Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 48

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Question 1
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The direction ratios of the joining $$A(1,\,2,\ 1)$$ and $$(2,\ 1,\ 2)$$ are

A
$$3,\ 3,\ 3$$

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

C
$$3,\ 1,\ 3$$

D
$$\dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}}$$

Solution

We have given points are:-
$$A\left( {1,2,1} \right)$$
$$B\left( {2,1,2} \right)$$
Now,
$$\overrightarrow {BA} = \left( {1 - 2} \right)\widehat i + \left( {2 - 1} \right)\widehat j + \left( {1 - 2} \right)\widehat k$$
$$ = - \widehat i + \widehat j - \widehat k$$
direction ratio are
$$a=-1$$
$$b=1$$
$$c=-1$$
Hence,
option $$B$$ is correct.
Question 2
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The direction ratios of $$AB$$ are $$- 2, 2, 1$$ . If coordinates of A are $$( 4,1,5 )$$ and $$l( A B ) = 6$$ , then coordinates of $$ B $$ ?

A
$$( 0,5 , - 7 )$$

B
$$( 8 , - 3,3 )$$

C
$$( 0,7,5 )$$

D
$$( 8,3,3 )$$

Solution
Question 3
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If $$\bar {a}, \bar {b}$$ and $$\bar {c}$$ are non-zero non collinear vectors and $$\theta(\neq 0 , \pi)$$ is the angle between $$\bar {b}$$ and $$\bar {c}$$ if $$(\bar {a}\times \bar {b}) \times \bar {c}=\dfrac {1}{2} |\bar {b}|\bar {c}|\bar {a}$$. then $$\sin \theta =$$

A
$$\sqrt{\dfrac{2}{3}}$$

B
$$\dfrac{\sqrt{3}}{2}$$

C
$$\dfrac{4\sqrt{2}}{3}$$

D
$$\dfrac{2\sqrt{2}}{3}$$

Solution

We have
$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$ - \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right] = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\overrightarrow c .\overrightarrow a = 0$$
$$\overrightarrow c .\overrightarrow a = \frac{{ - 1}}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$$
$$\cos \theta = \frac{{ - 1}}{2}$$
$$ \Rightarrow \theta = \frac{{2\pi }}{3}$$
$$\therefore \sin \theta = \frac{{\sqrt 3 }}{2}$$
Hence, $$B$$is the correct answer.
Question 4
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A line d.c's proportional to $$(2,1,2)$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$. Then the coordinates of each of the points of intersection are given by

A
$$(3a,2a,3a); (a,a,2a)$$

B
$$(3a,2a,3a); (a,a,a)$$

C
$$(3a,3a,3a); (a,a,a)$$

D
$$(2a,3a,3a); (2a,a,a)$$

Solution

$$L_1 : \dfrac{x}{1} = \dfrac{y + a}{1} = \dfrac{z}{1}$$
$$L_2 : \dfrac{x + a}{2} = \dfrac{y}{1} = \dfrac{z}{1}$$
Any point on $$L_1 \, A (k_1, k_1 - 1, k_1)$$
Any point on $$L_2 \, B (2k_2 - 1, k_2, k_2)$$
Drs of $$AB = (k_1 - 2k_2 + a , k_1 - k_2 - a , k_1 - k-2)$$
$$= \dfrac{k_1 - 2k_2 + a}{2} = \dfrac{k_1 - k_2 - a}{1} = \dfrac{k_1 - k_2}{2}$$
$$\Rightarrow k_1 - 2k_2 + a = 2k_1 - 2k_2 - 2a$$
$$\Rightarrow 3a = k_1$$
$$\therefore 2k_1 - 2k_2 - 2a = k_1 - k_2$$
$$\Rightarrow 2k_1 - k_1 - k_2 = 2a$$
$$\Rightarrow k_1 - k_2 = 2a$$
$$\Rightarrow k_2 = a$$
$$\therefore A (3a , 2a , 3a)$$
$$B (a, a, a)$$
Question 5
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If $$\frac{x-14}{l}=\frac{y-2}{m}=\frac{z+1}{n}$$ is the equation of the line through (1,2,-1) and (-1,0,1), then (l,m,n) is

A
(-1,0,1)

B
(1,1,-1)

C
(1,2,-1)

D
(0,1,0)

Solution
Question 6
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If $$A = (1,2,3) , B = (2,10,1), Q$$ are collinear points and $$Q_{x}=-1$$ then $$Q_{z}$$ is

A
$$-3$$

B
$$7$$

C
$$-14$$

D
$$-7$$

Solution
Question 7
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The direction cosines of a vector $$ \overrightarrow { A } $$ are $$ \cos \alpha = \frac {4} { 5 \sqrt {2}} , \cos\beta =\frac { 1 }{ \sqrt { 2 } } , \cos\gamma =\frac { 3 }{ 5\sqrt { 2 } } $$ then, the vector $$ \overrightarrow {A} $$ is

A
$$ 4\hat { i } +\hat { j } +3\hat { k } $$

B
$$ 4\hat { i } +5\hat { j } +3\hat { k } $$

C
$$ 4\hat { i } -5\hat { j } +3\hat { k } $$

D
$$ 4\hat { i } -\hat { j } -3\hat { k } $$

Solution
Question 8
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The direction cosines to two lines at right angles are (1,2,3) and (-2,$$\frac{1}{2}$$,$$\frac{1}{3}$$), then the direction cosine perpendicular to both given lines are:

A
$$\sqrt{\frac{25}{2198}}$$,$$\sqrt{\frac{19}{2198}}$$,$$\sqrt{\frac{729}{2198}}$$

B
$$\sqrt{\frac{24}{2198}}$$,$$\sqrt{\frac{38}{2198}}$$,$$\sqrt{\frac{730}{2198}}$$

C
$$\frac{1}{3}$$,-2,$$\frac{-7}{2}$$

D
None of the above

Solution

Direction cosines of $${ L }_{ 1 }=\left( 1,2,3 \right) $$
Direction cosines of $${ L }_{ 2 }=\left( -2,\cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 3 } \right) $$
Let direction ratios of required line be a,b,c then
$$a+2b+3c=0\rightarrow 1$$
and $$-2a+\cfrac { b }{ 2 } +\cfrac { c }{ 3 } =0$$
$$\Rightarrow -12a+3b+2c=0\rightarrow 2$$
Solving $$1$$ and $$2$$
$$a+2b+3c=0$$
$$-12a+3b+2c=0$$
$$\Rightarrow \cfrac { a }{ 4-9 } =\cfrac { b }{ -36-2 } =\cfrac { c }{ 3+24 } $$
$$\Rightarrow \cfrac { a }{ +5 } =\cfrac { b }{ +38 } =\cfrac { c }{ -27 } $$
$$\Rightarrow a=+5,b=+38,c=-27$$
$$\therefore $$Direction cosines are
$$\cfrac { 5 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( 27 \right) }^{ 2 } } } ,\cfrac { 38 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( -27 \right) }^{ 2 } } } ,\cfrac { -27 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( -27 \right) }^{ 2 } } } $$
$$\cfrac { 5 }{ \sqrt { 2198 } } ,\cfrac { 38 }{ \sqrt { 2198 } } ,\cfrac { -27 }{ \sqrt { 2198 } } $$
Question 9
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The vector equation of line passing through the point $$(-1,-1,2)$$ and parallel to the line $$2x-2=3y+1=6z-2$$

A
$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (3\hat { i } +2\hat { j } +\hat { k } )$$

B
$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +6\hat { k } )$$

C
$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (\hat { i } +2\hat { j } +3\hat { k } )$$

D
$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +\hat { k } )$$

Solution

Given
$$L_1 : 2x - 2 = 3y + 1 = 6z - 2$$
$$\Rightarrow 2 (x - 1) = 3 ( y + \dfrac{1}{3}) = 6(z - \dfrac{1}{3})$$
$$\Rightarrow \dfrac{x - 1}{3} = \dfrac{y + \dfrac{1}{3}}{2} = \dfrac{z - \dfrac{1}{3}}{1}$$
Equation of line passing through $$(-1, -1, 2)$$ parallel to $$L_1$$
$$\dfrac{x + 1}{3} = \dfrac{y + 1}{2} = \dfrac{z - 2}{1}$$
vector equation is
$$(- \hat{i} - \hat{j} + 2 \hat{k}) + \lambda ( 3 \hat{i} + 2 \hat{j} + \hat{k})$$
Question 10
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The direction ratios of the line joining the points $$(4, 3, -5)$$ and $$(-2, 1, -8)$$ are

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

B
$$6, 2, 3$$

C
$$5,8,0$$

D
$$3,7,9$$

Solution

Given points $$P(4,3,-5)$$ $$Q(-2,1,-8)$$
Direction ratios of $$PQ$$
$$=$$Position vector of $$P-$$Position vector of $$Q$$
$$=4-(-2),3-1,-5-(-8)$$
$$=6,2,3$$