Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

If the projections of the line segment $$AB$$ on the coordinate axes are $$2, 3, 6$$, then the sum of the d.cs of the line $$AB$$ is

A
$$11$$

B
$$1$$

C
$$\displaystyle \frac{11}{49}$$

D
$$\displaystyle \frac{11}{7}$$

Solution

Let $$\overrightarrow { AB } =l\hat { i } +m\hat { j } +n\hat { k } $$
Projection of $$AB$$ on $$x,y,z$$ axis are $$l,m,n$$ respectively
$$\overrightarrow { AB } .\hat { i } =l=2$$
$$\overrightarrow { AB } .\hat { j } =m=3$$
$$\overrightarrow { AB } .\hat { k } =n=6$$
$$\overrightarrow { AB } =2\hat { i } +3\hat { j } +6\hat { k } $$
$$\left| \overrightarrow { AB } \right| =\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =7$$
DR's of $$\overrightarrow { AB } $$ are $$(2,3,6)$$
DR's of $$\left( \dfrac { 2 }{ 7 } ,\dfrac { 3 }{ 7 } ,\dfrac { 6 }{ 7 } \right) $$
Sum of DR's $$\dfrac { 2+3+6 }{ 7 } =\dfrac { 11 }{ 7 } $$
Question 2
1 / -0

If $$l,m,n$$ are the d.cs of a line and $$l=\displaystyle \dfrac{1}{3}$$, then the maximum value of $$l\times m\times n $$ is

A
$$4$$

B
$$\displaystyle \dfrac{4}{9}$$

C
$$\displaystyle \dfrac{27}{4}$$

D
$$\displaystyle \dfrac{4}{27}$$

Solution

Given that $$l=\dfrac {1}{3}$$
Now, $$l^2+m^2+n^2=1$$
$$m^2+n^2=\dfrac {8}{9}$$
Using A.M $$\geq$$ GM,
$$\dfrac { { m }^{ 2 }+{ n }^{ 2 } }{ 2 } \ge \sqrt { { m }^{ 2 }{ n }^{ 2 } } $$
$$\Rightarrow \dfrac { 4 }{ 9 } \ge mn$$
Therefore, $$lmn \leq \dfrac { 4 }{ 27 } $$
Question 3
1 / -0

lf the projections ofthe line segment$${A}{B}$$ on the $$yz$$-plane, $$zx$$-plane, $$xy$$-plane are $$\sqrt{160}, \sqrt{153},5$$ respectively, then the projection of $${A}{B}$$ on the $${z}$$-axis is

A
$$\sqrt{12}$$

B
$$\sqrt{13}$$

C
$$12$$

D
$$144$$

Solution

Let $$\overrightarrow { AB } =l\hat { i } +m\hat { j } +n\hat { k } $$
$$l,m,n$$ are projections of $$\overrightarrow { AB } $$ on $$x,y,z$$ respectively
Projection of $$\overrightarrow { AB } $$ on $$yz$$ plane $$=\sqrt { { m }^{ 2 }+{ n }^{ 2 } } =\sqrt { 160 } $$
Projection of $$\overrightarrow { AB } $$ on $$zx$$ plane $$=\sqrt { { n }^{ 2 }+{ l }^{ 2 } } =\sqrt { 153 } $$
Projection of $$\overrightarrow { AB } $$ on $$xy$$ plane $$=\sqrt { { l }^{ 2 }+{ m }^{ 2 } } =5$$
$$\left( { m }^{ 2 }+{ n }^{ 2 } \right) -\left( { n }^{ 2 }+{ l }^{ 2 } \right) =160-153$$
$$\left( m-l \right) \left( m+l \right) =7$$
$$\Rightarrow m=4,\quad l=3,\quad n=12$$
Projection of $$AB$$ on $$z-$$ axis $$=n=12$$
Question 4
1 / -0

A point $$P$$ lies on a line whose ends are $$A(1,2,3)$$ and $$B(2,10,1).$$ If $$z$$ component of $$P$$ is $$7,$$ then the coordinates of $$P$$ are

A
$$(-1,-14,7)$$

B
$$(1,-14,7)$$

C
$$(-1,14,7)$$

D
$$(1,14,7)$$

Solution

The equation of line passing through $$A$$ and $$B$$ is $$\displaystyle\frac { x-1 }{ 2-1 } =\frac { y-2 }{ 10-2 } =\frac { z-3 }{ 1-3 } \Rightarrow \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { z-3 }{ -2 } =r$$
Substitute $$z=7$$, we get
$$\Rightarrow \displaystyle \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { 7-3 }{ -2 } =-2$$
Solve it to get $$x=1-2=-1$$, $$y=2-16=-14$$
So $$P:$$ $$\left( -1,-14,7 \right) $$
Question 5
1 / -0

lf the direction ratios of two lines are given by the equations $$2l+2m-n=0$$ and $$ml+nl+lm=0$$, then the angle between the two lines is

A
$$60^{o}$$

B
$$30^{o}$$

C
$$45^{o}$$

D
$$90^{o}$$

Solution

$$2l+2m=n ...............(1)$$ and $$mn+nl+lm=0......(2)$$
$$\Rightarrow 2lm+2m^2+2l^2+2lm+lm=0$$
$$\Rightarrow 2m^2+5lm+2l^2=0$$
On factorisation, we get
$$(m+2l)(2m+l)=0$$
$$\Longrightarrow m=−2l$$
$$\therefore m=\dfrac{-l}{2}$$
Put $$l=1$$
then $$d_1=(1,-2,-2,)$$
and $$d_2=(1,\dfrac{-1}{2},1)$$
$$d_1.d_2=0$$ for right angle $$i.e. 90$$
Question 6
1 / -0

If OA is equally inclined to OX, OY and OZ and if A is $$\sqrt{3} $$ units from the origin, then A is

A
$$(3, 3, 3)$$

B
$$( 1, -1, -1)$$

C
$$( -1, 1, -1)$$

D
$$(1,1,1)$$

Solution

Let $$\alpha $$ be the angle inclined by $$OA$$ with $$OX$$ , $$\beta$$ be the angle inclined by $$OA$$ with $$OY$$ and $$\gamma$$ be the angle inclined by $$OA$$ with $$OZ$$
We have $$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =1$$
Given $$\alpha=\beta=\gamma$$
So, we get $$\cos\alpha=\cos\beta=\cos\gamma=\cfrac{1}{\sqrt3}$$
Therefore, $$A$$ is $$(OA\cos\alpha,OA\cos\beta,OA\cos\gamma) = (1,1,1)$$
Question 7
1 / -0

If the d.rs of two lines are given by the equations $$l+m+n=0$$ and $$2lm-mn+2nl=0$$, then the angle between the two lines is

A
$$120^{o}$$

B
$$45^{o}$$

C
$$90^{\mathrm{o}}$$

D
$$30^{\mathrm{o}}$$

Solution

Given, $$l+m+n=0$$ and $$2lm-mn+2nl=0$$
$$2lm+2nl-mn=0$$
$$2l(m+n)-mn=0$$
$$-2(m+n)(m+n)-mn=0$$
$$2(m+n)^{2}+mn=0$$
$$2m^2+5mn+2n^2=0$$
Hence by applying quadratic formula, we get
$$m=\dfrac{-5n\pm3n}{4}$$

$$m=\dfrac{-n}{2}$$ and $$m={-2n}$$
Therefore, the corresponding values of l are $$n$$ and $$\dfrac{-n}{2}$$
Hence, the drs of one line is (-2n,n,n)
And d.r.s of other is $$(\dfrac{-n}{2},n,\dfrac{-n}{2})$$

Hence, the drs will be in the ration
$$(-2,1,1)$$ and $$(-1,2,-1)$$
Taking dot product, we get
$$(-2)(-1)+(1)(2)+(1)(-1)=\sqrt{6}\sqrt{6}cos\theta$$
$$3=6\cos\theta$$
$$\theta=60^0$$
Hence, angle between the lines will be $$60^0$$ and $$180^0-60^0$$
$$=120^0$$
Question 8
1 / -0

lf the $$\mathrm{D.R}$$s of two lines are given by the equations $$l+m+n=0$$ and $$l^{2}+m^{2}-n^{2}=0$$, then the angle between the two lines is:

A
$$60^{o}$$

B
$$30^{o}$$

C
$$45^{o}$$

D
$$90^{o}$$

Solution

Given that $$l+m=-n$$
Substituting $$n$$ in $$l^2+m^2-n^2=0$$, we get
$$l^2+m^2-(l+m)^2=0$$
$$\Rightarrow lm=0$$
$$\Rightarrow l=0$$ or $$m=0$$
Substituting in $$l+m=-n$$, we get
$$m=-n$$ or $$l=-n$$
Now, $$(l,m,n)=(0,-n,n)$$ or $$(-n,0,n)$$
Therefore, d.r.s are $$(0,-1,1)$$ or $$(-1,0,1)$$
Now, angle between two lines $$\cos { \theta } =\dfrac { \left( 0,-1,1 \right) .\left( -1,0.1 \right) }{ \sqrt { { \left( -1 \right) }^{ 2 }+{ 1 }^{ 2 } } \sqrt { { \left( -1 \right) }^{ 2 }+{ 1 }^{ 2 } } } =\dfrac { 1 }{ 2 } $$
$$\Rightarrow \theta= \cos ^{-1}\left (\dfrac {1}{2}\right)$$
$$\Rightarrow \theta =\dfrac { \pi }{ 3 } $$
Ans: A
Question 9
1 / -0

I. If the d.cs of two non-parallel lines satisfy $$1+m+n=0$$ and $$l^{2}+m^{2}-n^{2}={0}$$, then the angle between the line is $$ \dfrac{\pi }{3}$$
II. If the d.rs of two non-parallel lines are $$(0,\lambda, -\lambda)$$ and $$(\mu, 0,-\mu)$$, then angle between the lines is $$\displaystyle \dfrac{\pi}{3} (\lambda>0, \mu>0)$$

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true and R is not the correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

Given that $$l+m=-n$$
Substituting $$n$$ in $$l^2+m^2-n^2=0$$
We get $$l^2+m^2-(l+m)^2=0$$
$$\Rightarrow lm=0$$
$$\Rightarrow l=0$$ or $$m=0$$
Substituting in $$l+m=-n$$
We get $$m=-n$$ or $$l=-n$$
Now, $$(l,m,n)=(0,-n,n)$$ or $$(-n,0,n)$$
Therefore, d.r.s are $$(0,-1,1)$$ or $$(-1,0,1)$$
Now, angle between two lines $$\cos { \theta } =\dfrac { \left( 0,-1,1 \right) .\left( -1,0.1 \right) }{ \sqrt { { \left( -1 \right) }^{ 2 }+{ 1 }^{ 2 } } \sqrt { { \left( -1 \right) }^{ 2 }+{ 1 }^{ 2 } } } =\dfrac { 1 }{ 2 } $$
$$\Rightarrow \theta =\dfrac { \pi }{ 3 } $$
Both A and R are correct but R is not the correct explanation for A.

Ans: B
Question 10
1 / -0

Assertion $$({A})$$ . The direction ratios of the line joining origin and point $$(x,y, z)$$ must be $$x, y, {z}$$

Reason (R): lf $$P(x, y, z)$$ is a point in space and $$|{OP}|={r},$$ then the direction cosines of $${O}{P}$$ are $$\displaystyle \dfrac{x}{r}$$ , $$\displaystyle \dfrac{y}{r}$$ , $$\displaystyle \dfrac{z}{r}$$

A
Both A and R are individually true and R is the correct explanation of A

B
Both A and R individually true but R is not the correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

Assertion: Let $$P(x,y,z)$$
$$\overrightarrow { OP } =x\hat { i } +y\hat { j } +z\hat { k } $$
DR of $$\overrightarrow { OP } $$ is any vector parallel to $$x\hat { i } +y\hat { j } +z\hat { k } $$
So, given assertion is false.
Reason:
$$\overrightarrow { OP } =\left( x\hat { i } +y\hat { j } +z\hat { k } \right) \lambda $$
$$\overrightarrow { OP } =\lambda \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =r$$
$$\alpha ,\beta ,\gamma $$ be angles made by $$\overrightarrow { OP } $$ with $$x,y,z$$ axes respectively
$$\cos { \alpha } =\dfrac { \overrightarrow { OP } .\hat { i } }{ \left| \overrightarrow { OP } \right| \left| \hat { i } \right| } =\dfrac { x }{ r } $$
$$\cos { \beta } =\dfrac { \overrightarrow { OP } .\hat { j } }{ \left| \overrightarrow { OP } \right| \left| \hat { j } \right| } =\dfrac { y }{ r } $$
$$\cos { \gamma } =\dfrac { \overrightarrow { OP } .\hat { k } }{ \left| \overrightarrow { OP } \right| \left| \hat { k } \right| } =\dfrac { z }{ r } $$
DC of $$\overrightarrow { OP } $$ are $$\dfrac { x }{ r } ,\dfrac { y }{ r } ,\dfrac { z }{ r } $$
Reason is true.

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 26

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Three Dimensional Geometry Test - 26

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SHARING IS CARING

Question 1

$$11$$

$$1$$

$$\displaystyle \frac{11}{49}$$

$$\displaystyle \frac{11}{7}$$

Solution

Question 2

$$4$$

$$\displaystyle \dfrac{4}{9}$$

$$\displaystyle \dfrac{27}{4}$$

$$\displaystyle \dfrac{4}{27}$$

Solution

Question 3

$$\sqrt{12}$$

$$\sqrt{13}$$

$$12$$

$$144$$

Solution

Question 4

$$(-1,-14,7)$$

$$(1,-14,7)$$

$$(-1,14,7)$$

$$(1,14,7)$$

Solution

Question 5

$$60^{o}$$

$$30^{o}$$

$$45^{o}$$

$$90^{o}$$

Solution

Question 6

$$(3, 3, 3)$$

$$( 1, -1, -1)$$

$$( -1, 1, -1)$$

$$(1,1,1)$$

Solution

Question 7

$$120^{o}$$

$$45^{o}$$

$$90^{\mathrm{o}}$$

$$30^{\mathrm{o}}$$

Solution

Question 8

$$60^{o}$$

$$30^{o}$$

$$45^{o}$$

$$90^{o}$$

Solution

Question 9

Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not the correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 10

Both A and R are individually true and R is the correct explanation of A

Both A and R individually true but R is not the correct explanation of A

A is true but R is false

A is false but R is true

Solution

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