Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 57

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Weekly Quiz Competition

Question 1
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$$O$$ is the origin and $$A$$ is the point $$\displaystyle \left ( a,b,c \right ).$$ Find the direction cosines of the join of $$OA$$ and deduce the equation of the plane through $$A$$ at right angles to $$OA$$.

A
$$\displaystyle ax+by+cz=a^{2}-b^{2}-c^{2}$$

B
$$\displaystyle ax+by+cz=a^{2}+b^{2}+c^{2}$$

C
$$\displaystyle ax-by-cz=a^{2}+b^{2}+c^{2}$$

D
$$\displaystyle ax-by-cz=a^{2}-b^{2}-c^{2}$$

Solution

D.R.'s of $$OA$$ are $$\displaystyle a-0, b-0, c-0$$ or a,b,c.
$$\displaystyle \therefore $$ D.C.'s of $$OA$$ are
$$\displaystyle

\frac{a}{\sqrt{\left ( a^{2}+b^{2}+c^{2} \right

)}},\frac{b}{\sqrt{\left ( \sum a^{2} \right )}},\frac{c}{\sqrt{\sum

a^{2}}}$$
Equation of any plane through $$A$$, i.e. $$\displaystyle \left ( a,b,c \right )$$ is
$$\displaystyle A\left ( x-a \right )+B\left ( y-b \right )+C\left ( z-c \right )=0.$$
But plane is at right angles to $$OA$$ which therefore is normal and whose D.R.'s are $$a, b, c$$.
Hence, the plane is
$$\displaystyle a\left ( x-a \right )+b\left ( y-b \right )+c\left ( z-c \right )=0$$
$$\displaystyle ax+by+cz=a^{2}+b^{2}+c^{2}$$
Question 2
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The direction cosines of the lines bisecting the internal angle $$\theta$$ between the lines whose direction cosines are $$l_{1},m_{1},n_{1}$$ and $$l_{2},m_{2},n_{2}$$ are

A
$$< l_{1}+l_{2},m_{1} +m_{2},n_{1}+n_{2}> $$

B
$$\displaystyle < \frac{l_{1}+l_{2}}{2\sin \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\sin \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\sin \frac{\theta}{2}}>$$

C
$$\displaystyle < \frac{l_{1}+l_{2}}{2\cos \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\cos \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\cos \frac{\theta}{2}}>$$

D
none of these

Solution

The direction cosines of the lines bisecting the internal angle $$\theta$$ between the lines
direction cosines are $$l_{1},m_{1},n_{1}$$ and $$l_{2},m_{2},n_{2}$$ are
$$\displaystyle < \frac{l_{1}+l_{2}}{2\cos \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\cos \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\cos \frac{\theta}{2}}>$$
Option C is correct answer
Question 3
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The angle between the line $$2x=3y=-z$$ and $$6x=-y=-4z$$ is

A
$$90^{0}$$

B
$$0^{0}$$

C
$$30^{0}$$

D
$$45^{0}$$

Solution

From given line
$$\displaystyle \frac{x}{3}=\frac{y}{2}=\frac{z}{-6} \:$$ and $$\: \dfrac{x}{2}=\dfrac{y}{-12}=\dfrac{z}{-3}$$

$$\Rightarrow \displaystyle \cos \theta =\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$$

$$\Rightarrow \displaystyle \cos \theta =\frac{6-24+18}{\sqrt{3^{2}+2^{2}+\left ( -6 \right )^{2}}\sqrt{2^{2}+\left ( -12 \right )^{2}+\left ( -3 \right )^{2}}}=0$$

$$\Rightarrow \theta =90^{0}$$
Question 4
1 / -0

The projection of a directed line segment on the co-ordinate axes are $$12, 4, 3$$, the DC's of the line are

A
$$\displaystyle \frac {-12}{13}, \frac {-4}{13}, \frac {-3}{13}$$

B
$$\displaystyle \frac {12}{13}, \frac {4}{13}, \frac {3}{13}$$

C
$$\displaystyle \frac {12}{13}, \frac {-4}{13}, \frac {3}{13}$$

D
$$\displaystyle \frac {12}{13}, \frac {4}{13}, \frac {-3}{13}$$

Solution

Let the line segment be represented in vector form as $$\vec a = a_1\vec i+a_2\vec j+a_3\vec k$$
Vector along coordinate axes are $$\vec i,\vec j,\vec k$$
Given that projection of $$\vec a$$ on x axis is 12, that with y axis is 4 and that with z axis is 3.
$$\Longrightarrow \vec a . \vec i = 12, \therefore \ a_1=12$$
similarily, $$a_2=4,a_3=3$$
$$\therefore \vec a=12\vec i +4\vec j+3\vec k$$
the length of the line segment=$$ |\vec a| =\sqrt{144+16+9}=13$$
$$\therefore |PQ|=13$$
Thus the direction cosines of PQ are $$ \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$$
Question 5
1 / -0

The projections of a line segment on $$x, y, z$$ axes are $$12, 4, 3$$. The length and the direction cosines of the line segments are

A
$$\displaystyle 13,< 12/13, 4/13, 3/13> $$

B
$$\displaystyle 19,< 12/19, 4/19, 19> $$

C
$$\displaystyle 11,< 12/11, 4/11, 3/11> $$

D
None of these

Solution

Let the line segment be represented in vector form as $$ \vec a =a_1\vec i+a_2\vec j+a_3\vec k$$
Vector along coordinate axes are $$ \vec i,\vec j,\vec k $$
Vector along Given that projection of $$\vec a $$ on x axis is 12, that with y axis is 4 and that with z axis is 3.
$$\Longrightarrow \vec a . \vec i =12 \Longrightarrow a_1=12.$$
Similarily $$a_2=4 ,a_3=3$$
$$\therefore \vec a =12 \vec i+4 \vec j +3 \vec k$$
The length of the line segment $$= | \vec a |= \sqrt{144+16+9}=13$$
and the direction cosines of the line segments are
$$<\dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}>$$
Question 6
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Equation of the plane through three points $$\displaystyle A,B,C$$ with position vectors $$\displaystyle -6\vec i+3\vec j+2\vec k, 3\vec i-2\vec j+4\vec k, 5\vec i+7\vec j+3\vec k$$ is

A
$$\displaystyle \vec{r}\cdot \left ( \vec i-\vec j-7\vec k \right )+23=0$$

B
$$\displaystyle \vec{r}\cdot \left (\vec i+\vec j+7\vec k \right )=23$$

C
$$\displaystyle \vec{r}\cdot \left (\vec i+\vec j-7\vec k \right )+23=0$$

D
$$\displaystyle\vec{r}\cdot \left (\vec i-\vec j-7\vec k \right )=23$$

Solution

We have $$\vec{A}=-6\vec i+3\vec j+2\vec k, \vec{B} =3\vec i-2\vec j+4\vec k, \vec{C} =5\vec i+7\vec j+3\vec k$$
Thus $$\vec{AB}=9\vec i-5\vec j+2\vec k, \vec{AC}=11\vec i+4\vec j+\vec k$$
Normal vector perpendicular to plane $$ABC$$ is $$\vec{n} =\vec{AB}\times \vec{AC}$$
$$\Rightarrow \vec{n} = \begin{vmatrix} \vec i&\vec j&\vec k\\9&-5&2\\11&4&1\end{vmatrix}=-13\vec i+13\vec j+91\vec k=13(-\vec i+\vec j+7\vec k)$$
Thus equation of plane $$ABC$$ is $$(\vec{r}-\vec{A})\cdot \vec{n}=0$$
$$\Rightarrow (x+6)(-1)+(y-3)(1)+(z-2)(7)=0$$
$$\Rightarrow x+6-y+3-7z+14=0$$
$$\Rightarrow (x-y-7z)+23=0$$
$$\Rightarrow \vec{r}\cdot (\vec i-\vec j-7\vec k)+23=0$$
Hence, option 'A' is correct choice.
Question 7
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If the position vectors of the points $$A$$, $$B$$, and $$C$$ be $$i + j $$ , $$i - j$$ and $$ai + bj+ ck$$ respective;y , then the points $$A$$, $$B$$ and $$C$$ are collinear if:

A
$$a = b = c = 1$$

B
$$a= 1$$ , $$b$$ and $$c$$ are arbitrary scalars

C
$$a =b= c= 0$$

D
$$c = 0$$ , $$a =1$$ and $$b$$ is a arbitrary scalar.

Solution

Given points are $$A(i+j)$$ , $$B(i-j)$$ and $$C(ai+bj+ck)$$

These three points are collinear if drs of lines $$AB$$ and $$CB$$ are proportional

We have $$AB=2j$$ and $$CB = (a-1)i+(b+1)j+ck$$

Since their drs are proportional, we get $$\dfrac{a-1}{0}=\dfrac{b+1}{2}=\dfrac{c}{0}$$

$$\Rightarrow c=0 , a=1$$ and $$b$$ can be a arbitrary scalar
Therefore option $$D$$ is correct
Question 8
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The angle between the lines whose direction cosines are given by the equations $${l}^{2}+{m}^{2}-{n}^{2}=0,l+m+n=0$$ is

A
$$\displaystyle\frac{\pi}{6}$$

B
$$\displaystyle\frac{\pi}{4}$$

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

D
$$\displaystyle\frac{\pi}{2}$$

Solution

We also have $${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
So that $$\displaystyle { l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0\Rightarrow 2{ n }^{ 2 }=1\Rightarrow n=\pm \frac { 1 }{ \sqrt { 2 } } $$
and $$\displaystyle l+m+n=0\Rightarrow { \left( l+m \right) }^{ 2 }={ n }^{ 2 }=\frac { 1 }{ 2 } ={ l }^{ 2 }+{ m }^{ 2 }$$
$$\Rightarrow 2lm=0$$
Either $$l=0$$ or $$m=0$$, if $$\displaystyle l=0,m+n=0$$
$$\Rightarrow m=-n=\pm \dfrac { 1 }{ \sqrt { 2 } } $$
So direction ratios of one of the lines are
$$\displaystyle 0,\pm \frac { 1 }{ \sqrt { 2 } } ,\mp \frac { 1 }{ \sqrt { 2 } } $$
and if $$\displaystyle m=0,l+n=0\Rightarrow l=-n=\pm \frac { 1 }{ \sqrt { 2 } } $$
So the direction ratios of the other line are $$\displaystyle \pm \frac { 1 }{ \sqrt { 2 } } ,0,\mp \frac { 1 }{ \sqrt { 2 } } $$
Thus the required angle is
$$\displaystyle \cos ^{ -1 }{ \left[ 0\times \left( \pm \frac { 1 }{ \sqrt { 2 } } \right) +\left( \pm \frac { 1 }{ \sqrt { 2 } } \right) \left( 0 \right) +\left( \pm \frac { 1 }{ \sqrt { 2 } } \right) \left( \pm \frac { 1 }{ \sqrt { 2 } } \right) \right] } $$
$$\displaystyle =\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } =\frac { \pi }{ 3 } $$
Question 9
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Find the unit vectors perpendicular to the following pair of vectors:
$$2i+j+k$$, $$i-2j+k$$

A
$$\displaystyle \frac{1}{\sqrt{35}}(3i-j-5k)$$

B
$$\displaystyle \frac{1}{\sqrt{35}}(3i+j-5k)$$

C
$$\displaystyle \frac{1}{\sqrt{27}}(i-j-5k)$$

D
$$\displaystyle \frac{1}{\sqrt{27}}(i-j+5k)$$

Solution

$$\vec { a } = 2\hat { i } + \hat { j } + \hat { k }$$

$$ \vec { b } =\hat { i } -\hat { 2j } +\hat { k }$$

Vector perpendicular to Given vectors $$\vec { a } ,\vec { b }$$

$$\Rightarrow\vec { a } \times \vec { b } = \left| \begin{matrix} \hat i & \hat j & \hat k \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{matrix} \right|$$

$$\Rightarrow\vec { a } \times \vec { b } = \hat i(1-(-2))-\hat j(2-1)+\hat k(-4-1)$$

$$\Rightarrow \vec { a } \times \vec { b } = 3\hat { i } -\hat { j } -\hat { 5k }$$

Unit vector required is $$ \dfrac { \hat { 3i } -\hat { j } -5\hat { k } }{ \left| 3\hat { i } -\hat { j } -5\hat { k } \right| } $$$$=\dfrac { 1 }{ \sqrt { 35 } } (3\hat { i } -\hat { j } -\hat { 5k } )$$
Question 10
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If direction cosines of two lines are proportional to $$(2,3,-6)$$ and $$(3,-4,5)$$, then the acute angle between them is

A
$$\cos ^{ -1 }{ \left( \cfrac { 49 }{ 36 } \right) } $$

B
$$\cos ^{ -1 }{ \left( \cfrac { 18\sqrt { 2 } }{ 35 } \right) } $$

C
$${96}^{o}$$

D
$$\cos ^{ -1 }{ \left( \cfrac { 18 }{ 35 } \right) } $$

Solution

We know that
$$\cos { \theta } =\cfrac { \left| { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 } \right| }{ \sqrt { { a }_{ 1 }^{ 2 }+{ b }_{ 1 }^{ 2 }+{ c }_{ 1 }^{ 2 } } \sqrt { { a }_{ 2 }^{ 2 }+{ b }_{ 2 }^{ 2 }+{ c }_{ 2 }^{ 2 } } } $$
$$\therefore \cos { \theta } =\cfrac { \left| (2)(3)+(3)(-4)+(-6)(-5) \right| }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ \left( -6 \right) }^{ 2 } } \sqrt { { 3 }^{ 2 }+{ \left( -4 \right) }^{ 2 }+{ 5 }^{ 2 } } } $$
$$=\cfrac { \left| 6-12-30 \right| }{ \sqrt { 4+9+36 } \sqrt { 9+16+25 } } $$
$$\Rightarrow \cos { \theta } =\cfrac { 36 }{ 7.5\sqrt { 2 } } =\cfrac { 18\sqrt { 2 } }{ 35 } $$
$$\Rightarrow \theta =\cos ^{ -1 }{ \left( \cfrac { 18\sqrt { 2 } }{ 35 } \right) } $$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 57

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Three Dimensional Geometry Test - 57

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SHARING IS CARING

Question 1

$$\displaystyle ax+by+cz=a^{2}-b^{2}-c^{2}$$

$$\displaystyle ax+by+cz=a^{2}+b^{2}+c^{2}$$

$$\displaystyle ax-by-cz=a^{2}+b^{2}+c^{2}$$

$$\displaystyle ax-by-cz=a^{2}-b^{2}-c^{2}$$

Solution

Question 2

$$< l_{1}+l_{2},m_{1} +m_{2},n_{1}+n_{2}> $$

$$\displaystyle < \frac{l_{1}+l_{2}}{2\sin \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\sin \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\sin \frac{\theta}{2}}>$$

$$\displaystyle < \frac{l_{1}+l_{2}}{2\cos \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\cos \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\cos \frac{\theta}{2}}>$$

none of these

Solution

Question 3

$$90^{0}$$

$$0^{0}$$

$$30^{0}$$

$$45^{0}$$

Solution

Question 4

$$\displaystyle \frac {-12}{13}, \frac {-4}{13}, \frac {-3}{13}$$

$$\displaystyle \frac {12}{13}, \frac {4}{13}, \frac {3}{13}$$

$$\displaystyle \frac {12}{13}, \frac {-4}{13}, \frac {3}{13}$$

$$\displaystyle \frac {12}{13}, \frac {4}{13}, \frac {-3}{13}$$

Solution

Question 5

$$\displaystyle 13,< 12/13, 4/13, 3/13> $$

$$\displaystyle 19,< 12/19, 4/19, 19> $$

$$\displaystyle 11,< 12/11, 4/11, 3/11> $$

None of these

Solution

Question 6

$$\displaystyle \vec{r}\cdot \left ( \vec i-\vec j-7\vec k \right )+23=0$$

$$\displaystyle \vec{r}\cdot \left (\vec i+\vec j+7\vec k \right )=23$$

$$\displaystyle \vec{r}\cdot \left (\vec i+\vec j-7\vec k \right )+23=0$$

$$\displaystyle\vec{r}\cdot \left (\vec i-\vec j-7\vec k \right )=23$$

Solution

Question 7

$$a = b = c = 1$$

$$a= 1$$ , $$b$$ and $$c$$ are arbitrary scalars

$$a =b= c= 0$$

$$c = 0$$ , $$a =1$$ and $$b$$ is a arbitrary scalar.

Solution

Question 8

$$\displaystyle\frac{\pi}{6}$$

$$\displaystyle\frac{\pi}{4}$$

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

$$\displaystyle\frac{\pi}{2}$$

Solution

Question 9

$$\displaystyle \frac{1}{\sqrt{35}}(3i-j-5k)$$

$$\displaystyle \frac{1}{\sqrt{35}}(3i+j-5k)$$

$$\displaystyle \frac{1}{\sqrt{27}}(i-j-5k)$$

$$\displaystyle \frac{1}{\sqrt{27}}(i-j+5k)$$

Solution

Question 10

$$\cos ^{ -1 }{ \left( \cfrac { 49 }{ 36 } \right) } $$

$$\cos ^{ -1 }{ \left( \cfrac { 18\sqrt { 2 } }{ 35 } \right) } $$

$${96}^{o}$$

$$\cos ^{ -1 }{ \left( \cfrac { 18 }{ 35 } \right) } $$

Solution

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