Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 32

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Weekly Quiz Competition

Question 1
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Let the direction - cosines of the line which is equally inclined to the axis be $$\displaystyle \pm \frac{1}{\sqrt{k}}$$. Find $$k$$ ?

A
$$2$$

B
$$3$$

C
$$5$$

D
$$6$$

Solution

We know sum of the squares of the direction cosines is one.
But it is given that $$\alpha=\beta=\gamma$$
$$\therefore cos^2\alpha+cos^\alpha+cos^2\alpha=1$$
$$3cos^2\alpha=1$$
$$cos^2\alpha=\dfrac{1}{3}$$
Thus $$\cos\alpha=\dfrac{1}{\sqrt3}$$
$$\therefore \dfrac{1}{\sqrt k}=\dfrac{1}{\sqrt 3}$$
Taking squares on both sides we get
$$\therefore k=3$$
Question 2
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If a line makes an angle $$\displaystyle \theta_1, \theta_2, \theta_3$$ which the axis respectively, then $$\displaystyle cos 2\theta_1 + cos 2 \theta_2 + cos 2 \theta_3 = ?$$

A
$$-4$$

B
$$2$$

C
$$3$$

D
$$-1$$

Solution

Let $$P(x,y,z)$$ be any point on the line OP, D-Ratio of OP are x,y,z. D.R. of normal plane$$ i.e.,z = 0 $$are$$ 0, 0, 1 $$
$$\sin\theta_1=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$$
similarily for $$ sin\theta_2$$ and $$sin \theta_3$$
$$\therefore sin^2\theta_1 + sin^2\theta_2 + sin^2\theta_3$$
$$=\dfrac{x^2+y^2+z^2}{x^2+y^2+z^2}=1$$
$$\Longrightarrow 1-cos^2\theta_1+ 1-cos^2\theta_2+ 1-cos^2\theta_3=1$$
$$\therefore cos^\theta_1+cos^\theta_2+cos^\theta_3=3-1=2$$
Question 3
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The direction ratios of a normal to the plane through $$\left( 1,0,0 \right) ,\left( 0,1,0 \right) ,$$ which makes an angle of $$\displaystyle \frac { \pi }{ 4 } $$ with the plane $$x+y=3$$ are

A
$$\displaystyle 1,\sqrt { 2 } ,1$$

B
$$\displaystyle 1,1,\sqrt { 2 } $$

C
$$\displaystyle 1,1,2$$

D
$$\displaystyle \sqrt { 2 } ,1,1$$

Solution

Any plane through $$\left( 1,0,0 \right) $$ is

$$A\left( x-1 \right) +B\left( y-0 \right) +C\left( z-0 \right) =0$$ ...$$(1)$$

It contains $$\left( 0,1,0 \right) $$ if $$-A+B=0$$ ....$$(2)$$

Also, $$(1)$$ makes an angle of $$\displaystyle \dfrac { \pi }{ 4 } $$ with the plane $$x+y=3$$

Therefore, $$\displaystyle \cos { \dfrac { \pi }{ 4 } } =\dfrac { \left| A+B \right| }{ \sqrt { { A }^{ 2 }+{ B }^{ 2 }+{ C }^{ 2 } } \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } $$

$$\Rightarrow { \left( A+B \right) }^{ 2 }={ A }^{ 2 }+{ B }^{ 2 }+{ C }^{ 2 }\Rightarrow 2AB={ C }^{ 2 }$$ ....$$(3)$$

From (2) and (3), $${ C }^{ 2 }=2{ A }^{ 2 }\Rightarrow C=\pm \sqrt { 2 } A$$

Hence, $$A:B:C::A:A:\pm \sqrt { 2 } A$$

$$\therefore$$ direction ratios are $$1:1:\pm \sqrt { 2 } $$
Question 4
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let $$P(4,1,\lambda)$$ and $$Q(2,-1,\lambda)$$ be two points. A line having direction ratios $$1,-1,6$$ is perpendicular to the plane passing through the origin, $$P$$ and $$Q$$, then $$\lambda$$ equals

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

B
$$\dfrac{1}{2}$$

C
$$1$$

D
none of these

Solution

By given point $$P\left( 4,1,\lambda \right) $$ and $$Q\left( 2,-1,\lambda \right) $$
$$\begin{vmatrix} 1 & \lambda \\ -1 & \lambda \end{vmatrix},\begin{vmatrix} \lambda & 4 \\ \lambda & 2 \end{vmatrix},\begin{vmatrix} 4 & 1 \\ 2 & -1 \end{vmatrix}\\ \Rightarrow \left( \lambda +\lambda \right) ,\left( 2\lambda -4\lambda \right) ,\left( -4-2 \right) \\ \Rightarrow -2\lambda ,2\lambda ,6\\ \Rightarrow -2\lambda =1,2\lambda =-1$$
$$\displaystyle \Rightarrow \lambda =-\frac { 1 }{ 2 } $$
Question 5
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$$\bar a,\bar b,\bar c$$ are three non-zero vectors such that any two of them are non-collinear. If $$\bar a+\bar b$$ is collinear with $$\bar c$$ and $$\bar b+\bar c$$ is collinear with $$\bar a$$, then what is their sum?

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

We have
$$\bar a+\bar b =t\bar c$$ ----$$(1)$$
$$\bar b+\bar c =s\bar a$$ ----$$(2)$$
From $$(1)$$ and $$(2)$$
$$\bar a+\bar b=t(s\bar a-\bar b)$$
Since no two of them are collinear, comparing coeffficients gives
$$st=1$$ and $$t=-1$$
$$\Rightarrow s=-1$$ and $$t=-1$$
From $$(1)$$
$$\therefore \bar a+\bar b+\bar c=0$$
Hence, option $$B$$.
Question 6
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The equation of plane through $$(1, 2, 3)$$ and parallel to the plane $$\bar{r}.\left ( \hat{3i}+\hat{4j}+\hat{5k} \right )=0$$

A
$$\bar{r}.\left ( \hat{3i}+\hat{4j}+\hat{5k} \right )=26$$

B
$$\bar{r}.\left ( \hat{3i}-\hat{4j}+\hat{5k} \right )+4$$

C
$$\bar{r}.\left ( \hat{3i}+\hat{4j}-\hat{5k} \right )+4=0$$

D
None of these

Solution

Equation of plane passing through $$(1,2,3)$$ and parallel to plane $$\vec{r} \cdot (3\hat{i}+4\hat{j}+5\hat{k} = 0$$ is given by,
$$ (\vec{r}-(\hat{i}+2\hat{j}+3\hat{k}) )\cdot (3\hat{i}+4\hat{j}+5\hat{k}) = 0$$
$$\Rightarrow \vec{r} \cdot (3\hat{i}+4\hat{j}+5\hat{k}) -(\hat{i}+2\hat{j}+3\hat{k}) \cdot (3\hat{i}+4\hat{j}+5\hat{k} ) = 0$$
$$\Rightarrow \vec{r}\cdot (3\hat{i}+4\hat{j}+5\hat{k} )- 26 = 0$$
Hence, option 'A' is correct.
Question 7
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The acute angle between two lines whose direction ratios are $$2,3,6$$ and $$1,2,2$$ is

A
$$\displaystyle \cos ^{ -1 }{ \left( \frac { 20 }{ 21 } \right) } $$

B
$$\displaystyle \cos ^{ -1 }{ \left( \frac { 18 }{ 21 } \right) } $$

C
$$\displaystyle \cos ^{ -1 }{ \left( \frac { 8 }{ 21 } \right) } $$

D
None of these

Solution

We have $${ a }_{ 1 }=2,{ b }_{ 1 }=3,{ c }_{ 1 }=6,{ a }_{ 2 }=1,{ b }_{ 2 }=2,{ c }_{ 2 }=2$$
If $$\theta$$ is the angle between two lines whose direction ratios are given, then
$$\cos { \theta } =\displaystyle\frac { { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 } }{ \sqrt { { \left( { a }_{ 1 } \right) }^{ 2 }+{ \left( { b }_{ 1 } \right) }^{ 2 }+{ \left( { c }_{ 1 } \right) }^{ 2 } } \sqrt { { \left( { a }_{ 2 } \right) }^{ 2 }+{ \left( { b }_{ 2 } \right) }^{ 2 }+{ \left( { c }_{ 2 } \right) }^{ 2 } } } $$
$$\Rightarrow \cos { \theta } =\displaystyle \frac { 2\times 1+3\times 2+6\times 2 }{ \sqrt { { \left( 2 \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 } } \sqrt { { \left( 1 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 } } } =\frac { 20 }{ 21 }$$
$$ \Rightarrow \theta =\cos ^{ -1 }{ \dfrac { 20 }{ 21 } } $$
Question 8
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Directions For Questions

Consider the two planes $$ p_{1}\::\:2x-3y+z=4\: $$ & $$ \:p_{2}\::\: x-y+z=-1 $$
On the basis of above information answer the following question:

...view full instructions

Equation of the plane which passes through the point (-1, 3, 2) & is $$ \perp $$ to each of the planes $$ p_{1} $$ & $$ p_{2} $$ is

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

B
$$ 3x+8y-2z-17= 0$$

C
$$ 2x+y-z+1= 0 $$

D
$$ x-2y+z+1= 0 $$

Solution

Normal of $$p1 = [ 2 -3 1 ]$$

Normal of $$p2 = [ 1 -1 1 ]$$

If planes are perpendiculars then dot products of their normals must be 0,

So consider normal of desired plane is $$[ a b c]$$

Hence,
$$ 2a - 3b + c = 0 $$ (Dot product with normal of p1)

$$ a - b + c = 0 $$ (Dot product with normal of p2)

Solving above 2 we get ,
$$a - 2b = 0 $$

So if we use normals of planes in option b & d,

we notice that they don't satisfy the above equation

We observe that option a does not pass through (-1, 3, 2)

Since on substitution ,

$$ 2.(-1) + 3 + 2 + 1 \neq 0 $$

The correct answer is $$2x+y-z+1=0$$
Question 9
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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

Equation of the plane passing through a point with position vector $$\displaystyle 3\hat{i}-3\hat{j}+\hat{k} $$ & normal to the line joining the points with position vectors $$\displaystyle 3\hat{i}+4\hat{j}-\hat{k} $$ & $$\displaystyle 2\hat{i}-\hat{j}=5\hat{k} $$ is

A
$$\displaystyle \overline{r}.\left ( -\hat{i}-5\hat{j}+6\hat{k} \right )+18=0$$

B
$$\displaystyle \overline{r}.\left ( \hat{i}-5\hat{j}+6\hat{k} \right )=22$$

C
$$\displaystyle \overline{r}.\left ( \hat{i}+5\hat{j}-6\hat{k} \right )+18=0$$

D
$$\displaystyle \overline{r}.\left ( -\hat{i}+5\hat{j}+6\hat{k} \right )+12=0$$

Solution

$$\displaystyle \left ( \overline{r} - \overline{a} \right )\cdot \overline{n}=0$$
$$\displaystyle \Rightarrow \overline{r} \cdot \overline{n}= \overline{a} \cdot \overline{n}$$ where $$ \displaystyle \overline{n}=AB= -\hat{i}-5\hat{j}+6\hat{k}$$
$$\displaystyle \Rightarrow \overline{r} \cdot \left ( -\hat{i}-5\hat{j}+6\hat{k} \right )=\left ( 3\hat{i}-3\hat{j}+\hat{k} \right ) \cdot \left ( -\hat{i}-5\hat{j}+6\hat{k} \right )$$
$$\displaystyle \Rightarrow \overline{r} \cdot \left ( \hat{i}+5\hat{j}-6\hat{k} \right )+18=0$$
Hence the correct option becomes c.
Question 10
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If a point $$P$$ in the space such that $$\overline{OP}$$ is inclined to $$OX$$ at $$45$$ and $$OZ$$ to $$60$$ then $$\overline{OP}$$ inclined to $$OY$$ is

A
$$75^{\circ}$$

B
$$75^{\circ} \text{or} 105^{\circ}$$

C
$$60^{\circ} \text{or} 120^{\circ}$$

D
None of these

Solution

$$\displaystyle l=\cos { { 45 }^{ O } } =\frac { 1 }{ \sqrt { 2 } } $$ and $$\displaystyle m=\cos { { 60 }^{ O } } =\frac { 1 }{ 2 } $$
$$\displaystyle \Rightarrow { n }^{ 2 }=1-\frac { 1 }{ 2 } -\frac { 1 }{ 4 } =\frac { 1 }{ 4 } \Rightarrow n=\pm \frac { 1 }{ 2 } $$
So $$\alpha ={ 60 }^{ O }$$ or $${ 120 }^{ O }$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 32

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Three Dimensional Geometry Test - 32

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$5$$

$$6$$

Solution

Question 2

$$-4$$

$$2$$

$$3$$

$$-1$$

Solution

Question 3

$$\displaystyle 1,\sqrt { 2 } ,1$$

$$\displaystyle 1,1,\sqrt { 2 } $$

$$\displaystyle 1,1,2$$

$$\displaystyle \sqrt { 2 } ,1,1$$

Solution

Question 4

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

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

$$1$$

none of these

Solution

Question 5

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 6

$$\bar{r}.\left ( \hat{3i}+\hat{4j}+\hat{5k} \right )=26$$

$$\bar{r}.\left ( \hat{3i}-\hat{4j}+\hat{5k} \right )+4$$

$$\bar{r}.\left ( \hat{3i}+\hat{4j}-\hat{5k} \right )+4=0$$

None of these

Solution

Question 7

$$\displaystyle \cos ^{ -1 }{ \left( \frac { 20 }{ 21 } \right) } $$

$$\displaystyle \cos ^{ -1 }{ \left( \frac { 18 }{ 21 } \right) } $$

$$\displaystyle \cos ^{ -1 }{ \left( \frac { 8 }{ 21 } \right) } $$

None of these

Solution

Question 8

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

$$ 3x+8y-2z-17= 0$$

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

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

Solution

Question 9

$$\displaystyle \overline{r}.\left ( -\hat{i}-5\hat{j}+6\hat{k} \right )+18=0$$

$$\displaystyle \overline{r}.\left ( \hat{i}-5\hat{j}+6\hat{k} \right )=22$$

$$\displaystyle \overline{r}.\left ( \hat{i}+5\hat{j}-6\hat{k} \right )+18=0$$

$$\displaystyle \overline{r}.\left ( -\hat{i}+5\hat{j}+6\hat{k} \right )+12=0$$

Solution

Question 10

$$75^{\circ}$$

$$75^{\circ} \text{or} 105^{\circ}$$

$$60^{\circ} \text{or} 120^{\circ}$$

None of these

Solution

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