Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 16

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Question 1
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The vector equation of line passing through two points $$A(x_1,y_1,z_1),B(x_2,y_2,z_2) $$ is

A
$$\vec r=\vec a-\vec b$$

B
$$\vec r=\vec a +\lambda(\vec b-\vec a)$$

C
$$\vec r=\vec a+\lambda\vec b$$

D
$$\vec r=\vec a+\vec b$$

Solution

Given points
$$A(x_{1},y_{1},z_{1})$$ and $$B(x_{2},y_{2},z_{2})$$
position vector of point A
$$\vec{a}=x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$$
position vector of point B
$$\vec{b}=x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$$
normal vector of line be
$$\vec{n}=\vec{b}-\vec{a}$$
eq of line
$$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$$
Question 2
1 / -0

A line passes through the points (6, -7, -1) and (2, -3, 1). What are the direction ratios of the line ?

A
$$(4, -4, 2)$$

B
$$( 4, 4, 2 )$$

C
$$( -4, 4, 2 )$$

D
$$( 2, 1, 1 )$$

Solution

Direction ratios of a line passing through points $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2)$$ are represented by $$\pm (x_1-x_2, \ y_1-y_2, \ z_1-z_2)$$

Hence for the given line, direction ratios are $$(6-2,-7-(-3),-1-1)$$
$$\Rightarrow \pm (4,-4,-2)$$
$$\Rightarrow (-4,4,2)$$ or $$(4,-4,-2)$$
Question 3
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Find vector equation for the line passing through the points $$3\overline i+4\overline j-7\overline k,\overline i-\overline j+6\overline k$$.

A
$$\overline r=(3-2\lambda)\overline i+(4-5\lambda)\overline j+(-7+13\lambda)\overline k$$

B
$$\overline r=(2\lambda)\overline i+(4+5\lambda)\overline j+(-7-13\lambda)\overline k$$

C
$$\overline r=(3-2\lambda)\overline i-(4-5\lambda)\overline j+(-7+13\lambda)\overline k$$

D
$$\overline r=(3-2\lambda)\overline i+(4-5\lambda)\overline j-(-7+13\lambda)\overline k$$

Solution

Given points
$$\vec{a}=3\hat{i}+4\hat{j}-7\hat{k}$$
$$\vec{b}=\hat{i}-\hat{j}+6\hat{k}$$
normal vector of line be
$$\vec{n}=\vec{b}-\vec{a}$$
$$\vec{n}=\hat{i}-\hat{j}+6\hat{k}-(3\hat{i}+4\hat{j}-7\hat{k})$$
$$\vec{n}=\hat{i}-\hat{j}+6\hat{k}-3\hat{i}-4\hat{j}+7\hat{k}$$
$$\vec{n}=-2\hat{i}-5\hat{j}+13\hat{k}$$
vector eq of line
$$\vec{r}=\vec{a}+\lambda(\vec{n})$$
$$\vec{r}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda(-2\hat{i}-5\hat{j}+13\hat{k})$$
$$\vec{r}=3\hat{i}+4\hat{j}-7\hat{k}-2\lambda\hat{i}-5\lambda\hat{j}+13\lambda\hat{k})$$
$$\vec{r}=(3-2\lambda)\hat{i}+(4-5\lambda)\hat{j}+(-7+13\lambda)\hat{k}$$
Question 4
1 / -0

The projections of a directed line segment on the coordinate axes are $$12, 4, 3$$ respectively.
What are the direction cosines of the line segment?

A
$$(12/13, 4/13, 3/13)$$

B
$$(12/13, -4/13, 3/13)$$

C
$$(12/13, -4/13, -3/13)$$

D
$$(-12/13, -4/13, 3/13)$$

Solution

Let the direction cosines of the line segment in x,y,z are 12k, 4k, 3k respectively
But $${ cos }^{ 2 }\alpha +{ cos }^{ 2 }\beta +{ cos }^{ 2 }\gamma =1$$
$$\Rightarrow 144{ k }^{ 2 }+16{ k }^{ 2 }+9{ k }^{ 2 }=1\\ \Rightarrow { k }^{ 2 }=\dfrac { 1 }{ 169 } \\ \Rightarrow k=\dfrac { 1 }{ 13 } $$
$$\therefore$$ The direction cosines of the line segment are $$\left(\dfrac { 12 }{ 13 } ,\dfrac { 4 }{ 13 } ,\dfrac { 3 }{ 13 } \right)$$
Question 5
1 / -0

From the point $$P(3, -1, 11)$$, a perpendicular is drawn on the line $$L$$ given by the equation $$\dfrac {x}{2} = \dfrac {y - 2}{3} = \dfrac {z - 3}{4}$$. Let $$Q$$ be the foot of the perpendicular.
What are the direction ratios of the line segment $$PQ$$?

A
$$(1, 6, 4)$$

B
$$(-1, 6, -4)$$

C
$$(-1, -6, 4)$$

D
$$(2, -6, 4)$$

Solution

The general point on the line $$\dfrac { x }{ 2 } =\dfrac { y-2 }{ 3 } =\dfrac { z-3 }{ 4 } $$ is $$(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$$
Let $$Q(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$$
The directional cosines of $$PQ= 2\lambda -3,3\lambda +3,4\lambda -8 $$
Since the line $$L$$ and $$PQ$$ are perpendicular
$$(2\lambda -3)(2)+(3\lambda +3)(3)+(4\lambda -8)4=0\\ \Longrightarrow \lambda =1\\ \Longrightarrow Q=(2,5,7)$$
The direction ratios of $$PQ=(-1,6,-4)$$
Question 6
1 / -0

Find the vector equation of line joining the points $$ (2,1,3)$$ and $$(-4,3,-1)$$

A
$$\overline r=2(1-3\lambda)\overline i-(1+2\lambda)\overline j-(3-4\lambda)\overline k$$

B
$$\overline r=2(1-3\lambda)\overline i-(1+2\lambda)\overline j+(3-4\lambda)\overline k$$

C
$$\overline r=2(1-3\lambda)\overline i+(1+2\lambda)\overline j+(3-4\lambda)\overline k$$

D
$$\overline r=2(1+3\lambda)\overline i+(1+2\lambda)\overline j+(3+4\lambda)\overline k$$

Solution

given Points
A(2,1,3) and B(-4,3,-1)
$$\vec{b}=-4\hat{i}+3\hat{j}-\hat{k}$$
Position vector of line
$$A=\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$$
normal vector can be calculated by
$$\vec{b}-\vec{a}=\vec{n}=-4\hat{i}+3\hat{j}-\hat{k}-(2\hat{i}+\hat{j}+3\hat{k})$$
$$\vec{n}=-4\hat{i}+3\hat{j}-\hat{k}-2\hat{i}-\hat{j}-3\hat{k}$$
$$\vec{n}=-6\hat{i}+2\hat{j}-4\hat{k}$$
eq of line
$$\vec{r}=\vec{a}+\lambda\vec{n}$$
$$\vec{r}=2\hat{i}+\hat{j}+3\hat{k}+\lambda(-6\hat{i}+2\hat{j}-4\hat{k})$$
$$\vec{r}=2\hat{i}+\hat{j}+3\hat{k}-6\lambda\hat{i}+2\lambda\hat{j}-4\lambda\hat{k}$$
$$\vec{r}=(2-6\lambda)\hat{i}+(1+2\lambda)\hat{j}+(3-4\lambda)\hat{k}$$
$$\vec{r}=2(1-3\lambda)\hat{i}+(1+2\lambda)\hat{j}+(3-4\lambda)\hat{k}$$
Question 7
1 / -0

If a line makes the angles $$ \alpha , \beta$$ and $$\gamma$$ with the axes, then what is the value of $$1+\cos 2\alpha +\cos 2\beta+\cos 2\gamma$$ equal to ?

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

Since, the line makes the angles $$ \alpha , \beta$$ and $$\gamma$$ with the axes
So, $$\cos\alpha, \cos\beta, \cos\gamma $$ will be the direction cosines of the line.
And we know that the sum of squares of the direction cosines of any line is $$1$$.
Now,
$$1+\cos 2\alpha +\cos 2\beta +\cos 2\gamma$$
$$ =1+(2\cos ^{ 2 }\alpha -1)+(2\cos ^{ 2 }\beta -1)+(2\cos ^{ 2 }\gamma -1)$$
$$ =1+2(\cos ^{ 2 }\alpha +\cos ^{ 2 }\beta +\cos ^{ 2 }\gamma )-3$$
$$ =1+2-3=0$$
Hence, B is correct.
Question 8
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Which of the following represents direction cosines of the line:

A
$$0,\cfrac { 1 }{ \sqrt { 2 } } ,\cfrac { 1 }{ 2 } $$

B
$$0,\cfrac { -\sqrt { 3 } }{ 2 } ,\cfrac { 1 }{ \sqrt 2 } $$

C
$$0,\cfrac { \sqrt { 3 } }{ 2 } ,\cfrac { 1 }{ { 2 } } $$

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

Solution

If direction cosine of a line is $$l,m,n$$, then
$${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
$${ 0 }^{ 2 }+{ \left( \cfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }=1$$
Hence, the correct answer from the given alternative is (c) $$0,\cfrac { \sqrt { 3 } }{ 2 } ,\cfrac { 1 }{ \sqrt { 2 } } $$.
Question 9
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The length of perpendicular from the origin to the plane which makes intercepts $$\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5}$$ respectively on the coordinate axes is

A
$$\dfrac{1}{\sqrt[5]{2}}$$

B
$$\dfrac{1}{10}$$

C
$$\displaystyle\sqrt[5]{2}$$

D
5

Solution

Equation of plane $$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$$

$$3x+4y+5z-1=0$$

diatance from origin $$\dfrac{1}{\sqrt{50}}=\dfrac{1}{\sqrt[5]{2}}$$
Question 10
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If the normal of the plane makes an angles $$\dfrac {\pi}{4}, \dfrac {\pi}{4}$$ and $$\dfrac {\pi}{2}$$ with positive X-axis, Y-axis and Z-axis respectively and the length of the perpendicular line segment from origin to the plane is $$\sqrt {2}$$, then the equation of the plane is ________.

A
$$x + y + z = \sqrt {2}$$

B
$$x + y + z = 1$$

C
$$x + y = 2$$

D
$$x = \sqrt {2}$$

Solution

Normal of the plane makes an angle of $$\dfrac{\pi}{4},\dfrac{\pi}{4},\dfrac{\pi}{2}$$ with X-axix,Y-axis, and Z-axis respectively,

Direction ratio of normal to the plane is$$(\cos\alpha,\cos\beta,\cos\gamma)$$
Direction ratio of normal to the plane is
$$\left (\cos \dfrac {\pi}{4}, \cos \dfrac {\pi}{4}, \cos \dfrac {\pi}{2}\right )$$
$$\left (\dfrac {1}{\sqrt {2}}, \dfrac {1}{\sqrt {2}}, 0\right )$$
Equation of the plane is given by
$$x\cos\alpha+y\cos\beta+z\cos\gamma=\perp\ distance$$
$$\therefore \dfrac {x}{\sqrt {2}} + \dfrac {y}{\sqrt {2}} = \sqrt {2}$$
$$x + y = 2$$.

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 16

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Three Dimensional Geometry Test - 16

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SHARING IS CARING

Question 1

$$\vec r=\vec a-\vec b$$

$$\vec r=\vec a +\lambda(\vec b-\vec a)$$

$$\vec r=\vec a+\lambda\vec b$$

$$\vec r=\vec a+\vec b$$

Solution

Question 2

$$(4, -4, 2)$$

$$( 4, 4, 2 )$$

$$( -4, 4, 2 )$$

$$( 2, 1, 1 )$$

Solution

Question 3

$$\overline r=(3-2\lambda)\overline i+(4-5\lambda)\overline j+(-7+13\lambda)\overline k$$

$$\overline r=(2\lambda)\overline i+(4+5\lambda)\overline j+(-7-13\lambda)\overline k$$

$$\overline r=(3-2\lambda)\overline i-(4-5\lambda)\overline j+(-7+13\lambda)\overline k$$

$$\overline r=(3-2\lambda)\overline i+(4-5\lambda)\overline j-(-7+13\lambda)\overline k$$

Solution

Question 4

$$(12/13, 4/13, 3/13)$$

$$(12/13, -4/13, 3/13)$$

$$(12/13, -4/13, -3/13)$$

$$(-12/13, -4/13, 3/13)$$

Solution

Question 5

$$(1, 6, 4)$$

$$(-1, 6, -4)$$

$$(-1, -6, 4)$$

$$(2, -6, 4)$$

Solution

Question 6

$$\overline r=2(1-3\lambda)\overline i-(1+2\lambda)\overline j-(3-4\lambda)\overline k$$

$$\overline r=2(1-3\lambda)\overline i-(1+2\lambda)\overline j+(3-4\lambda)\overline k$$

$$\overline r=2(1-3\lambda)\overline i+(1+2\lambda)\overline j+(3-4\lambda)\overline k$$

$$\overline r=2(1+3\lambda)\overline i+(1+2\lambda)\overline j+(3+4\lambda)\overline k$$

Solution

Question 7

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 8

$$0,\cfrac { 1 }{ \sqrt { 2 } } ,\cfrac { 1 }{ 2 } $$

$$0,\cfrac { -\sqrt { 3 } }{ 2 } ,\cfrac { 1 }{ \sqrt 2 } $$

$$0,\cfrac { \sqrt { 3 } }{ 2 } ,\cfrac { 1 }{ { 2 } } $$

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

Solution

Question 9

$$\dfrac{1}{\sqrt[5]{2}}$$

$$\dfrac{1}{10}$$

$$\displaystyle\sqrt[5]{2}$$

5

Solution

Question 10

$$x + y + z = \sqrt {2}$$

$$x + y + z = 1$$

$$x + y = 2$$

$$x = \sqrt {2}$$

Solution

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