Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 45

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Question 1
1 / -0

A line passes through the points $$(6,-7,-1)$$ and $$(2,-3, 1)$$. If the angle a which the line makes with the positive direction of x-axis is acute, the direction cosines of the line are.

A
$$(2/3),(-2/3),(-1/3)$$

B
$$(2/3),(2/3), (-1/3)$$

C
$$(2/3), (-7/3), (1/3)$$

D
$$(8/3), (2/3), (1/3)$$

Solution

Direction Ratios of line joining $$(6,-7,-1)$$ and $$(2,-3,1)$$ are

$$\rightarrow (6,-2,-7-(-3)-(-1))$$

$$\rightarrow (4,-4,-2)$$

if it make acute angle with positive x-axis then $$D.R(x)>0$$

D.C are $$\left(\dfrac{4}{\sqrt{4^2+4^2+2^2}},\dfrac{-4}{4^2+4^2+2^2},\dfrac{-2}{4^2+4^2+2^2}\right)$$

$$\left(\dfrac{4}{6},\dfrac{-4}{6},\dfrac{-2}{6}\right)$$

$$\left(\dfrac{2}{3},\dfrac{-2}{3},\dfrac{-1}{3}\right)$$
$$A$$ is correct.
Question 2
1 / -0

Prove that the points $$A=(1,2,3),B(3,4,7),C(-3,-2,-5)$$ are collinear & find the ratio in which $$B$$ divides $$AC$$.

A
$$2:5$$

B
$$2:3$$

C
$$2:8$$

D
$$2:7$$

Solution

$$A(1,2,3),B(3,4,7),C(-3,-2,-5)$$
$$\overrightarrow { AB } =\,\,2\hat{i}+2\hat{j}+4\hat{k}$$
$$\overrightarrow {BC}\,\,-6\hat{i}-6\hat{j}-12\hat{k}$$
If points are collinear then $$|(\overrightarrow {AB } ).(\overrightarrow { BC } )|=|\overrightarrow {AB } ||\overrightarrow {BC } |$$
$$\overrightarrow {AB } .\overrightarrow {AC } =-12-12-24=-48$$
$$|\overrightarrow { AB } .\overrightarrow {AC } |=48$$
$$|\overrightarrow { AB } |.|\overrightarrow {AC } |=\sqrt{24}\sqrt{216}=2\sqrt{24}\sqrt{24}=48$$
Points are collinear
$$|\overrightarrow {AC } |=|-4\hat{i}-4\hat{j}-8\hat{k}|=4\sqrt{6}$$
$$|\overrightarrow {AB } |=2\sqrt{6}$$
$$B$$ divides $$AC$$ in the ration $$2:3$$
Question 3
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Equation of pair of lines passing through origin and making and angle $${\tan ^{ - 1}}2$$ with the lines $$4x-3y+7=0$$.

A
$$
{\left( {4x - 3y} \right)^2} - 4{\left( {3x + 4y} \right)^2} = 0$$

B
$${\left( {4x - 3y} \right)^2} - {\left( {3x + 4y} \right)^2} = 0$$

C
$${\left( {4x - 3y} \right)^2} - 3{\left( {3x + 4y} \right)^2} = 0$$

D
$$4{\left( {4x - 3y} \right)^2} - {\left( {3x + 4y} \right)^2} = 0$$

Solution

Slope of the line $$4x-3y+7=0$$ is $${m}_{1}=-\dfrac{coeff.\,of\,x}{coeff.\,of\,y}=\dfrac{4}{3}$$
$$\tan{\theta}=\left|\dfrac{m-{m}_{1}}{1+m{m}_{1}}\right|$$

Given: $${\tan}^{-1}{2}=\theta\Rightarrow\,\tan{\theta}=2$$
$$\Rightarrow\,2=\left|\dfrac{m-\dfrac{4}{3}}{1+\dfrac{4m}{3}}\right|$$
$$\Rightarrow\,\left|\dfrac{3m-4}{3+4m}\right|=2\\$$
$$\Rightarrow\,\dfrac{3m-4}{3+4m}=\pm\,2\\$$
$$\Rightarrow\,3m-4=\pm\,2\left(4m+3\right)$$
$$\Rightarrow\,3m-4=8m+6$$ and $$3m-4=-8m-6$$
$$\Rightarrow\,-5m=10$$ and $$11m=-2$$
$$\Rightarrow\,m=-2$$ and $$m=\dfrac{-2}{11}$$
Equation is $$y=-2x$$ and $$y=\dfrac{-2}{11}x$$
$$\Rightarrow\,2x+y=0$$ and $$2x+11y=0$$

Pair of lines is $$\left(2x+y\right)\left(2x+11y\right)=0$$
or $$4{x}^{2}+22xy+2xy+11{y}^{2}=0$$
or $$4{x}^{2}+24xy+11{y}^{2}=0$$ is the required equation.
Which is equivalent to equation given in option A
$${\left( {4x - 3y} \right)^2} - 4{\left( {3x + 4y} \right)^2} = 0$$
Question 4
1 / -0

If $$\vec { a } ,\vec { b } ,\vec { c } $$ are three non-zero vectors, no two of which are collinear and the vector $$\vec { a } +\vec { b } $$ is collinear with $$\vec { c }, \vec { b } +\vec { c } $$ is collinear with $$\vec {a},$$ then $$\vec { a } +\vec { b } +\vec { c }$$ is equal to -

A
$$\vec {a}$$

B
$$\vec {b}$$

C
$$\vec {c}$$

D
$$none\ of\ these$$

Solution

$$\bar a+\bar b=K_{1}\bar c $$

$$\bar b+\bar c=K_{2}\bar a $$

$$\bar a- \bar c=K_{1}c-K_{2}\bar a$$

$$(k_{2}+1) \bar a-\bar c(1+k_{1})$$=0

$$k_{2}=-1 $$ and $$k_{1}=-1 $$

$$\bar{a}+\bar{b}+\bar{c}=0$$
Question 5
1 / -0

Find the equation of the plane if the foot of the perpendicular from origin to the plane is $$ (2, 3, -5 ) $$

A
$$2x+3y+5y=38$$

B
$$2x+3y-5y=38$$

C
$$2x-3y-5y=38$$

D
None of these

Solution

$$\textbf{Step -1: Find a,b,c}$$
$$\text{The line is passing through }(2,3,-5)\text{ and let the direction ratios of the line be a,b,c}$$
$$\text{Then the equation of the line is}$$
$$\dfrac{x-2}{a}=\dfrac{y-3}{b}=\dfrac{z-(-5)}{c}\Rightarrow{\dfrac{x-2}{a}=\dfrac{y-3}{b}=\dfrac{z+4}{c}=k~(\text{let})}$$
$$\dfrac{x-2}{a}=k\Rightarrow{x-2=ak}\Rightarrow{x=ak+2}$$
$$\dfrac{y-3}{b}=k\Rightarrow{y-3=bk}\Rightarrow{y=bk+3}$$
$$\dfrac{z+5}{c}=k\Rightarrow{z+5=ck}\Rightarrow{z=ck-5}$$
$$\text{Since the line passes through the origin i.e. }(0,0,0)\text{ then,}$$
$$\text{ak+2=0}\Rightarrow{a=-\dfrac{2}{k}}$$
$$\text{bk+3=0}\Rightarrow{b=-\dfrac{3}{k}}$$
$$\text{ck-5=0}\Rightarrow{c=\dfrac{5}{k}}$$
$$\textbf{Step -2: Find the equation of plane.}$$
$$\text{Then the equation of required plane is,}$$
$$ax+by+cz=d\Rightarrow{\left(-\dfrac{2}{k}\right)x+\left(-\dfrac{3}{k}\right)y+\left(\dfrac{5}{k}\right)z=d}\Rightarrow{-2x-3y+5z=dk}$$
$$\text{The foot of the perpendicular is }(2,3,-5)\text{ then it satisfies the equation of the plane.}$$
$$\therefore{-2\times2+-3\times3+5\times-5=dk}\Rightarrow{dk=-4-9-25=-38}$$
$$\text{Then the equation of the required plane is,}$$
$$-2x-3y+5z=dk\Rightarrow{-2x-3y+5z=-38}$$
$$\therefore{2x+3y-5z=38}$$
$$\textbf{Hence, option (B) 2x+3y-5z=38}\textbf{ is correct answer.}$$
Question 6
1 / -0

If the points with position vectors $$10\hat { i } +\lambda \hat { j } ,3\hat { i } -\hat { j } $$ and $$4\hat { i } +5\hat { j } $$ are collinear then $$\lambda$$ is

A
$$41$$

B
$$-41$$

C
$$42$$

D
$$-42$$

Solution

position vectors : $$ (10\hat{i}+\lambda \hat{j}) a, (3\hat{i}-\hat{j})b, (4\hat{i}+5\hat{j})c$$
since they are collinear, so
$$ \vec{AB}$$ must parallel to $$ \vec{BC}$$
$$ \therefore \vec{AB} = b-a = (3-10)\hat{i}+(-1-\lambda)\hat{j})$$
$$ = (-7\hat{i}+(-1-\lambda )\hat{j})$$
$$ \vec{BC} = \bar{C}-a = (4-3)\hat{i}+(5-(-1)) \hat{j}$$
$$ = \hat{i}+6\hat{j}$$
$$ \therefore \dfrac{-7}{1} = \dfrac{-1-\lambda }{6}$$
$$ \therefore -42 = -1-\lambda $$
$$ \lambda = -1+42 $$
$$ \lambda = 41 $$ (option A)
Question 7
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Point $$\left(\alpha,\beta,\gamma\right)$$ lies on the plane $$x+y+z=2$$. Let $$\overrightarrow{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$$ and $$\hat{k}\times \left(\hat{k}\times \overrightarrow{a}\right)=0$$ then $$\gamma=$$

A
$$0$$

B
$$1$$

C
$$2$$

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

Solution

$$\hat{k}\times \left(\hat{k}\times \overrightarrow{a}\right)=\left(\hat{k}.\overrightarrow{a}\right)\hat{k}-\left(\hat{k}.\hat{k}\right)\overrightarrow{a}$$
$$ =\gamma\hat{k}-\left(\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}\right)=0$$
$$\Rightarrow \alpha\hat{i}+\beta\hat{j}=0$$
$$\therefore \alpha=\beta=0$$
$$\left(\alpha,\beta,\gamma\right)$$ lies on the plane $$x+y+z=2 \left(given\right)$$
$$\therefore \alpha+\beta+\gamma=2 \Rightarrow \gamma=2$$ where $$\alpha=\beta=0$$
Question 8
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If the points with position vectors $$60\hat{i}+3\hat{j}, 40\hat{i}-8\hat{j}$$ and $$a\hat{i}-52j$$ are collinear, then $$a=?$$

A
$$-40$$

B
$$-20$$

C
$$20$$

D
$$40$$

Solution

Suppose, position vector $$A=60\widehat i+3\widehat j$$

position vector $$B=40\widehat i-8\widehat j$$

position vector $$C=a\widehat i-52\widehat j$$

Now, find vector AB and BC

$$AB= -20\widehat i-11\widehat j$$

$$BC= (a-40)\widehat i-44\widehat j$$

To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.

That’s why the cross product of the vectors should be zero

$$ABXBC=(-20\widehat i-11\widehat j)X(a-40)\widehat i-44\widehat j$$

$$0\widehat i+0\widehat j+(880+11(a-40))=0$$

$$a-40= -80$$

$$a=-40$$

Therefore, a should be $$-40$$ to be the given positions vectors collinear.
Question 9
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If $$A ( 2 \overline{i} - \overline{j} - 3 \overline{k} , B ( 4 \overline {i} + \overline{j} - \overline{k} )$$ and $$ D( \overline{i} - \overline{j} - 2 \overline{k})$$ then the vector equation of the plane parallel to $$ \overline{ABC} $$ and passing through the centroid of the tetrahedron $$ABCD$$ is :

A
$$ \overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} + s ( 2 \overline{i} + 2 \overline{j} + 2 \overline{k} ) + t (\overline{i} - \overline{k} )$$

B
$$ \overline{r} = ( 2 \overline{i} - \overline{j} -3 \overline{k})+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} - \overline{k} )$$

C
$$ \overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} )+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} + \overline{j} -5 \overline{k} )$$

D
$$ \overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} )+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} + \overline{j} +5 \overline{k} )$$

Solution
Question 10
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The distance of the point $$3\hat{i}+5\hat{k}$$ from the line parallel to the vector $$6\hat{i}+\hat{j}-2\hat{k}$$ and passing through the point $$8\hat{i}+3\hat{j}+\hat{k}$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

The distance of the point $$\overrightarrow{c}$$ from the line
$$\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}$$ is $$\dfrac{\left|\left(\overrightarrow{c}-\overrightarrow{a}\right)\times \overrightarrow{b}\right|} {\left|\overrightarrow{b}\right|}$$
$$\overrightarrow{c}=3\hat{i}+5\hat{k},\overrightarrow{a}=8\hat{i}+3\hat{j}+\hat{k},\overrightarrow{b}=6\hat{i}+\hat{j}-2\hat{k}$$
$$\left(\overrightarrow{c}-\overrightarrow{a}\right)\times \overrightarrow{b}=\left|\begin{matrix} \hat{i}&\hat{j}&\hat{k} \\-5&-3&4 \\6&1&-2\end{matrix}\right|$$
$$=\hat{i}\left(6-4\right)-\hat{j}\left(10-24\right)+\hat{k}\left(-5+18\right)$$
$$ =\hat{i}\left(2\right)-\hat{j}\left(-14\right)+\hat{k}\left(13\right)$$
$$ =2\hat{i}+14\hat{j}+13\hat{k}$$
Distance $$ =\dfrac{\sqrt{{2}^{2}+{14}^{2}+{13}^{2}}} {\sqrt{{6}^{2}+{1}^{2}+{2}^{2}}}=\sqrt{\dfrac{369}{41}}=3$$ on simplification

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 45

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Three Dimensional Geometry Test - 45

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SHARING IS CARING

Question 1

$$(2/3),(-2/3),(-1/3)$$

$$(2/3),(2/3), (-1/3)$$

$$(2/3), (-7/3), (1/3)$$

$$(8/3), (2/3), (1/3)$$

Solution

Question 2

$$2:5$$

$$2:3$$

$$2:8$$

$$2:7$$

Solution

Question 3

$${\left( {4x - 3y} \right)^2} - 4{\left( {3x + 4y} \right)^2} = 0$$

$${\left( {4x - 3y} \right)^2} - {\left( {3x + 4y} \right)^2} = 0$$

$${\left( {4x - 3y} \right)^2} - 3{\left( {3x + 4y} \right)^2} = 0$$

$$4{\left( {4x - 3y} \right)^2} - {\left( {3x + 4y} \right)^2} = 0$$

Solution

Question 4

$$\vec {a}$$

$$\vec {b}$$

$$\vec {c}$$

$$none\ of\ these$$

Solution

Question 5

$$2x+3y+5y=38$$

$$2x+3y-5y=38$$

$$2x-3y-5y=38$$

None of these

Solution

Question 6

$$41$$

$$-41$$

$$42$$

$$-42$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

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

Solution

Question 8

$$-40$$

$$-20$$

$$20$$

$$40$$

Solution

Question 9

$$ \overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} + s ( 2 \overline{i} + 2 \overline{j} + 2 \overline{k} ) + t (\overline{i} - \overline{k} )$$

$$ \overline{r} = ( 2 \overline{i} - \overline{j} -3 \overline{k})+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} - \overline{k} )$$

$$ \overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} )+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} + \overline{j} -5 \overline{k} )$$

$$ \overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} )+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} + \overline{j} +5 \overline{k} )$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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$$
{\left( {4x - 3y} \right)^2} - 4{\left( {3x + 4y} \right)^2} = 0$$