Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

If $$A(p,q,r)$$ and $$B=(p\prime ,q\prime ,r\prime )$$ are two points on the line $$\lambda x=\mu y=yz$$ such that $$OA=3,OB=4$$ then $$pp\prime +qq\prime +rr\prime $$ is equal to

A
$$7$$

B
$$12$$

C
$$5$$

D
$$None$$ $$of$$ $$these$$

Solution
Question 2
1 / -0

The angle between the lines whose de's satisfy the equation $$l+m+m=0$$ and $$l^2+m^{2}-n^{2}=0$$ is

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

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

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

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

Solution

Given that the equations
$$l+m+n=0$$ ………..$$(1)$$
$$l+m=-n$$
$$\Rightarrow -(l+m)=n$$
and
$$l^2+m^2+n^2=0$$ ……….$$(2)$$
Put the value of n in equation $$(2)$$
$$l^2+m^2+n^2=0$$
$$\Rightarrow l^2+m^2-(-(l+m))^2=0$$
$$\Rightarrow l^2+m^2-(l^2+m^2-2ml)=0$$
$$\Rightarrow l^2+m^2-l^2-m^2+2ml=0$$
$$\Rightarrow 2ml0$$
$$\Rightarrow ml=0$$
$$\Rightarrow m=0, l=0$$
Let us put $$m=0$$ in equation $$(3)$$
$$l+o+n=0$$
$$l=-n$$
Hence, direction rates$$(l, m, o)=(1, 0, -1)$$
Let us put $$l=0$$, we get $$m=-n$$
Here, direction ratios $$(l, m, n)=(0, 1, -1)$$
we know that,
$$\cos\theta =\dfrac{\vec{b_1}\cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|}$$
$$=\dfrac{(1, 0, -1)\cdot (0, 1, -1)}{\sqrt{1^2+0^2+(-1)^2}\sqrt{0^2+1^2+(-1)^2}}$$
$$=\dfrac{1}{\sqrt{2}\sqrt{2}}$$
$$\cos\theta =\dfrac{1}{2}$$
$$\cos\theta =\cos\dfrac{\pi}{3}$$
$$\theta =\dfrac{\pi}{3}$$
Hence, this is the answer.
Question 3
1 / -0

The angle between the lines, whose direction ratios are $$1,1,2$$ and $$\sqrt { 3 } - 1 , - \sqrt { 3 } - 1,4 ,$$ is

A
$${45} ^ { \circ }$$

B
$${30} ^ { \circ }$$

C
$${60} ^ { \circ }$$

D
$${90} ^ { \circ }$$

Solution

The direction ratios are $$(1, 1, 2)$$ and $$(\sqrt{3}-1, -\sqrt{3}-1, 4)$$
Angle between them is given by

$$\cos\theta =\dfrac{1(\sqrt{3}-1)+1(-\sqrt{3}-1)+2/4)}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2+4^2}}$$
$$=\dfrac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6}\sqrt{8+16}}$$
$$=\dfrac{6}{\sqrt{6}\sqrt{24}}=\dfrac{6}{\sqrt{144}}$$

$$=\dfrac{6}{12}$$
$$=\dfrac{1}{2}$$

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

$$\Rightarrow \theta =\dfrac{\pi}{3}=60^o$$.
Question 4
1 / -0

The directions cosines of the line which is perpedicular to the lines whose direction cosines are proportional to (1, -1, 2) and (2,-1,-1) are:

A
$$\dfrac { 1 }{ \sqrt { 35 } } ,-\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } } $$

B
$$-\dfrac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } \dfrac { 3 }{ \sqrt { 35 } } $$

C
$$\dfrac { 1 }{ \sqrt { 35 } } ,\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } } $$

D
None of these

Solution
Question 5
1 / -0

If a plane passes through the point $$(1, 1, 1)$$ and is perpendicular to the line $$\dfrac{x-1}{3}=\dfrac{y-1}{0}=\dfrac{z-1}{4}$$ then its perpendicular distance from the origin is

A
$$\dfrac{3}{4}$$

B
$$\dfrac{4}{3}$$

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

D
$$1$$

Solution

$$\begin{array}{l} Let\, eq{ u^{ n } }\, of\, plane\, be\, :\, a\left( { x-1 } \right) +b\left( { y-1 } \right) +\left( { z-1 } \right) =0 \\ 3\left( { x-1 } \right) +0\left( { y-1 } \right) +4\left( { z-1 } \right) =0 \\ 3x+4z=7 \\ perpendicular\, \, dis\tan ce\, from\, \, origin\, \, d=\frac { 7 }{ 5 } \\ Hence,\, the\, option\, C\, is\, correct\, answer. \end{array}$$
Question 6
1 / -0

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as $$l_{1},m_{1},n_{1};l_{2},m_{2},n_{2}$$ and $$l_{3},m_{3},n_{3}$$ are

A
$$l_{1}+ l_{2}+ l_{3},m_{1}+m_{2}+m_{3},n_{1}+n_{2}+n_{3}$$

B
$$\dfrac{l_{1}+l_{2}+l_{3}}{\sqrt{3}},\dfrac{m_{1}+m_{2}+m_{3}}{\sqrt{3}},\dfrac{n_{1}+n_{2}+n_{3}}{\sqrt{3}}$$

C
$$\dfrac{l_{1}+l_{2}+l_{3}}{3},\dfrac{m_{1}+m_{2}+m_{3}}{3},\dfrac{n_{1}+n_{2}+n_{3}}{3}$$

D
$$None\ of\ these$$
Question 7
1 / -0

$$l=m=n=1$$ represent the direction cosines of the

A
$$x-$$ axis

B
$$y-$$ axis

C
$$z-$$ axis

D
$$none\ of\ these$$

Solution

$$l=m=n=1$$
$$(1,1,1)$$ represent direction ratio of line equally inclined to $$x,y\, \&\,zaxis$$
Hence,
option $$D$$ is correct answer.
Question 8
1 / -0

If the points (p. 0), (0, q) and (1, 1) are collinear then $$\dfrac { 1 }{ p } +\dfrac { 1 }{ q } $$ is equal to

A
-1

B
1

C
2

D
0

Solution

If the area of triangle is zero, then the points are collinear.
Points are collinear.
Point are $$(p, o)(o, q)(i,q)$$

$$\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]$$
$$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$$
$$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$$

$$\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o$$
$$\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1$$

$$\Rightarrow \dfrac{1}{p}+\dfrac{1}{q}=1$$
Question 9
1 / -0

The direction ratios of the line
$$x-y+z-5=\quad 0\quad =\quad x-3y-6\quad are$$

A
$$3,1,-2$$

B
$$2,-4,1$$

C
$$\frac { 3 }{ \sqrt { 14 } } ,\frac { 1 }{ \sqrt { 14 } } ,\frac { -2 }{ \sqrt { 14 } } $$

D
$$\frac { 2 }{ \sqrt { 41 } } ,\frac { -4 }{ \sqrt { 41 } } ,\frac { 1 }{ \sqrt { 41 } } $$

Solution

$$x-y+z-5=0=x-3y-6$$

$$x-y+z-5=0$$--(I)
$$x-3y-6=0$$--(II)
The Direction ratios of the line is $$(\hat{i} - \hat{j} + \hat{k})\times(\hat{i} -3\hat{j}) $$
=$$3\hat{i}+\hat{j}-2\hat{k}$$
Hence,the direction ratios is (3,1,-2).
Hence, Option A is correct
Question 10
1 / -0

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as $$l_1$$, $$m_1$$, $$n_1$$ : $$l_2$$, $$m_2$$, $$n_2$$ and $$l_3$$, $$m_3$$, $$n_3$$ are

A
$$l_1$$+$$l_2$$+$$l_3$$, $$m_1$$+$$m_2$$+$$m_3$$, $$n_1$$+$$n_2$$+$$n_3$$

B
$$\dfrac{l_1+l_2+l_3}{\sqrt{3}} $$, $$\dfrac{m_1+m_2+m_3}{\sqrt{3}} $$, $$\dfrac{n_1+n_2+n_3}{\sqrt{3}} $$

C
$$\dfrac{l_1+l_2+l_3}{3} $$, $$\dfrac{m_1+m_2+m_3}{3} $$, $$\dfrac{n_1+n_2+n_3}{3} $$

D
None of these

Solution

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 50

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Three Dimensional Geometry Test - 50

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Question 1

$$7$$

$$12$$

$$5$$

$$None$$ $$of$$ $$these$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$${45} ^ { \circ }$$

$${30} ^ { \circ }$$

$${60} ^ { \circ }$$

$${90} ^ { \circ }$$

Solution

Question 4

$$\dfrac { 1 }{ \sqrt { 35 } } ,-\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } } $$

$$-\dfrac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } \dfrac { 3 }{ \sqrt { 35 } } $$

$$\dfrac { 1 }{ \sqrt { 35 } } ,\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } } $$

None of these

Solution

Question 5

$$\dfrac{3}{4}$$

$$\dfrac{4}{3}$$

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

$$1$$

Solution

Question 6

$$l_{1}+ l_{2}+ l_{3},m_{1}+m_{2}+m_{3},n_{1}+n_{2}+n_{3}$$

$$\dfrac{l_{1}+l_{2}+l_{3}}{\sqrt{3}},\dfrac{m_{1}+m_{2}+m_{3}}{\sqrt{3}},\dfrac{n_{1}+n_{2}+n_{3}}{\sqrt{3}}$$

$$\dfrac{l_{1}+l_{2}+l_{3}}{3},\dfrac{m_{1}+m_{2}+m_{3}}{3},\dfrac{n_{1}+n_{2}+n_{3}}{3}$$

$$None\ of\ these$$

Question 7

$$x-$$ axis

$$y-$$ axis

$$z-$$ axis

$$none\ of\ these$$

Solution

Question 8

-1

1

2

0

Solution

Question 9

$$3,1,-2$$

$$2,-4,1$$

$$\frac { 3 }{ \sqrt { 14 } } ,\frac { 1 }{ \sqrt { 14 } } ,\frac { -2 }{ \sqrt { 14 } } $$

$$\frac { 2 }{ \sqrt { 41 } } ,\frac { -4 }{ \sqrt { 41 } } ,\frac { 1 }{ \sqrt { 41 } } $$

Solution

Question 10

$$l_1$$+$$l_2$$+$$l_3$$, $$m_1$$+$$m_2$$+$$m_3$$, $$n_1$$+$$n_2$$+$$n_3$$

$$\dfrac{l_1+l_2+l_3}{\sqrt{3}} $$, $$\dfrac{m_1+m_2+m_3}{\sqrt{3}} $$, $$\dfrac{n_1+n_2+n_3}{\sqrt{3}} $$

$$\dfrac{l_1+l_2+l_3}{3} $$, $$\dfrac{m_1+m_2+m_3}{3} $$, $$\dfrac{n_1+n_2+n_3}{3} $$

None of these

Solution

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