Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The projection of a directed line segment on the co-ordinate axes are $$12,4,3$$, then the direction cosines of the line are

A
$$\displaystyle \dfrac{-12}{13},\dfrac{-4}{13},\dfrac{-3}{13}$$

B
$$\displaystyle \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$$

C
$$\displaystyle \dfrac{12}{13},\dfrac{-4}{13},\dfrac{3}{13}$$

D
$$\displaystyle \dfrac{12}{13},\dfrac{4}{13},\dfrac{-3}{13}$$

Solution

$$x=12,y=4,z=3$$

Direction cosines $$\displaystyle =\dfrac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} ,\displaystyle \dfrac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\dfrac{z}{\sqrt{x^{2}+y^{2}+z^{2}}} $$

$$\displaystyle = \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$$

Hence, option 'B' is correct.
Question 2
1 / -0

A line OP where O $$=$$ $$(0, 0, 0)$$ makes equal angles with ox, oy, oz. The point on OP, which is at a distance of $$6$$ units from O is:

A
$$(\displaystyle \dfrac{6}{\sqrt{3}}, \dfrac{6}{\sqrt{3}}, \dfrac{6}{\sqrt{3}})$$

B
$$(2\sqrt{3}, -2\sqrt{3},2\sqrt{3})$$

C
$$-(2\sqrt{3},2\sqrt{3},2\sqrt{3})$$

D
$$(6\sqrt{3},6\sqrt{3},6\sqrt{3})$$

Solution

Let $$l,m,n$$ be the direction cosines.
We have $$l^2+m^2+n^2=1$$.
Given that $$l=m=n$$ $$\Rightarrow 3l^2=1$$.
Solving we get $$l=\dfrac{1}{\sqrt{3}}$$.
Any point on the line is a multiple of direction cosines.
Let the point P is given by $$(\dfrac{k}{\sqrt{3}},\dfrac{k}{\sqrt{3}},\dfrac{k}{\sqrt{3}})$$.
Using the distance formula we get, $$k=6$$.
Question 3
1 / -0

The direction ratios of a normal to the plane through $$(1, 0, 0)$$ and $$(0, 1, 0)$$, which makes an angle of $$\dfrac {\pi}{4}$$ with the plane $$x+y=3$$, are:

A
$$< 1, \sqrt 2, 1 >$$

B
$$< 1, 1, \sqrt 2 >$$

C
$$< 1, 1, 2 >$$

D
$$< \sqrt 2, 1, 1 >$$

Solution

Let $$A(1,0,0)$$ & $$B(0,1,0)$$ be two points on the plane and direction ratio of its normal be $$ai+bj+ck$$, then $$(ai+bj+ck). \vec{BA}=0$$
$$\Rightarrow (ai+bj+ck).(i-j)=0$$
$$\Rightarrow a=b$$ ....(1)
Since, $$ai+bj+ck$$ makes an angle $$\dfrac {\pi}{4}$$ with $$x+y-3=0$$
Therefore, $$\cos { \dfrac { \pi }{ 4 } } =\dfrac { \left( ai+bj+ck \right) .\left( i+j \right) }{ \sqrt { \left( a^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) \left( { 1 }^{ 2 }+{ 1 }^{ 2 } \right) } } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 2 } } =\dfrac { a+b }{ \sqrt { 2\left( a^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) } } =\dfrac { 2a }{ \sqrt { 2\left( 2a^{ 2 }+{ c }^{ 2 } \right) } } $$ ....[ from (1) ]
$$\Rightarrow 2a^{ 2 }+{ { c }^{ 2 } }=4{ a }^{ 2 }$$
$$\Rightarrow c=\pm \sqrt { 2 } a$$
Therefore, direction ratios are $$\left( a,b,c \right) =\left( a,a,\pm \sqrt { 2 } a \right) =\left( 1,1,\pm \sqrt { 2 } \right) $$

Ans: B
Question 4
1 / -0

What is the equation of the plane which passes through the z-axis and its perpendicular to the line $$\dfrac {x-a}{cos\theta}=\dfrac {y+2}{sin\theta}=\dfrac {z-3}{0} ?$$

A
$$x+y tan\theta=0$$

B
$$y+xtan\theta=0$$

C
$$x cos\theta-y sin\theta=0$$

D
$$x sin\theta-y cos\theta=0$$

Solution

Given plane is passing through z-axis, $$i.e$$ it also passes through origin
Hence equation of plane passing through origin and perpendicular to line
$$\dfrac{x-a}{cos\theta}=\dfrac{y+2}{sin\theta}=\dfrac{z-3}{0}$$ is given by,
$$cos\theta (x)+sin\theta(y)+0(z)=0$$
$$\Rightarrow x+y tan\theta =0$$
Question 5
1 / -0

If a line makes an angle of $$\dfrac {\pi}{4}$$ with the positive direction of each of $$x$$-axis and $$y$$-axis, then the angle that the line makes with the positive direction of $$z$$-axis is-

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

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

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

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

Solution

let $$\alpha$$, $$\beta$$ and $$\gamma$$ be the angle made by the line with co-ordinate axes.
Therefore, $$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =1$$
$$\Rightarrow \cos ^{ 2 }{ \dfrac { \pi }{ 4 } } +\cos ^{ 2 }{ \dfrac { \pi }{ 4 } } +\cos ^{ 2 }{ \gamma } =1$$
$$\Rightarrow \cos ^{ 2 }{ \gamma } =0$$

$$\Rightarrow \gamma =\dfrac { \pi }{ 2 } $$

Ans: C
Question 6
1 / -0

The equation of the plane passing through the lines $$\frac {x-4}{1}=\frac {y-3}{1}=\frac {z-2}{2}$$ and $$\frac {x-3}{1}=\frac {y-2}{-4}=\frac {z}{5}$$ is-

A
$$11x-y-3z=35$$

B
$$11x+y-3z=35$$

C
$$11x-y+3z=35$$

D
None of these.

Solution

Normal vector of the plane is given by,
$$\vec{n} = \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\1&1&2\\1&-4&5 \end{vmatrix}=9\hat{i}-3\hat{j}-5\hat{k}$$
Hence equation of plane is,
$$ (\vec{r} - (3\hat{i}+2\hat{j}))\cdot \vec{n}=0$$
$$\Rightarrow 9x-3y-5z-21=0$$
Question 7
1 / -0

A line makes an angle $$\theta$$ with each of the $$x$$- and $$z$$- axes. If the angle $$\beta$$, which it makes with the $$y$$-axis, is such that $$\sin^2\beta=3 \sin^2\theta$$, then $$\cos^2\theta$$ equals-

A
$$\dfrac {2}{3}$$

B
$$\dfrac {1}{5}$$

C
$$\dfrac {3}{5}$$

D
$$\dfrac {2}{5}$$

Solution

Given that $$\sin^2\beta=3 \sin^2\theta$$ ...$$(1)$$
Let $$\theta$$, $$\theta$$ and $$\beta$$ be the angle made by the line with co-ordinate axes.
Therefore, $$\cos ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +\cos ^{ 2 }{ \beta } =1$$
$$\Rightarrow 2\cos ^{ 2 }{ \theta }=\sin ^{ 2 }{ \beta }=3\sin ^{ 2 }{ \theta }=3-3\cos ^{ 2 }{ \theta }$$ from $$(1)$$
Therefore, $$\cos ^{ 2 }{ \theta }=\dfrac {3}{5}$$

Ans: C
Question 8
1 / -0

For waht value of $$\lambda$$ , the three numbers $$2\lambda - 1 , \frac{1}{4}, \lambda -\frac{1}{2}$$ can be the direction cosines of a straight line?

A
$$\displaystyle \frac{1}{2} \pm \frac{{\sqrt 3 }}{4}$$

B
$$\displaystyle \frac{3}{4}$$

C
$$\displaystyle \pm \frac{3}{4}$$

D
$$\displaystyle \frac{{\sqrt 3 }}{2} \pm \frac{1}{4}$$

Solution

for being direction cosines of straight line,
$$(2\lambda -1 )^2 + \left ( \dfrac{1}{4} \right )^2+ \left ( \lambda -\dfrac{1}{2} \right )^2 =1$$
$$4\lambda^2 -4 \lambda + y+\dfrac{1}{16} + \lambda^2-\lambda +\dfrac{1}{4} =\lambda$$
$$5\lambda^2 -5 \lambda + \dfrac{5}{16}=0$$
$$16 \lambda^2 -16 \lambda +1=0$$
$$\displaystyle \lambda =\dfrac{16 \pm \sqrt{256 - 64}}{32}$$
$$\displaystyle =\dfrac{16 \pm \sqrt{13}}{32}=\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{4}$$
Question 9
1 / -0

If the foot of the perpendicular from the origin to a plane is $$P(a, b, c)$$, the equation of the plane is-

A
$$\dfrac {x}{a}+\dfrac {y}{b}+\dfrac {z}{c}=3$$

B
$$ax+by+cz=3$$

C
$$ax+by+cz=a^2+b^2+c^2$$

D
$$ax+by+cz=a+b+c$$

Solution

$$ {\textbf{Step - 1: Find direction ratio}} $$

$$ {\text{Let the foot of the perpendicular be P(a,b,c)}}{\text{.}} $$

$$ {\text{Then, direction ratios of OP are }} {\text{a - 0,b - 0,c - 0 }}\;{\text{i}}{\text{.e}}{\text{ a,b,c}} $$

$$ {\textbf{Step - 2: Make equation}} $$

$$ {\text{So, the equation of the plane passing through P(a,b,c), }} $$

$$ {\text{the direction ratios of the normal to which are a,b,c is}} $$

$$ {\text{a(x - a) + b(y - b) + c(z - c) = 0}} $$

$$ \Rightarrow {\text{ax + by + cz = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{.}} $$

$$ {\textbf{Hence,correct answer is option C}}{\text{.}} $$
Question 10
1 / -0

If the points $$(-1, 3, 2), (-4, 2, -2)$$ and $$(5, 5, \lambda)$$ are collinear, then $$\lambda$$ is equal to

A
$$-10$$

B
$$5$$

C
$$-5$$

D
$$10$$

Solution

Let the points be $$A(-1,3,2) B(-4,2,-2) C(5,5,\lambda)$$
Direction ratio of $$AB{=}(-3,-1,-4)$$
If $$A,B,C$$ are collinear points then direction ratio of $$AB$$ and $$BC$$ must be proportional.
Direction ratio of $$BC{=}(9,3,\lambda+2)$$
$$\therefore 9{=}\alpha (-3)$$
$$\therefore 3{=}\alpha(-1)$$
$$\therefore \lambda+2{=}\alpha(-4)$$ and $$\alpha{=} -3$$
$$\therefore \dfrac{\lambda+2}{-3}{=}-4$$
$$\therefore \lambda{=} 10$$

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Three Dimensional Geometry Test - 27

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Three Dimensional Geometry Test - 27

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Question 1

$$\displaystyle \dfrac{-12}{13},\dfrac{-4}{13},\dfrac{-3}{13}$$

$$\displaystyle \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$$

$$\displaystyle \dfrac{12}{13},\dfrac{-4}{13},\dfrac{3}{13}$$

$$\displaystyle \dfrac{12}{13},\dfrac{4}{13},\dfrac{-3}{13}$$

Solution

Question 2

$$(\displaystyle \dfrac{6}{\sqrt{3}}, \dfrac{6}{\sqrt{3}}, \dfrac{6}{\sqrt{3}})$$

$$(2\sqrt{3}, -2\sqrt{3},2\sqrt{3})$$

$$-(2\sqrt{3},2\sqrt{3},2\sqrt{3})$$

$$(6\sqrt{3},6\sqrt{3},6\sqrt{3})$$

Solution

Question 3

$$< 1, \sqrt 2, 1 >$$

$$< 1, 1, \sqrt 2 >$$

$$< 1, 1, 2 >$$

$$< \sqrt 2, 1, 1 >$$

Solution

Question 4

$$x+y tan\theta=0$$

$$y+xtan\theta=0$$

$$x cos\theta-y sin\theta=0$$

$$x sin\theta-y cos\theta=0$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$11x-y-3z=35$$

$$11x+y-3z=35$$

$$11x-y+3z=35$$

None of these.

Solution

Question 7

$$\dfrac {2}{3}$$

$$\dfrac {1}{5}$$

$$\dfrac {3}{5}$$

$$\dfrac {2}{5}$$

Solution

Question 8

$$\displaystyle \frac{1}{2} \pm \frac{{\sqrt 3 }}{4}$$

$$\displaystyle \frac{3}{4}$$

$$\displaystyle \pm \frac{3}{4}$$

$$\displaystyle \frac{{\sqrt 3 }}{2} \pm \frac{1}{4}$$

Solution

Question 9

$$\dfrac {x}{a}+\dfrac {y}{b}+\dfrac {z}{c}=3$$

$$ax+by+cz=3$$

$$ax+by+cz=a^2+b^2+c^2$$

$$ax+by+cz=a+b+c$$

Solution

Question 10

$$-10$$

$$5$$

$$-5$$

$$10$$

Solution

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