2 }{ \sqrt { { 2 }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ 2 }^{ 2 } } } ,\cfrac { -1 }{ \sqrt { { 2 }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ 2 }^{ 2 } } } ,\cfrac { 2 }{ \sqrt { { 2 }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ 2 }^{ 2 } } } $$
or $$\left( \cfrac { 2 }{ 3 } ,\cfrac { -1 }{ 3 } ,\cfrac { 2 }{ 3 } \right) $$
Question 7
1 / -0

The points, whose position vectors are $$60i + 3j, 40i - 8j$$ and $$ai - 52j$$ collinear, if

A
$$a = 40$$

B
$$a = -40$$

C
$$a = 20$$

D
$$a = -20$$

Solution

Let $$60i +3j,40i-8j,ai-52j$$ be the position vectors of $$A(\overline a)$$,$$B(\overline b)$$ and $$C(\overline c)$$
For points A,B,C to be collinear
$$\overline{AB}= \lambda \overline{BC}$$,For some scalar $$\lambda$$
$$\therefore$$ $$\overline b - \overline a =\lambda ( \overline c - \overline b)$$
$$(40i - 8j) - (60i + 3j) = \lambda[(ai - 52j) - (40i - 8j)]$$
$$-20 = \lambda (a - 40)$$ and $$-11 = -44\lambda$$
$$\Rightarrow a - 40 = \dfrac {-20}{\lambda}$$ and $$\lambda = \dfrac {1}{4}$$
$$\Rightarrow a = 40 - \dfrac {20}{1} \times 4 = -40$$.
Question 8
1 / -0

The direction ratios of the line perpendicular to the lines with direction ratios $$ 1, -2, -2 $$ and $$ 0, 2, 1 $$ are

A
$$2, -1, 2 $$

B
$$ -2, 1, 2 $$

C
$$ 2, 1, -2 $$

D
$$ -2, -1, -2 $$

Solution

Direction ratios of two lines: $$<1,-2,-2>$$ and $$<0,2,1>$$

Let direction ratio of required line is $$<l,m,n>$$

Since, lines are perpendicular,

So, $$l-2m-2n=0$$ and $$2m+n=0$$
$$\cfrac{l}{-2+4}=\cfrac{m}{0-1}=\cfrac{n}{2-0}$$

So, Reuired direction ratio is $$<2,-1,2>$$

Hence, A is the correct option.
Question 9
1 / -0

The equation of the plane perpendicular to the $$yz-$$ plane and passing through the point $$(1,-2,4)$$ and $$(3,-4,5)$$ is

A
$$y+2z=5$$

B
$$2y+z=5$$

C
$$y+2z=6$$

D
$$2y+z=6$$

Solution

Plane is perpendicular to yz plane
So it have x-axis as parallel to it
Find a vector lying in the plane using two given point as-
$$\vec { 5 }= (3-1)\hat { i } +(-4-(-2)\hat { j } +(5-A)\hat { k } $$
$$\vec { 5 } = 2\hat { i } -2\hat { j } +\hat { k } $$
parallel vector, $$\vec { p } =1.\hat { i } +0.\hat { j } +0.\hat { k } $$
So normal to the plane,
$$\vec { N } =\vec { S } \times \vec { P } $$
$$=\begin{vmatrix} \hat { i } & \hat { \quad j } & \hat { \quad k } \\ 2 & \quad -2 & \quad 1 \\ 1 & \quad 0 & \quad 0 \end{vmatrix}$$
$$=\hat { j } +2\hat { k } $$
equation of plane using a point and normal,
$$\Rightarrow 0.(x-1)+1.(y-(-2)+2.(z-4)=0$$
$$\Rightarrow y+2+2z-8=0$$
$$\Rightarrow y+2z=6$$
Question 10
1 / -0

The direction cosines $$l,m,n$$ of two lines are connected by the relations $$l+m+n=0$$, $$lm=0$$, then the angle between them is

A
$$\cfrac { \pi }{ 3 } $$

B
$$\cfrac { \pi }{ 4 } $$

C
$$\cfrac { \pi }{ 2 } $$

D
$$0$$

Solution

Given, $$l+m+n=0.....(i)$$

And $$lm=0$$

Either, $$m=0$$ or $$l=0$$

If $$l=0$$, then put in eq (i) we get $$m=-n$$

$$\therefore$$ Direction ratios are $$0.-n,n$$ i.e., $$0,-1,1$$

If $$m=0$$, then put in eq.(i) we get $$l=-n$$

$$\therefore$$ Direction ratios are $$-n,0,n$$ i.e.,$$-1,0,1$$

$$\cos { \theta } =\cfrac { 0(-1)+(-1)\times 0+1\times 1 }{ \sqrt { { (0) }^{ 2 }+{ (-1) }^{ 2 }+{ (1) }^{ 2 } } \sqrt { { (-1) }^{ 2 }+{ (0) }^{ 2 }+{ (1) }^{ 2 } } } =\cfrac { 1 }{ 2 } $$

$$\Rightarrow \theta =\cfrac { \pi }{ 3 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 39

Result

Three Dimensional Geometry Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$< 2, 3, -1 >$$

$$< 2, 3, 1 >$$

$$< -1, 2, 3 >$$

None of the above

Solution

Question 2

$$(1,2,3)$$

$$(2,1,3)$$

$$(3,2,1)$$

$$(1,3,2)$$

Solution

Question 3

0

1

3

$$\frac{2}{\sqrt{26}}$$

Solution

Question 4

$$1,1,2$$

$$2,-1,2$$

$$-2,1,2$$

$$2,1,-2$$

Solution

Question 5

$$1,0,0$$

$$0,0,1$$

$$1,1,0$$

$$0,1,0$$

$$0,1,1$$

Solution

Question 6

$$\left( \dfrac{1}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$$

$$\left( \dfrac{2}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$$

$$\left( -\dfrac{2}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$$

$$\left( \dfrac{2}{\sqrt{14}}, - \dfrac{1}{\sqrt{14}} , \dfrac{3}{\sqrt{14}}\right )$$

Solution

Question 7

$$a = 40$$

$$a = -40$$

$$a = 20$$

$$a = -20$$

Solution

Question 8

$$2, -1, 2 $$

$$ -2, 1, 2 $$

$$ 2, 1, -2 $$

$$ -2, -1, -2 $$

Solution

Question 9

$$y+2z=5$$

$$2y+z=5$$

$$y+2z=6$$

$$2y+z=6$$

Solution

Question 10

$$\cfrac { \pi }{ 3 } $$

$$\cfrac { \pi }{ 4 } $$

$$\cfrac { \pi }{ 2 } $$

$$0$$

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.