Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 58

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Weekly Quiz Competition

Question 1
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If the direction cosines of two lines are given by $$l+m+n=0$$ and $$l^2-5m^2+n^2=0$$, then the angle between them is

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

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

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

D
$$\dfrac{\pi}{3}$$

Solution

$$l+m+n=0$$
$$\Rightarrow l+n=-m$$...(i)
$$l^{2}-5m^{2}+n^{2}=0$$

Substituting the equation from eq (i)

$$l^{2}+n^{2}-5(-l-n)^{2}=0$$

$$l^{2}+n^{2}-5(l^{2}+n^{2}+2ln)=0$$

$$-4l^{2}-4n^{2}-10ln=0$$

$$2l^{2}+2n^{2}+5ln=0$$

$$2l^{2}+5ln+2n^{2}=0$$

$$l=\dfrac{-5n\pm n\sqrt{25-16}}{4}$$

$$l=\dfrac{-5n\pm3n}{4}$$

$$l=-2n$$ and $$l=\dfrac{-n}{2}$$

Hence $$l=\dfrac{-n}{2}, m=\dfrac{-n}{2}$$
If $$l=-2n, m=n$$

Hence the Dr's are

$$(\dfrac{-n}{2},\dfrac{-n}{2},n)$$ and $$(-2n,n,n)$$

Therefore Dc's are
$$(\dfrac{-1}{\sqrt{6}},\dfrac{-1}{\sqrt{6}},\dfrac{2}{\sqrt{6}})$$ and $$(\dfrac{-2}{\sqrt{6}},\dfrac{1}{\sqrt{6}},\dfrac{1}{\sqrt{6}})$$

Hence $$\cos\theta=(\dfrac{-2}{\sqrt{6}}.\dfrac{-1}{\sqrt{6}})+(\dfrac{-1}{\sqrt{6}}.\dfrac{1}{\sqrt{6}})+(\dfrac{2}{\sqrt{6}}.\dfrac{1}{\sqrt{6}})$$

$$=\dfrac{2-1+2}{6}$$

$$=\dfrac{3}{6}$$

$$=\dfrac{1}{2}$$

Hence $$\theta=\dfrac{\pi}{3}$$
Question 2
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Directions For Questions

Let $$A$$ be the given point whose position vectors with reference to origin $$O$$ be $$ \overrightarrow{a}$$ and $$ \overrightarrow{ON}= \overrightarrow{n}$$ Let $$P$$ be any point such that $$\overline{OP}= \overrightarrow{r}$$ lies on the plane & passes through $$A$$ and orthogonal to $$ON$$. Then for any point $$P$$ lies on the plane then.$$\overrightarrow{AP} \perp \overrightarrow{n}$$
$$\displaystyle \therefore \overrightarrow{AP}.\ \overrightarrow{n}=0$$
$$\displaystyle \Rightarrow \left ( \overrightarrow{OP}-\overrightarrow{OA} \right ).\overrightarrow{n}=0$$
$$\displaystyle \Rightarrow \overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$$ $$($$Knows as scalar product form$$)$$
$$\displaystyle \Rightarrow \overrightarrow{r}\cdot \overrightarrow{n}=p,$$ where $$p$$ is the $$\perp$$er distance from origin to the plane.
On the bases of above information answer the following questions

...view full instructions

The, position vector of the foot of the $$\perp$$er drawn from origin to the plane is $$\displaystyle 4\hat{i}-2\hat{j}-5\hat{k} $$ then equation of the plane is

A
$$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}+5\hat{k} \right )=45$$

B
$$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=45$$

C
$$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )+45=0$$

D
$$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}-5\hat{k} \right )=37$$

Solution

Equation of the plane passes through the point $$B$$ with position vector
$$\displaystyle 4\hat{i}-2\hat{j}-5\hat{k} \ and \ \perp \ to \ \overline{OB}=\overline{n}$$
$$\displaystyle \therefore \overline{n}=\overline{OB}=4\hat{i}-2\hat{j}-5\hat{k} $$
$$\displaystyle \therefore$$ Equation of the plane is $$\displaystyle \overline{r} \cdot \overline{n}= \overline{a} \cdot \overline{n}$$
$$\displaystyle \Rightarrow \overline{r} \cdot \left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )\cdot \left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )$$
$$\displaystyle \Rightarrow \overline{r} \cdot \left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=45$$
Hence (B) is correct choice.
Question 3
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Directions For Questions

Consider the lines:

$$\displaystyle L_{1}=\dfrac{x+1}{3}=\dfrac{y+2}{1}=\dfrac{z+1}{2}$$ and $$\displaystyle L_{2}=\dfrac{x-2}{1}=\dfrac{y+2}{2}=\dfrac{z-3}{3}$$

On the basis of the above information, answer the following question:

...view full instructions

The vector equation of the line $$\displaystyle L_{1}$$ is $$\displaystyle a+\lambda \overline{b}$$ then $$\displaystyle \overline{a}$$ equals

A
$$\displaystyle -\hat{i}+2\hat{j}-\hat{k}$$

B
$$\displaystyle 2\hat{i}-\hat{j}+\hat{k}$$

C
$$\displaystyle 2\hat{i}-2\hat{j}+3\hat{k}$$

D
$$\displaystyle \hat{i}+2\hat{j}+3\hat{k}$$

Solution

The equations of the given lines in vectors form may be written as
$$ \displaystyle L_{1}:\overline{r}=\overline{a}+\lambda \overline{b} $$ where $$ \overline{a}=-\hat{i}-2\hat{j}-\hat{k},\overline{b}=3\hat{i}+\hat{j}+2\hat{k}$$
$$\displaystyle L_{2}:\overline{r}=\overline{c}+\mu \overline{d} $$
Where
$$\displaystyle \overline{c} = 2\hat{i} - 2\hat{j} + 3\hat{k}, \overline{d} = 2\hat{i} + 2\hat{j} + 3\hat{k}$$
$$\overline {n} = \overline {b} \times \overline {b} = \begin {vmatrix}i & j & k\\ 3 &1 & 2 \\1 & 2 & 3 \end{vmatrix} = -\hat{i}-7\hat{j}+5\hat{k}$$

Equation of the line passes through the point $$\displaystyle A\left ( \overline{a} \right )$$ and parallel to $$ \overline{b}$$ is given by

$$\displaystyle \overline{r}=\overline{a}+\lambda \overline{b}$$

Now line $$\displaystyle L_{1}=\dfrac{x+1}{3}=\dfrac{y+2}{1}=\dfrac{z+1}{2}$$ can be written as $$\displaystyle L_{1}:\dfrac{x+1}{3}-1=\dfrac{y+2}{1}-1=\dfrac{z+1}{2}-1 $$

$$\displaystyle L_{1}:\dfrac{x-2}{3}=\dfrac{y+1}{1}=\dfrac{z-1}{2}$$
which passes through the point whose position vector is $$\displaystyle \overline{a}=2\hat{i}-\hat{j}+\hat{k}$$.
Question 4
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If $$A$$ , $$B$$ and $$C$$ are three collinear points, where $$A= i + 8 j - 5k $$, $$ B = 6i-2j$$ and $$C= 9i + 4j - 3 k$$, then $$B$$ divides $$AC$$ in the ratio of :

A
$$\dfrac{5}{7}$$

B
$$\dfrac{5}{3}$$

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

D
None of these

Solution

Given points are $$A(i+8j-5k)$$ , $$B(6i-2j)$$ and $$C(9i+4j-3k)$$
$$AB = -5i+10j-5k$$ and $$CB = 3i+6j-3k$$
$$\Rightarrow \dfrac{|AB|}{|CB|}=\dfrac{\sqrt{150}}{\sqrt{54}}=\dfrac{5\sqrt6}{3\sqrt6}=\dfrac{5}{3}$$
Therefore correct option is $$B$$
Question 5
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If the points $$a(cos \alpha + i sin \alpha)$$ , $$b(cos \beta + i sin \beta)$$ and $$c(cos \gamma + isin \gamma)$$ are collinear then the value of $$|z|$$ is:
( where $${z = bc \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$$ )

A
$$2$$

B
$$5$$

C
$$1$$

D
None of these.

Solution

Given $$a=cos\alpha+i\sin\alpha=e^{i\alpha}$$ , $$b=cos\beta+isin\beta=e^{i\beta}$$ and $$c=cos\gamma+isin\gamma=e^{i\gamma}$$
Consider $$bcsin(\beta-\gamma)=e^{i(\beta+\gamma)}sin(\beta-\gamma)=\frac{1}{2i}e^{i(\beta+\gamma)}(e^{i(\beta-\gamma)}-e^{-i(\beta-\gamma)}) = \frac{1}{2i}(e^{i(2\beta)}-e^{i(2\gamma)})$$
Similarly we get $$casin(\gamma-\alpha) = \frac{1}{2i}(e^{i(2\gamma)}-e^{i(2\alpha)})$$ and $$absin(\alpha-\beta) = \frac{1}{2i}(e^{i(2\alpha)}-e^{i(2\beta)})$$
Therefore we get $$bcsin(\beta-\gamma)+casin(\gamma-\alpha)+absin(\alpha-\beta)=0$$
So we get $$z=3i-4i$$
$$\Rightarrow |z|=5$$
Question 6
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The angle between two diagonals of a cube is

A
$$\cos^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$$

B
$$\cos^{-1}\left (\dfrac {1}{3}\right )$$

C
$$30^{\circ}$$

D
$$45^{\circ}$$

Solution

Let $$ OABCDEFG $$ be a cube with vertices as below

$$O(0,0,0), A(a,0,0), B(a,a,0), C(0,a,0), D(0,a,a), E(0,0,a), F(a,0,a), G(a,a,a)$$

There are four diagonals $$OG,CF,AD$$ and $$BE$$ for the cube.

Let us consider any two say $$OG$$ and $$AD$$

We know that if $$A(x_1,y_1,z_1)$$ and $$B(x_2,y_2,z_2)$$ are two points in space then
$$\vec{AB}=(x_2-x_1)i + (y_2-y_1)j + (z_2-z_1)k$$

$$\Rightarrow \vec{OG}=(a-0)i+(a-0)j+(a-0)k=ai+aj+ak$$

and $$\vec{AD} = (0-a)i+(a-0)j+(a-0)k=-ai+aj+ak$$.

Therefore $$|\vec{OG}|=\sqrt{a^2+a^2+a^2}=a\sqrt{3}$$ and $$|\vec{AD}|=\sqrt{(-a)^2+a^2+a^2}=a\sqrt{3}$$

$$\vec{OG} \cdot \vec{AD}=-a^2+a^2+a^2=a^2$$

We know that angle between two vectors $$\vec{a},\vec{b}$$ is given by $$\theta=cos^{-1}\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$$

Thus angle between two diagonals $$\vec{OG}$$ and $$\vec{AD}$$ is

$$\theta = cos^{-1}\dfrac{a^2}{a\sqrt{3} \times a\sqrt{3}}=cos^{-1}\left(\dfrac{1}{3}\right)$$
Question 7
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Equation of the plane through the mid-point of the line segment joining the points $$P(4, 5, -10), \,Q(-1, 2, 1)$$ and perpendicular to $$PQ$$ is

A
$$r. \left( \dfrac{}3{}2 \widehat i + \dfrac{7}{2} \widehat j - \dfrac{9}{2} \widehat k\right ) =45$$

B
$$r. (- \widehat i + 2 \widehat j + \widehat k) = \dfrac{135}{2}$$

C
$$r. (5 \widehat i + 3 \widehat j - 11 \widehat k) + \dfrac{135}{2} = 0$$

D
$$r. (4 \widehat i + 5 \widehat j - 10 \widehat k) = 85$$

E
$$r. (5\widehat i + 3 \widehat j - 11\widehat k) = \dfrac{135}{2}$$

Solution

Coordinate the mid-point of $$P(4, 5, -10)$$ and $$Q(-1, 2, 1)$$ is
$$\left( \dfrac{4 - 1}{2}, \dfrac{5 + 2}{2}, \dfrac{-10 + 1}{2}\right )$$. i.e. $$\left ( \dfrac{3}{2}. \dfrac{7}{2} , \dfrac{-9}{2} \right )$$
Now, DR's of $$PQ$$ is $$(-1 -4, 2 -5, 1 + 10)$$
i.e., $$(-5, -3, 11)$$ or $$(5, 3, -11)$$
Therefore, euation of plane passing through
$$\left( \dfrac{3}{2}, \dfrac{7}{2}, \dfrac{-9}{2} \right )$$ and having DR's $$(5, 3, -11)$$ is
$$5 \left( x - \dfrac{3}{2} \right ) + 3 \left( y - \dfrac{7}{2} \right ) - 11 \left( z - \dfrac{9}{2} \right ) = 0$$
$$\Rightarrow 5x + 3y - 11z = \dfrac{15}{2} + \dfrac{21}{2} + \dfrac{99}{2}$$
$$\Rightarrow 5x + 3y - 11z = \dfrac{135}{2}$$
It is written by vector form
$$r. (5 \widehat i + 3 \widehat j - 11 \widehat k) = \dfrac{135}{2}$$
Question 8
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If $$(2, -1, 3)$$ is the foot of the perpendicular drawn from the origin to the plane, then the equation of the plane is

A
$$2x + y - 3z + 6 = 0$$

B
$$2x - y + 3z - 14 = 0$$

C
$$2x - y + 3z - 13 = 0$$

D
None of these

Solution

Let $$P(2,-1,3)$$ be the foot of the perpendicular from the origin to the plane.

$$\therefore \vec{OP}=\vec n=2\vec i-\vec j+3\vec k$$ and $$p=|2\vec i-\vec j+3\vec k|=\sqrt{2^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}$$

$$\therefore \hat n=\dfrac{2\vec i-\vec j+3\vec k}{\sqrt{14}}$$

The vector equation of the plane is $$\vec r \cdot \hat n=p$$

$$\Rightarrow (x\vec i+y\vec j+z\vec k)\cdot \dfrac{2\vec i-\vec j+3\vec k}{\sqrt{14}}=\sqrt{14}$$

$$\Rightarrow (x\vec i+y\vec j+z\vec k)\cdot (2\vec i-\vec j+3\vec k)=14$$

$$\Rightarrow 2x-y+3z=14$$

$$\Rightarrow 2x-y+3z-14=0$$
Question 9
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What is the angle between the lines $$\cfrac { x-2 }{ 1 } =\cfrac { y+1 }{ -2 } =\cfrac { z+2 }{ 1 } $$ and $$\cfrac { x-1 }{ 1 } =\cfrac { 2y+3 }{ 3 } =\cfrac { z+5 }{ 2 } =?$$

A
$$\cfrac { \pi }{ 2 } $$

B
$$\cfrac { \pi }{ 3 } $$

C
$$\cfrac { \pi }{ 6 } $$

D
None of the above

Solution

The direction ratios of the line are:
$$B=(a,b,c) = (1,-2,1)$$
The normal to the plane has the coordinates:
$$N=(-2, 1, 2)$$
Let the angle between the line and the plane be $$'x'$$
$$\therefore, sinx= \dfrac{(B.N)}{( |B|\times|N| )}$$
By substuting the values we get
$$sin x =1$$
$$ i.e.x=\dfrac {\pi}{2}$$.
Question 10
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A plane mirror is placed at the origin so that the direction ratios of its normal are $$(1,-1,1)$$. A ray of light, coming along the positive direction of the x-axis, strikes the mirror. The direction cosines of the reflected ray are

A
$$\cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 } ,\cfrac { 2 }{ 3 } $$

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

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

D
$$-\cfrac { 1 }{ 3 } ,-\cfrac { 2 }{ 3 } ,\cfrac { 2 }{ 3 } $$

Solution

Let the source of light be situated at A (a,0,0) where a is not equal to 0.
let OA be the incident ray and OB the reflected ray

ON is the normal to the mirror at O

D.R's of OA are $$0,0,1$$
and its D.R's are $$1,0,0$$

D.R's of ON are $$\dfrac {1}{\sqrt 3} , \dfrac {-1}{\sqrt 3} , \dfrac {1}{\sqrt 3}$$

therefore cos $$\dfrac {\theta}{2}$$ = $$\dfrac {1}{\sqrt 3}$$

let l,m,n be the D.R's of the reflected ray OA then,

$$l$$ =$$\dfrac {2}{ 3}-1,$$ $$m $$= $$\dfrac {-2}{ 3}$$, $$n$$ = $$\dfrac {2}{ 3}$$

l= $$\dfrac {-1}{ 3}$$, m = $$\dfrac {-2}{ 3},$$ n = $$\dfrac {2}{3}$$

Hence the D.R's of the reflected ray are $$\dfrac {-1}{ 3}$$, $$\dfrac {-2}{ 3},$$, $$\dfrac {2}{3}$$.

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Three Dimensional Geometry Test - 58

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Three Dimensional Geometry Test - 58

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Question 1

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

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

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

$$\dfrac{\pi}{3}$$

Solution

Question 2

$$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}+5\hat{k} \right )=45$$

$$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=45$$

$$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )+45=0$$

$$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}-5\hat{k} \right )=37$$

Solution

Question 3

$$\displaystyle -\hat{i}+2\hat{j}-\hat{k}$$

$$\displaystyle 2\hat{i}-\hat{j}+\hat{k}$$

$$\displaystyle 2\hat{i}-2\hat{j}+3\hat{k}$$

$$\displaystyle \hat{i}+2\hat{j}+3\hat{k}$$

Solution

Question 4

$$\dfrac{5}{7}$$

$$\dfrac{5}{3}$$

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

None of these

Solution

Question 5

$$2$$

$$5$$

$$1$$

None of these.

Solution

Question 6

$$\cos^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$$

$$\cos^{-1}\left (\dfrac {1}{3}\right )$$

$$30^{\circ}$$

$$45^{\circ}$$

Solution

Question 7

$$r. \left( \dfrac{}3{}2 \widehat i + \dfrac{7}{2} \widehat j - \dfrac{9}{2} \widehat k\right ) =45$$

$$r. (- \widehat i + 2 \widehat j + \widehat k) = \dfrac{135}{2}$$

$$r. (5 \widehat i + 3 \widehat j - 11 \widehat k) + \dfrac{135}{2} = 0$$

$$r. (4 \widehat i + 5 \widehat j - 10 \widehat k) = 85$$

$$r. (5\widehat i + 3 \widehat j - 11\widehat k) = \dfrac{135}{2}$$

Solution

Question 8

$$2x + y - 3z + 6 = 0$$

$$2x - y + 3z - 14 = 0$$

$$2x - y + 3z - 13 = 0$$

None of these

Solution

Question 9

$$\cfrac { \pi }{ 2 } $$

$$\cfrac { \pi }{ 3 } $$

$$\cfrac { \pi }{ 6 } $$

None of the above

Solution

Question 10

$$\cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 } ,\cfrac { 2 }{ 3 } $$

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

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

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

Solution

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