Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$\overline{a}$$ and $$\overline{b}$$ are two non-collinear vectors, then the points $$l_{1}\overline{a}+m_{1}\overline{b}$$, $$ l_{2}\overline{a}+m_{2}\overline{b}$$ and $$l_{3}\overline{a}+m_{3}\overline{b}$$ are collinear if

A
$$\displaystyle \sum l_{1}(m_{2}-m_{3}) =0$$

B
$$\displaystyle \sum l_{1}(m_{1}-m_{3})=0$$

C
$$\displaystyle \sum l_{1}(m_{1}+m_{3})=0$$

D
$$\displaystyle \sum l_{1}(m_{2}+m_{3})=0$$

Solution

$$\bar { a } $$ and $$\bar { b }$$ are non-collinear

$$\triangle \quad =\quad \left| \begin{matrix} 1\quad & 1\quad & 1 \\ { l }_{ 1 }\quad & { l }_{ 2 }\quad & { l }_{ 3 } \\ { m }_{ 1 }\quad & { m }_{ 2 }\quad & { m }_{ 3 } \end{matrix} \right| =0$$
$$ { c }_{ 1 }\rightarrow { c }_{ 1 }-{ c }_{ 2 }\\ { c }_{ 3 }\rightarrow { c }_{ 1 }-{ c }_{ 3 }\\ \triangle \quad =\quad \left| \begin{matrix} 0 & 1 & 0 \\ { { l }_{ 1 }-{ l }_{ 2 }\quad } & { l }_{ 2 }\quad & { l }_{ 1 }-{ l }_{ 3 } \\ { m }_{ 1 }-{ m }_{ 2 }\quad & { m }_{ 2 }\quad & { m }_{ 1 }-{ m }_{ 3 } \end{matrix} \right|=0$$
$$\triangle \quad =\quad -({ l }_{ 1 }-{ l }_{ 2 })({ m }_{ 1 }-{ m }_{ 3 })+({ l }_{ 1 }-{ l }_{ 3 })({ m }_{ 1 }-{ m }_{ 2 })=0$$
$$\Rightarrow \displaystyle\sum{ l }_{ 1 }({ m }_{ 2 }-{ m }_{ 3 })\quad =\quad 0$$
Question 2
1 / -0

A line makes an angle $$\alpha,\beta,\gamma$$ with the $$X,Y,Z$$ axes. Then $$\sin^2\alpha+\sin^2\beta+\sin^2\gamma=$$

A
$$1$$

B
$$2$$

C
$$\dfrac 32$$

D
$$4$$

Solution

For a vector
$$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$$
$$1-\sin^2(\alpha)+1-\sin^2(\beta)+1-\sin^2(\gamma)=1$$
$$\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma)=3-1=2$$

Option B
Question 3
1 / -0

The angle made by line $$r[cos \theta - \sqrt{3} sin \theta ]=5$$ with initial line is

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

Given equation
$$r[cos \theta - \sqrt{3} sin \theta ]=5$$
$$x-\sqrt{3}y=5$$
Slope of the line is $$\tan\theta=\frac{1}{\sqrt{3}}$$
$$\Rightarrow \theta={30}^{0}$$
Hence, the line makes an angle of $$30^{0}$$ with the initial line.
Question 4
1 / -0

The line passing through the points $$10\hat{i}+3\hat{j}$$, $$ 12\hat{i}+5\hat{j}$$ also passes through the point $$a\hat{i}+11 \hat{j}$$, then $$a=$$

A
$$-8$$

B
$$4$$

C
$$18$$

D
$$12$$

Solution

Let $$\vec{v}=10\hat{i}+3\hat{j}$$
$$\vec{u}=12\hat{i}+5\hat{j}$$
$$\vec{w}=a\hat{i}+11\hat{j}$$
As $$\vec{u},\vec{v},\vec{w}$$ are on a line then $$\vec{u}-\vec{v}$$ is collinear with $$\vec{w}-\vec{u}$$
$$2\hat{i}+2\hat{j}=k((a-12)\hat{i}+6\hat{j})$$
$$2 =6k$$, $$2$$$$=k(a-12)$$
$$k=\dfrac{1}{3}$$, $$a=18$$
Question 5
1 / -0

The position vectors of three points are $$2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } ,\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } $$ and $$\mu \overrightarrow { a } -5\overrightarrow { b } $$, where $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are non-coplanar vectors. The points are coliinear when

A
$$\displaystyle\lambda =-2,\mu =\frac { 9 }{ 4 } $$

B
$$\displaystyle\lambda =-\frac { 9 }{ 4 } ,\mu =2$$

C
$$\displaystyle\lambda =\frac { 9 }{ 4 } ,\mu =-2$$

D
None of these

Solution

Let the points $$A,B$$ and $$C$$ respectively, then
$$\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } \\ =\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } \right) \\ =-\overrightarrow { a } -\overrightarrow { b } +\left( \lambda -3 \right) \overrightarrow { c } $$
$$\overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } \\ =\mu \overrightarrow { a } -5\overrightarrow { b } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } \right) \\ =\left( \mu -2 \right) \overrightarrow { a } -4\overrightarrow { b } -3\overrightarrow { c } $$
The points are collinear if $$\overrightarrow { AB } =t\overrightarrow { AC } $$
$$\Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$$
$$\displaystyle \Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$$ and $$\displaystyle -1=\frac { -1 }{ 4 } \left( \mu -2 \right) ,\lambda -3=\frac { 3 }{ 4 } $$
$$\displaystyle \Rightarrow \mu =6,\lambda =\frac {15 }{ 4 } $$
Question 6
1 / -0

The vectors $$2\hat{i}+3\hat{j};5\hat{i}+6\hat{j};8\hat{i}+\lambda\hat{j}$$ have their initial points at $$(1,1 )$$. The value of $$\lambda$$ so that the vectors terminate on one straight line is

A
$$0$$

B
$$3$$

C
$$6$$

D
$$9$$

Solution

$$\vec{V}= 2\hat{i}+3\vec{j}$$
$$\vec{U}= 5\hat{i}+6\vec{j}$$
$$\vec{W}= 8\hat{i}+\lambda \vec{j}$$
$$\vec{V}-\vec{U}= k(\vec{W}-\vec{U})$$ (as they are collnear)
$$(-3\hat{i}-3\hat{j})= k(3\vec{i}+(\lambda -6)\hat{j})$$)
as $$\hat{i}$$ & $$j $$ are non collienar
$$-3=3k$$
$$k=-1$$
$$k(\lambda -6)= -3$$
$$\lambda = 9$$
Question 7
1 / -0

The point collinear with $$(1, -2, -3)$$ and $$(2, 0, 0)$$ among the following is

A
$$(0,4,6)$$

B
$$(0,-4,-5)$$

C
$$(0, -4, -6)$$

D
$$(0,-4,6)$$

Solution

Any point lying on the line joining $$(1,-2,-3)$$ and $$(2,0,0)$$ can be written in the form of $$(1+k, -2+2k, -3+3k)$$.
Since, all the options have their $$x$$ coordinate as $$0$$,
$$1+k = 0$$
$$k = -1$$
Thus, $$y=-2 + 2k = -4$$ and $$z=-3+3k = -6$$
Hence, option C is correct.
Question 8
1 / -0

If the points $$(h, 3, -4), (0, -7, 10)$$ and $$(1, k, 3)$$ are collinear, then $$h + k$$ is

A
$$4$$

B
$$0$$

C
$$-4$$

D
$$14$$

Solution

Any point lying on the line joining $$(h,3,-4)$$ and $$(0,-7,10)$$ can be written in the form of $$(h-hl,3-10l,-4+14l)$$, where $$l$$ is a constant.
Since, $$(1,k,3)$$ is also a point on this line, it should satisfy this form.
$$\Rightarrow -4+14l = 3$$.
$$\therefore l =$$ $$ \dfrac{1}{2} $$
$$\Rightarrow k = 3-10l = -2 $$
$$\Rightarrow h(1-l) = 1 $$
$$\therefore h = 2$$
Hence, $$h + k = 2-2 = 0$$
Question 9
1 / -0

Assertion ($$A$$): The points with position vectors $$\overline{a},\overline{b},\overline{c}$$ are collinear if $$2\overline{a}-7\overline{b}+5\overline{c}=0$$.
Reason ($$R$$): The points with position vectors $$\overline{a},\overline{b},\overline{c}$$ are collinear if $$l\overline{a}+m\overline{b}+n\overline{c}=\overline{0}$$.

A
Both $$A$$ and $$R$$ are true and $$R$$ is correct reason of $$A$$

B
Both $$A$$ and $$R$$ are true and $$R$$ is not correct reason of $$A$$

C
$$A$$ is true $$R$$ is false

D
$$A$$ is false $$R$$ is true

Solution

$$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are collinear
$$\Rightarrow \ni l,m,n$$ all zeros such that
$$l\overrightarrow { a } +m\overrightarrow { b } +n\overrightarrow { c } =0$$ if $$2\overrightarrow { a } -7\overrightarrow { b } +5\overrightarrow { c } =0$$
then $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are collinear since $$2-7+5=0$$
Question 10
1 / -0

If the points whose position vectors are $$2\overline{i}+\overline{j}+\overline{k},\ 6\overline{i}-\overline{j}+2\overline{k}$$ and $$14\overline{i}-5\overline{j}+p\overline{k}$$ are collinear then the value of $$\mathrm{p}$$ is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

$$2\hat{i}+\hat{j}+\hat{k}=\lambda (6\hat{i}-\hat{j}+2\hat{k})+\mu (14\hat{i}-5\hat{j}+p\hat{k})$$ (Property of Collinear)
$$2=b\lambda +14 \mu$$
$$1=-\lambda -5\mu$$
$$0=8\lambda+24\mu$$
$$\dfrac{-\lambda}{3}= \mu$$
$$\lambda =\dfrac{3}{2}\ \ \ \ \mu=\dfrac{-1}{2}$$
$$1=2\lambda + p \mu$$
$$-2= \dfrac{-p}{2}$$
$$p= 4$$

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Three Dimensional Geometry Test - 20

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Three Dimensional Geometry Test - 20

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Question 1

$$\displaystyle \sum l_{1}(m_{2}-m_{3}) =0$$

$$\displaystyle \sum l_{1}(m_{1}-m_{3})=0$$

$$\displaystyle \sum l_{1}(m_{1}+m_{3})=0$$

$$\displaystyle \sum l_{1}(m_{2}+m_{3})=0$$

Solution

Question 2

$$1$$

$$2$$

$$\dfrac 32$$

$$4$$

Solution

Question 3

$$30^{\circ}$$

$$45^{\circ}$$

$$60^{\circ}$$

$$90^{\circ}$$

Solution

Question 4

$$-8$$

$$4$$

$$18$$

$$12$$

Solution

Question 5

$$\displaystyle\lambda =-2,\mu =\frac { 9 }{ 4 } $$

$$\displaystyle\lambda =-\frac { 9 }{ 4 } ,\mu =2$$

$$\displaystyle\lambda =\frac { 9 }{ 4 } ,\mu =-2$$

None of these

Solution

Question 6

$$0$$

$$3$$

$$6$$

$$9$$

Solution

Question 7

$$(0,4,6)$$

$$(0,-4,-5)$$

$$(0, -4, -6)$$

$$(0,-4,6)$$

Solution

Question 8

$$4$$

$$0$$

$$-4$$

$$14$$

Solution

Question 9

Both $$A$$ and $$R$$ are true and $$R$$ is correct reason of $$A$$

Both $$A$$ and $$R$$ are true and $$R$$ is not correct reason of $$A$$

$$A$$ is true $$R$$ is false

$$A$$ is false $$R$$ is true

Solution

Question 10

$$2$$

$$4$$

$$6$$

$$8$$

Solution

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