Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 49

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Question 1
1 / -0

If $$\left(\dfrac {1}{2},\dfrac {1}{3},n\right)$$ are the direction cosines of a line then the value of $$n$$ is

A
$$\dfrac {\sqrt {23}}{6}$$

B
$$\dfrac {23}{6}$$

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

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

Solution

Given direction cosines are $$\cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 3 } ,n$$
we have that if $$l,m,n$$ are direction cosines of a line then $${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
$$\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 9 } +{ n }^{ 2 }=1\\ \Rightarrow { n }^{ 2 }=1-(\cfrac { 13 }{ 36 } )=\cfrac { 23 }{ 36 } \\ n=\cfrac { \sqrt { 23 } }{ 6 } $$
Question 2
1 / -0

The angle between the pair of lines with direction ratios (1, 1, 2) and $$(\sqrt{3} - 1, -\sqrt{3} - 1, 4)$$ is

A
$$30^o$$

B
$$45^o$$

C
$$60^o$$

D
$$90^o$$

Solution

$$cos \theta = \dfrac{1(\sqrt{3} - 1) + 1 (-\sqrt{3} - 1) + 2(4)}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2}}$$
$$= \dfrac{\sqrt{3} - 1 - \sqrt{3} - 1 + 8}{\sqrt{6}. \sqrt{4 - 2 \sqrt{3} + 4 + 2\sqrt{3} + 16}}$$
$$= \dfrac{6}{\sqrt{6} . \sqrt{24}}$$
$$= \dfrac{6}{12}$$
$$= \dfrac{1}{2}$$
$$\theta = 60^o$$
Question 3
1 / -0

A line makes angles $$\alpha,\beta,\gamma,\delta$$ with the four diagonals of a cube then $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta$$ is equal to

A
$$1$$

B
$$4/3$$

C
$$3/4$$

D
$$4/5$$

Solution

REF.Image
Take 'O' as a corner
$$OA,OB,OC$$ are 3 edges through the axes
Let $$OA=OB=OC=a$$
coordinates of $$O=(o,o,o)$$
$$A(a,o,o) B(o,a,o) C(o,o,a)$$
$$P(a,a,o) L(o,a,a) M(a,o,a) N(a,a,o)$$
The four diagonals $$OP, AL, BM, CN$$
Direction cosine of $$OP : a-o,a-o,a-o = a,a,a = 1,1,1$$
Direction cosine of $$AL : o-a,a-o,a-o = -a,a,a = -1,1,1$$
Direction cosine of $$BM : a-o,o-a,a-o = a,-a,a = 1,-1,1$$
Direction cosine of $$CN : a-o,a-o,o-a = a,a,-a = 1,1,-1$$
$$\therefore $$ DC's of OP are $$\displaystyle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$
DC's of AL are $$\displaystyle\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$
DC's of BM are $$\displaystyle\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$
DC's of CN are $$\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}}$$
Let l,m,n be dc's of line and line makes angle $$\alpha $$
with OP :- $$\displaystyle \cos \alpha = l(\frac{1}{\sqrt{3}})+m(\frac{1}{\sqrt{3}})+n(\frac{1}{\sqrt{3}})=\frac{l+m+n}{\sqrt{3}}$$
Similarly $$\cos\, \beta =\dfrac{-l+m+n}{\sqrt{3}}$$
$$\cos \delta = \dfrac{l+m-n}{\sqrt{3}}$$
$$\cos\gamma =\dfrac{l-m+n}{\sqrt{3}}$$
suaring and adding all the four
i.e ; $$\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta $$
$$= \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}]$$
$$=\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2})$$
$$[\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3}$$
$$\therefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}$$
Question 4
1 / -0

The direction ratios of the line, given by the planes x - y + z - 5 = 0, x - 3y - 6 = 0 are

A
(3, 1, -2)

B
(2, -4, 1)

C
(1,-1, 1)

D
(0,2,1)

Solution

Given,

$$x – y + z – 5 = 0 = x – 3y – 6$$

$$\Rightarrow x – y + z – 5 = 0$$

$$x – 3y – 6 = 0$$

$$\Rightarrow x – y + z – 5 = 0 $$ (1)

$$x = 3y + 6$$ (2)

From (1) and (2) we get,

$$3y + 6 – y + z – 5 = 0$$

$$2y + z + 1 = 0$$

$$y =\dfrac{-z-1}{2}$$

Also, $$y=\dfrac{x-6}{3}$$ From (2)

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

So, the given equation can be re-written as

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

Hence the direction ratios of the given line are proportional to $$3,1,-2$$
Question 5
1 / -0

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

A
$$4i+j+3k$$

B
$$4i+5j+3k$$

C
4i-5j-3k

D
none

Solution

$$\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}$$

$$\cos \alpha = \dfrac{a}{\sqrt{a^2+ b^2+c^2}}$$ $$\cos \beta = \dfrac{b}{\sqrt{a^2 + b^2 + c^2}}$$ $$\cos \gamma = \dfrac{c}{\sqrt{a^2 + b^2 + c^2}}$$

$$\sqrt{4^2 + 5^2 + 3^2} = \sqrt{16 +25+ 9} = \sqrt{50} = 5\sqrt{2}$$

$$\vec{A} = 4\hat{i} + 5\hat{j} + 3\hat{k}$$
Question 6
1 / -0

If $$\overline { O A } = 3 \overline { i } + \overline { j } - \overline { k }$$, $$| \overline { A B } | = 2 \sqrt { 6 }$$ and AB has the direction ratios 1, -1 , 2 then $$| O B | =$$

A
$$\sqrt { 35 }$$

B
$$\sqrt { 41 }$$

C
$$\sqrt { 26 }$$

D
$$\sqrt { 55 }$$

Solution

$$\vec{OA}=3\hat{i}+\hat{j}-\hat{k}$$
$$|\vec{AB}|=2\sqrt{6}$$ has dr's $$(1, -1, 2)$$
Unit vector with above dr's are
$$\vec{AB}=\left(\dfrac{1}{\sqrt{6}}\hat{i}-\dfrac{1}{\sqrt{6}}\hat{j}+\dfrac{2}{\sqrt{6}}\hat{k}\right)(2\sqrt{6})$$
$$\vec{AB}=2\hat{i}-2\hat{j}+4\hat{k}$$

$$\vec{AB}=\vec{OB}-\vec{OA}$$
$$\vec{OB}=(2\hat{i}-2\hat{j}+4\hat{k})+(3\hat{i}+\hat{j}-\hat{k})$$
$$\vec{OB}=(5\hat{i}-\hat{j}+3\hat{k})$$
$$|OB|=\sqrt{25+1+9}=\sqrt{35}$$.
Question 7
1 / -0

The vector $$a = \alpha 1 + 2 j + \beta k$$ lies in the plane of the vectors $$b = i + jt$$ and $$c = j + k$$ and bisects the angle between $$b$$ and $$c$$. Then which one of the following gives possible values $$\alpha$$ and $$\beta$$.

A
$$\alpha = 1 , \beta = 2$$

B
$$\alpha = 2 , \beta = 1$$

C
$$\alpha = 1 , \beta = 1$$

D
$$\alpha = 2 , \beta = 2$$
Question 8
1 / -0

Let $$l_{1},\ m_{1},\ n_{1};\ l_{2},\ m_{2},\ n_{2};\ l_{3},\ m_{3},\ n_{3}$$ be the direction cosines of three mutually perpendicular line then $$\begin{vmatrix} { l }_{ 1 } & m_{ 1 } & n_{ 1 } \\ { l }_{ 2 } & m_{ 2 } & n_{ 2 } \\ { l }_{ 3 } & m_{ 3 } & n_{ 3 } \end{vmatrix}$$

A
$$0$$

B
$$\pm 1$$

C
$$\pm 2$$

D
$$\pm \dfrac{1}{2}$$

Solution
Question 9
1 / -0

The point where $$\vec{ x }$$ which is perpendicular to $$(2,-3,1)$$ and $$(1,-2,3)$$ and which satisfies the condition $$\vec { x } \cdot ( \hat { i } + 2 \hat { j } - 7 \hat{ k } ) = 10$$

A
$$\left(3,5,1\right)$$

B
$$\left(7,-5,1\right)$$

C
$$\left(3,-5,1\right)$$

D
$$\left(7,5,1\right)$$

Solution

$$\bar{x} = (a , b, c)$$
$$\bar{x} \bot (2, -3, 1) \, \& \, x \bot (1, -2 , 3)$$
$$\Rightarrow x .(2, -3, 1) = 0$$ & $$x. (1, -2, 3) = 0$$
$$\Rightarrow 2a - 3b + c = 0$$__(1) & $$ a - 2b + 3c = 0$$__(2)
$$\bar{x} . (\hat{i} + 2 \hat{j} - 7 \hat{k} ) = 10$$
$$\Rightarrow a + 2b - 7c = 10$$ __(3)
$$\left.\begin{matrix}(1) \Rightarrow c = 3b - 2a\\(2) \Rightarrow a = 2b - 3c \end{matrix}\right\}$$ Put in (3)
$$a + 2b - 7c = 10$$
$$2b - 3c + 2b - 7c = 10$$
$$\Rightarrow 4b - 10 c = 10$$ __(4)
(1) $$\Rightarrow a = \dfrac{-c + 3b}{2} $$
$$ a = \dfrac{3b - c}{2} = 2b - 3c$$
$$\Rightarrow 3b - c = 4b - 6c$$
$$\Rightarrow b = 5c$$ __(5)
from (4) & (5)
$$c = 1 , b = 5, a = 7$$
$$\Rightarrow \bar{x} = (7, 5 , 1)$$
Question 10
1 / -0

The equation of the plane through $$\left(0,-5,1\right)$$ which is perpendicular to the planes $$2x+4y+2z+3=0$$,$$2x+5y+3z+4=0$$ is

A
$$x+y+z=6$$

B
$$x-y+z=6$$

C
$$x-y-z=6$$

D
$$x+y+z+6=0$$

Solution

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Three Dimensional Geometry Test - 49

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Three Dimensional Geometry Test - 49

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SHARING IS CARING

Question 1

$$\dfrac {\sqrt {23}}{6}$$

$$\dfrac {23}{6}$$

$$\dfrac {2}{3}$$

$$\dfrac {3}{2}$$

Solution

Question 2

$$30^o$$

$$45^o$$

$$60^o$$

$$90^o$$

Solution

Question 3

$$1$$

$$4/3$$

$$3/4$$

$$4/5$$

Solution

Question 4

(3, 1, -2)

(2, -4, 1)

(1,-1, 1)

(0,2,1)

Solution

Question 5

$$4i+j+3k$$

$$4i+5j+3k$$

4i-5j-3k

none

Solution

Question 6

$$\sqrt { 35 }$$

$$\sqrt { 41 }$$

$$\sqrt { 26 }$$

$$\sqrt { 55 }$$

Solution

Question 7

$$\alpha = 1 , \beta = 2$$

$$\alpha = 2 , \beta = 1$$

$$\alpha = 1 , \beta = 1$$

$$\alpha = 2 , \beta = 2$$

Question 8

$$0$$

$$\pm 1$$

$$\pm 2$$

$$\pm \dfrac{1}{2}$$

Solution

Question 9

$$\left(3,5,1\right)$$

$$\left(7,-5,1\right)$$

$$\left(3,-5,1\right)$$

$$\left(7,5,1\right)$$

Solution

Question 10

$$x+y+z=6$$

$$x-y+z=6$$

$$x-y-z=6$$

$$x+y+z+6=0$$

Solution

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