Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 41

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Question 1
1 / -0

If $$\alpha ,\beta ,\gamma $$ are the angles which a directed line makes with the positive directions of the coordinate axes, then $$\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } $$ is equal to

A
$$1$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

The direction cosines of the line are
$${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
Now, $$\Rightarrow \cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =1$$
$$\Rightarrow 1-\sin ^{ 2 }{ \alpha } +1-\sin ^{ 2 }{ \beta } +1-\sin ^{ 2 }{ \gamma } =1$$
$$\Rightarrow \sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2$$
Question 2
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The acute angle between the lines whose direction ratios are given by $$l+m-n=0$$ and $${ l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0$$, is

A
$$0$$

B
$$\pi /6$$

C
$$\pi /4$$

D
$$\pi /3$$

Solution

We have
$$l+m-n=0$$
and $${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=0$$
$$\Rightarrow { l }^{ 2 }+{ m }^{ 2 }={ n }^{ 2 }$$ and $$l+m=n$$
$$\Rightarrow { l }^{ 2 }+{ m }^{ 2 }={ (l+m) }^{ 2 }$$
$$\Rightarrow 2lm=0$$
$$\Rightarrow l=0,m=0$$
If $$l=0$$, then $$m=n$$
$$\therefore \cfrac { l }{ 0 } =\cfrac { m }{ 1 } =\cfrac { n }{ 1 } $$
If $$m=0$$, then
$$l=n$$
$$\therefore \cfrac { l }{ 1 } =\cfrac { m }{ 0 } =\cfrac { n }{ 1 } $$
thus, the dr's of the lines are proportional to $$0,1,1$$ and $$1,0,1$$
Therefore, angle $$\theta$$ between them is given by
$$\cos { \theta } =\cfrac { 0\times 1+1\times 0+\left| 1\times 1 \right| }{ \sqrt { 0+1+1 } \sqrt { 1+0+1 } } =\cfrac { 1 }{ 2 } $$
$$\Rightarrow \theta =\cfrac { \pi }{ 3 } \quad $$
Question 3
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ABCD is a trapezium in which AB and CD are parallel sides. If $$l(AB) = 3 l (CD) $$ and $$\bar {DC} = 2 \hat {i} - 5 \hat {k}$$. Then vector $$ \bar {AB} $$ is

A
$$ \frac{3}{\sqrt{29} } (2 \hat{i} - 5 \hat {k}) $$

B
$$ \frac {\sqrt{29}} {3} (5 \hat{i} - 2 \hat {k}) $$

C
$$ -6 \hat {i} + 15 \hat {k} $$

D
A or B

Solution

Given
$$\vec{DC}=2\hat{i}-5\hat{k}$$
$$\vec{CD}=-2\hat{i}+5\hat{k}$$
AB is parallel to CD
$$\vec{AB}=\lambda\vec{CD}$$
$$\vec{AB}=\lambda(-2\hat{i}+5\hat{k})$$
$$\vec{AB}=-2\lambda\hat{i}+5\lambda\hat{k}$$

$$l (AB)=3 l (CD)$$

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

$$\sqrt{4\lambda^2+25\lambda^2}=3\sqrt{4+25}$$

$$\sqrt{29\lambda^2}=3\sqrt{29}$$

$$\sqrt{29}\lambda=3\sqrt{29}$$

on comparing both sides
$$\lambda=3$$
putting value of $$\lambda$$ in AB
$$\vec{AB}=-2\times3\hat{i}+5\times3\hat{k}$$
$$\vec{AB}=-6\hat{i}+15\hat{k}$$
Hence option C is correct
Question 4
1 / -0

If a line makes $${ 45 }^{ o },{ 60 }^{ o }$$ with positive direction of axes $$x$$ and $$y$$ then the angle it makes with the z-axis is:

A
$${ 30 }^{ o }\quad $$

B
$${ 90 }^{ o }\quad $$

C
$${ 45 }^{ o }\quad $$

D
$${ 60 }^{ o }\quad $$

Solution

If $$\alpha, \beta, \gamma $$ are the angles made by a line with the positive direction of $$x$$-axis,$$y$$-axis and $$z$$-axis then $$cos\:\alpha, cos\:\beta, cos\:\gamma$$ are called the direction cosines of the line.

Also $$cos^2 \alpha+cos^2\beta+cos^2\gamma=1$$

Given $$\alpha=45^\circ, \beta=60^\circ$$

$$\Rightarrow cos^2 45^\circ+cos^2 60^\circ+cos^2\gamma=1$$

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

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

$$\Rightarrow cos^2 \gamma=1-\dfrac{1}{2}-\dfrac{1}{4}$$

$$\Rightarrow cos^2 \gamma=1-\dfrac{3}{4}$$

$$\Rightarrow cos^2 \gamma=\dfrac{1}{4}$$

$$\Rightarrow cos \gamma=\dfrac{1}{2}$$

$$\Rightarrow \gamma=cos^{-1}\left(\dfrac{1}{2}\right)$$

$$\Rightarrow \gamma=60^\circ$$
Question 5
1 / -0

The dc's $$(l,m,n)$$ of two lines are connected between the relation $$l + m + n = 0, \ lm = 0$$, then the angle between the lines is

A
$$\displaystyle\frac{\pi}{3}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{2}$$

D
$$\dfrac{\pi}{6}$$

Solution

We know that,
$$\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}.}\overrightarrow{{{b}_{2}}}}{\overrightarrow{\left| {{b}_{1}} \right|}\left| {{\overrightarrow{b}}_{2}} \right|}$$ ...... (1)

Given that,

$$ l+m+n=0 $$

$$ l+m=-n $$

$$ -\left( l+m \right)=n $$

And, $$lm=0$$

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

When, $$l=0$$, then
$$m= -n$$

And,
$$(l, m, n)= (0, 1, -1)$$

When, $$m=0$$, then
$$l= -n$$

And,
$$(I, m, n)= (1, 0, -1)$$

Calculate $$b_1\cdot b_2$$.

$$ {{b}_{1}}.{{b}_{2}}=(0,1,-1).(1,0,-1) $$

$$ =0+0+1 $$

$$ =1 $$

Therefore,
$$ \left| {{b}_{1}} \right|=\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2} $$

$$ \left| {{b}_{2}} \right|=\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2} $$

Now, substitute the values in equation (1).

$$ \cos \theta =\dfrac{\overrightarrow{{{b}_{1}}}.\overrightarrow{{{b}_{2}}}}{\left| {{\overrightarrow{b}}_{1}} \right|\left| \overrightarrow{{{b}_{2}}} \right|} $$

$$ \Rightarrow \cos \theta =\dfrac{1}{\sqrt{2}.\sqrt{2}}=\dfrac{1}{2} $$

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

Hence, the angle between the lines is $$\dfrac{\pi}{3}$$.
Question 6
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$$A = \begin{bmatrix} l_{1}& m_{1} & n_{1}\\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3}\end{bmatrix}$$ and $$B = \begin{bmatrix}p_{1} & q_{1} & r_{1}\\ p_{2} & q_{2} & r_{2}\\ p_{3} & q_{3} & r_{3}\end{bmatrix}$$ where $$p_{1}, q_{1}, r_{1}$$ are the co-factors of the elements $$l_{i},m_{i},n_{i}$$ for $$i = 1, 2, 3$$. If $$(l_{1}, m_{1}, n_{1}), (l_{2}, m_{2}, n_{2})$$ and $$(l_{3},m_{3}, n_{3})$$ are the direction cosines of three mutually perpendicular lines then $$(p_{1}, q_{1}, r_{1}), (p_{2}, q_{2}, r_{2})$$ and $$(p_{3}, q_{3}, r_{3})$$ are

A
The direction cosines of three mutually perpendicular lines

B
The direction ratios of three mutually perpendicular lines which are not direction cosines

C
The direction cosines of three lines which need not be perpendicular

D
The direction ratios but not the direction cosines of three lines which need not be perpendicular

Solution

Given $$p_1,q_1,r_1$$ are the co-factors of the elements $$l_i,m_i,n_i $$
For $$i=1,2,3$$
$$(l_2,m_2,n_2)$$ and $$(l_3,m_3,n_3)$$ are the direction cosines of three mutually perpendicular lines
Thus $$(p_1,q_1,r_1),(p_2,q_2,r_2) ,(p_3,q_3,r_3)$$ are also the direction cosines of three mutually perpendicular lines
Option A is Correct answer
Question 7
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If a lines makes angles $$ \alpha , \beta , \gamma , \delta $$ with four digonals of a cube. Then $$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos ^2 \delta $$ will be :

A
$$4/3$$

B
$$3/4$$

C
$$1/4$$

D
None of these

Solution

$$DC's$$ of the four diagonals of a cube on putting $$ a=b=c $$ are
$$ \left( \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} \right) \left( \dfrac {-1}{\sqrt3} , \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} \right) \left( \dfrac {1}{\sqrt3} , \dfrac {-1}{\sqrt3} , \dfrac {1}{\sqrt3} \right) \left( \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} , \dfrac {-1}{\sqrt3} \right) $$
Let $$l,m,n$$ be the dc's of the line which is inclined at angles $$ \alpha , \beta , \gamma , \delta $$ resp to the four diagonals then
$$ \cos \alpha = \dfrac {1}{\sqrt3} \cdot l + m \cdot \dfrac{1}{\sqrt3} + n \cdot \dfrac{1}{\sqrt3} = \dfrac {l+m+n}{\sqrt3} $$
Similarly $$ \cos \beta = \dfrac {-l+m-n}{\sqrt3} , \cos \gamma = \dfrac {l-m+n}{\sqrt3} , \cos \delta = \dfrac {l+m-n}{\sqrt3} $$
$$ \therefore \cos^2 \alpha , \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta $$
$$ = \dfrac {4 ( l^2 + m^2 + n^2)}{3} = \dfrac {4}{3} ( \because l^2 + m^2 + n^2 = 1) $$
Question 8
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ABC is a triangle where $$A(2,3,5), B(-1,3,2)$$ and $$C(\lambda , 5, \mu)$$. Let the median through $$A$$ is equally inclined to the axes.
The value of $$\mu - \lambda$$ is equal to:

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Now, the mid point of the side $$BC$$ be $$D\left(\dfrac{\lambda-1}{2},4,\dfrac{\mu+2}{2}\right)$$.

Let $$\alpha$$, $$\beta$$ and $$\gamma$$ be the angles made by the median $$AD$$ of $$\Delta$$ABC.

Then direction cosines of the side $$AD$$ are

$$\cos \alpha=\dfrac{\dfrac{\lambda-1}{2}-2}{\vec{AD}}$$, $$\cos \beta=\dfrac{4-3}{\vec{AD}}$$, $$\cos \gamma=\dfrac{\dfrac{\mu+2}{2}-5}{\vec{AD}}$$.

Since $$\alpha=\beta=\gamma$$

$$\Rightarrow \cos \alpha=\cos \beta=\cos \gamma$$......$$\because$$ (equally inclined to the axes)

$$\Rightarrow \dfrac{\lambda-5}{2}=1=\dfrac{\mu-8}{2}$$

$$\Rightarrow \lambda=7$$ and $$\mu=10$$.

$$\therefore \mu-\lambda=3$$
Question 9
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The point of intersection of the line joining the points $$(-3, 4, -8)$$ and $$(5, -6, 4)$$ with the $$XY$$-plane is

A
$$\left (\dfrac {7}{3}, -\dfrac {8}{3}, 0\right )$$

B
$$\left (-\dfrac {7}{3}, -\dfrac {8}{3}, 0\right )$$

C
$$\left (-\dfrac {7}{3}, \dfrac {8}{3}, 0\right )$$

D
$$\left (\dfrac {7}{3}, \dfrac {8}{3}, 0\right )$$

Solution

$$\overrightarrow { r } =\overrightarrow { a } +x(\overrightarrow { b } -\overrightarrow { a } )\\ \overrightarrow { a } =-3\hat { i } +4\hat { j } +(-8)\hat { k } \\ \overrightarrow { b } =5\hat { i } +(-6)\hat { j } +4\hat { k } \\ \overrightarrow { r } =-3\hat { i } +4\hat { j } +(-8)\hat { k } +x(8\hat { i } +(-10)\hat { j } +12\hat { k } )=(8x-3)\hat { i } +(4-10x)\hat { j } +(12x-8)\hat { k }$$
On $$XY$$ plane $$\hat { k }$$ will be $$0$$
So $$12x-8=0$$ it means $$x=\dfrac { 2 }{ 3 } $$

Putting the value of $$x$$ in component of $$\hat { i }$$ and $$\hat { j }$$ we get
$$8\times \dfrac { 2 }{ 3 } -3$$ and $$4-10\times \dfrac { 2 }{ 3 }$$
$$ \dfrac { 7 }{ 3 } \ and\ \dfrac { -8 }{ 3 } $$

So correct optiopn will be option A.
Question 10
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Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to 1, -2, -2 and 0, 2, 1.

A
$$\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{1}{3}$$

B
$$\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}$$

C
$$1, -\dfrac{1}{3}, \dfrac{2}{3}$$

D
None of these

Solution

Let $$l,m,n$$ be the direction cosines of the required line. Since it is perpendicular to the lines whose direction cosines are proportional to $$1, -2, -2$$ and $$0, 2, 1$$ respectively.

Thus,

$$l-2m-2n=0$$

$$0l+2m+n=0$$

On solving these equations, we get,

$$\dfrac{l}{2}=\dfrac{m}{-1}=\dfrac{n}{2}$$

Thus, the direction ratios of the required line are proportional to $$2,-1,2$$. Hence, its direction cosines are :

$$\dfrac{2}{\sqrt{2^2+(-1)^2+2^2}}, \dfrac{-1}{\sqrt{2^2+(-1)^2+2^2}}, \dfrac{2}{\sqrt{2^2+(-1)^2+2^2}}=\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}$$.

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 41

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Three Dimensional Geometry Test - 41

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SHARING IS CARING

Question 1

$$1$$

$$4$$

$$3$$

$$2$$

Solution

Question 2

$$0$$

$$\pi /6$$

$$\pi /4$$

$$\pi /3$$

Solution

Question 3

$$ \frac{3}{\sqrt{29} } (2 \hat{i} - 5 \hat {k}) $$

$$ \frac {\sqrt{29}} {3} (5 \hat{i} - 2 \hat {k}) $$

$$ -6 \hat {i} + 15 \hat {k} $$

A or B

Solution

Question 4

$${ 30 }^{ o }\quad $$

$${ 90 }^{ o }\quad $$

$${ 45 }^{ o }\quad $$

$${ 60 }^{ o }\quad $$

Solution

Question 5

$$\displaystyle\frac{\pi}{3}$$

$$\dfrac{\pi}{4}$$

$$\dfrac{\pi}{2}$$

$$\dfrac{\pi}{6}$$

Solution

Question 6

The direction cosines of three mutually perpendicular lines

The direction ratios of three mutually perpendicular lines which are not direction cosines

The direction cosines of three lines which need not be perpendicular

The direction ratios but not the direction cosines of three lines which need not be perpendicular

Solution

Question 7

$$4/3$$

$$3/4$$

$$1/4$$

None of these

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$\left (\dfrac {7}{3}, -\dfrac {8}{3}, 0\right )$$

$$\left (-\dfrac {7}{3}, -\dfrac {8}{3}, 0\right )$$

$$\left (-\dfrac {7}{3}, \dfrac {8}{3}, 0\right )$$

$$\left (\dfrac {7}{3}, \dfrac {8}{3}, 0\right )$$

Solution

Question 10

$$\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{1}{3}$$

$$\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}$$

$$1, -\dfrac{1}{3}, \dfrac{2}{3}$$

None of these

Solution

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