Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is

A
$$0^{0}$$

B
$$90^{0}$$

C
$$45^{0}$$

D
$$30^{0}$$

Solution

Direction ratios of line $$2x=3y=-z$$ which can be written as $$\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1}$$ are $$\left(\dfrac{1}{2},\dfrac{1}{3},-1\right)$$ and
Direction ratios of line $$6x=-y=-4z$$ which can be written as $$\frac{x}{\frac{1}{6}}=\frac{y}{-1}=\frac{z}{-\frac{1}{4}}$$ are $$\left(\dfrac{1}{6},-1,-\dfrac{1}{4}\right).$$
Dot product of the direction ratio of the two line $$=\dfrac{1}{2}\times\dfrac{1}{6}+\dfrac{1}{3}\times(-1)+(-1)\times(-\dfrac{1}{4})$$
$$=\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}$$
$$=0$$
So, angle between the lines is $$90^\circ.$$
Hence, B is the correct option.
Question 2
1 / -0

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

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

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

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

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

Solution

Since line makes an angle of $$\displaystyle \dfrac { \pi }{ 4 } $$ with positive direction of each of $$x$$-axis and $$y$$-axis, therefore $$\displaystyle \alpha =\dfrac { \pi }{ 4 } ,\beta =\dfrac { \pi }{ 4 } $$

We know that,

$$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =1$$

$$\displaystyle \\ \Rightarrow \cos ^{ 2 }{ \dfrac { \pi }{ 4 } } +\cos ^{ 2 }{ \dfrac { \pi }{ 4 } } +\cos ^{ 2 }{ \gamma } =1$$

$$\displaystyle \Rightarrow \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +\cos ^{ 2 }{ \gamma } =1$$

$$\Rightarrow \cos ^{ 2 }{ \gamma } =0$$

$$\Rightarrow \gamma ={ 90 }^{ 0 }$$
Question 3
1 / -0

The line passing through the points $$(5,1,a)$$ and $$(3,b,1)$$ crosses the $$yz$$-plane at the point $$\left (0,\dfrac {17}{2}, \dfrac {-13}{2}\right)$$, then

A
$$a=11,\ b=4$$

B
$$a=8,\ b=2$$

C
$$a=2,\ b=8$$

D
$$a=4,\ b=6$$

Solution

Equation of line passing through $$(5,1,a)$$ and $$(3,b,1)$$ is

$$ \displaystyle \dfrac { x-5 }{ 2 } =\dfrac { y-1 }{ 1-b } =\dfrac { z-a }{ a-b } =\lambda $$

If line crosses $$yz$$-plane $$i.e.,x=0$$

$$x=2\lambda +5=0$$

$$ \displaystyle \Rightarrow \lambda =-\dfrac { 5 }{ 2 } ,$$

Since, $$ \displaystyle y=\lambda \left( 1-b \right) +1\Rightarrow \dfrac { 17 }{ 2 } =\dfrac { -5 }{ 2 } \left( 1-b \right) +1\Rightarrow b=4$$

Also, $$ \displaystyle z=\lambda \left( a-1 \right) +a\Rightarrow-\dfrac { 13 }{ 2 } =\dfrac { -5 }{ 2 } \left( a-1 \right) +a\Rightarrow a=11$$
Question 4
1 / -0

A line $$AB$$ in three-dimensional space makes angles $$45^{\mathrm{o}}$$ and $$120^{\mathrm{o}}$$ with the positive $$\mathrm{x}$$-axis and the positive $$\mathrm{y}$$-axis respectively. lf $$AB$$ makes an acute angle $$e$$ with the positive $$\mathrm{z}$$-axis, then $$e$$ equals

A
$$45^{\mathrm{o}}$$

B
$$60^{\mathrm{o}}$$

C
$$75^{\mathrm{o}}$$

D
$$30^{\mathrm{o}}$$

Solution

The angle made by line $$AB$$ with $$x$$ axis is $$45^{0}$$
The angle made by line $$AB$$ with $$y$$ axis is $$120^{0}$$

Let the angle made by line $$AB$$ with $$z$$ axis is $$\theta$$
We have $$\cos^{2}{(45)} +\cos^{2}{(120)}+\cos^{2}{\theta}=1$$

$$\Rightarrow cos^{2}{\theta}=1-\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$$

Given that $$\theta$$ is accute
So we get $$cos\theta=\dfrac{1}{2}$$
$$\Rightarrow \theta=60^{0}$$
Therefore option $$B$$ is correct
Question 5
1 / -0

A line in the 3-dimensional space makes an angle $$\theta \left(0 < \theta \leq \dfrac {\pi}{2}\right)$$ with both the x and y axes, then the set of all values of $$\theta$$ is the interval :

A
$$\left[0, \dfrac {\pi}{4}\right]$$

B
$$\left[\dfrac {\pi}{6}, \dfrac {\pi}{3}\right]$$

C
$$\left[\dfrac {\pi}{4}, \dfrac {\pi}{2}\right]$$

D
$$\left[\dfrac {\pi}{3}, \dfrac {\pi}{2}\right]$$

Solution

Sum of square of direction cosines is $$1$$.
Lets assume $$\alpha $$ is the angle with $$x$$, $$\beta$$ with $$y$$ and $$\gamma$$ with $$z$$.

$$\alpha,\beta,\gamma \in \left[0,\dfrac\pi2\right]$$
Now,
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1$$
But it is given that, $$\alpha = \beta = \theta$$

$$\therefore \cos^2\theta + \cos^2\theta + \cos^2\gamma = 1$$
$$\Rightarrow 2\cos^2\theta = 1-\cos^2\gamma$$

$$\Rightarrow = \cos^2\theta = \dfrac{\sin^2\gamma}2$$

$$\Rightarrow \cos\theta = \dfrac{\sin\gamma}{\sqrt2}$$

As $$\gamma \in \left[0,\dfrac\pi2\right]$$, then $$\theta \in \left[\dfrac\pi4,\dfrac\pi2\right]$$
Question 6
1 / -0

A line makes the same angle $$\Theta$$, with each of the $$\mathrm{x}$$ and $$\mathrm{z}$$ axis. If the angle $$\beta$$, which it makes with $$\mathrm{y}$$-axis, is such that $$\sin^{2}\beta=3\sin_{\Theta}^{2}$$, then $$\cos^{2}\Theta$$ equals:

A
$$\displaystyle \frac{2}{3}$$

B
$$\displaystyle \frac{1}{5}$$

C
$$\displaystyle \frac{3}{5}$$

D
$$\displaystyle \frac{2}{5}$$

Solution

A line makes angle $$\theta $$ with x-axis and z axis and $$\beta $$ with y-axis
$$\therefore l=\cos { \theta } ,m=\cos { \beta } ,n=\cos { \theta } $$
we know that $${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
$$\Rightarrow \cos ^{ 2 }{ \theta } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \theta } =1\Rightarrow 2\cos ^{ 2 }{ \theta } =1-\cos ^{ 2 }{ \beta } $$
$$\Rightarrow 2\cos ^{ 2 }{ \theta } =\sin ^{ 2 }{ \beta } $$ ...(1)
But $$\sin ^{ 2 }{ \beta } =3\sin ^{ 2 }{ \theta } $$ ...(2)
From (1) and (2)
$$3\sin ^{ 2 }{ \theta } =2\cos ^{ 2 }{ \theta } \Rightarrow 3\left( 1-\cos ^{ 2 }{ \theta } \right) =2\cos ^{ 2 }{ \theta } \\ \Rightarrow 3-3\cos ^{ 2 }{ \theta } =2\cos ^{ 2 }{ \theta } \Rightarrow 3=5\cos ^{ 2 }{ \theta } $$
$$\displaystyle\Rightarrow\cos ^{ 2 }{ \theta } =\frac { 3 }{ 5 } $$
Question 7
1 / -0

A line with positive direction cosines passes through the point $$P(2, -1, 2)$$ and makes equal angles with the coordinate axes. The line meets the plane $$2x + y + z = 9$$ at point $$Q$$. The length of the line segment $$PQ$$ equals

A
1

B
$$\sqrt{2}$$

C
$$\sqrt{3}$$

D
2

Solution

$$D.\mathrm{C}$$ of the line are $$\displaystyle \dfrac{1}{\sqrt{3}}$$ , $$\displaystyle \dfrac{1}{\sqrt{3}}$$ , $$\displaystyle \dfrac{1}{\sqrt{3}}$$

Any point on the line at a distance $$t$$ from $${P}(2, -1,2)$$ is $$\left (2+\displaystyle \dfrac{{t}}{\sqrt{3}}, -1+\dfrac{{t}}{\sqrt{3}}, 2+\dfrac{{t}}{\sqrt{3}}\right)$$

which lies on $$2x+y+z=9$$.

$$\Rightarrow t=\sqrt 3$$
Question 8
1 / -0

If the lines $$\dfrac{x-2}{1}=\dfrac{y-3}{1}=\dfrac{z-4}{-k}$$ and $$\dfrac{x-1}{k}=\dfrac{y-4}{2}=\dfrac{z-5}{1}$$ are coplanar, then $$k$$ can have

A
exactly one value

B
exactly two values

C
exactly three values

D
any value

Solution

We have, $$
\left|\begin{array}{lll}
1 & -1 & -1\\
1 & 1 & -k\\
k & 2 & 1
\end{array}\right|=0
$$

$$\Rightarrow 1(1+2k)+1(1+k^{2})-1(2-k)=0$$

$$
\Rightarrow k^{2}+1+2k+1-2+k=0
$$

$$
\Rightarrow k^{2}+3k=0
$$

$$\Rightarrow (k)(k+3)=0$$

Thus $$k$$ has $$2$$ values.
Question 9
1 / -0

If the straight lines $$\displaystyle \dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2}$$ and $$\displaystyle \dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{K}$$ are coplanar, then the plane(s) containing these two lines is(are)

A
$$y+2z=-1$$

B
$$y+z =-1$$

C
$$y-z=-1$$

D
$$y-2z =-1$$

Solution

Two straight lines
$$\dfrac {x-1}{2}=\dfrac {y+1}{K}=\dfrac {z}{2}$$ and $$\dfrac {x+1}{5}=\dfrac {y+1}{2}=\dfrac {z}{K}$$ are co-planar, then we get

$$\begin{vmatrix} 2& K & 2 \\5& 2 & K\\ 2 & 0 & 0\end{vmatrix}=0$$

$$\Rightarrow K^2=4\Rightarrow K=\pm 2$$

For $$K=2$$, the plane $$y+1=z$$ is common in both lines.

For $$K=-2$$, family of plane containing first line is $$x+y+\lambda (x-z-1)=0$$

Point $$(-1, -1, 0)$$ must satisfy $$\lambda$$.

$$\Rightarrow -2+\lambda(-2)=0\Rightarrow \lambda=-1$$

$$\Rightarrow y+z+1=0$$

$$\Rightarrow y+z=-1$$
Question 10
1 / -0

The points with position vectors $$60 \widehat{i} + 3 \widehat{j}$$, $$40 \widehat{i} - 8 \widehat{j}$$, $$a\widehat{i} -52\widehat{j}$$ are collinear if

A
$$a = -40$$

B
$$a = 40$$

C
$$a = 20$$

D
None of these

Solution

Three points $$A(\vec{a}), B (\vec{b}), C(\vec{c})$$ are collinear if $$\vec{AB} || \vec{AC}$$

$$\vec{AB} = - 20 \widehat{i} - 11 \widehat{j}$$
$$\vec{AC} = (a - 60) \widehat{i} - 55 \widehat{j}$$

$$\Rightarrow \vec{AB} || \vec{AC} $$
$$\Rightarrow \displaystyle \dfrac{a - 60}{- 20 } $$$$= \dfrac{-55}{-11}$$
$$\Rightarrow a = - 40$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 13

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Three Dimensional Geometry Test - 13

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SHARING IS CARING

Question 1

$$0^{0}$$

$$90^{0}$$

$$45^{0}$$

$$30^{0}$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$$a=11,\ b=4$$

$$a=8,\ b=2$$

$$a=2,\ b=8$$

$$a=4,\ b=6$$

Solution

Question 4

$$45^{\mathrm{o}}$$

$$60^{\mathrm{o}}$$

$$75^{\mathrm{o}}$$

$$30^{\mathrm{o}}$$

Solution

Question 5

$$\left[0, \dfrac {\pi}{4}\right]$$

$$\left[\dfrac {\pi}{6}, \dfrac {\pi}{3}\right]$$

$$\left[\dfrac {\pi}{4}, \dfrac {\pi}{2}\right]$$

$$\left[\dfrac {\pi}{3}, \dfrac {\pi}{2}\right]$$

Solution

Question 6

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{1}{5}$$

$$\displaystyle \frac{3}{5}$$

$$\displaystyle \frac{2}{5}$$

Solution

Question 7

1

$$\sqrt{2}$$

$$\sqrt{3}$$

2

Solution

Question 8

exactly one value

exactly two values

exactly three values

any value

Solution

Question 9

$$y+2z=-1$$

$$y+z =-1$$

$$y-z=-1$$

$$y-2z =-1$$

Solution

Question 10

$$a = -40$$

$$a = 40$$

$$a = 20$$

None of these

Solution

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