Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

If direction cosines of a vector of magnitude $$3$$ are $$\dfrac {2}{3}, -\dfrac {1}{3}, \dfrac {2}{3}$$ and $$a > 0$$, then vector is ____

A
$$2i + j + 2k$$

B
$$2i - j + 2k$$

C
$$i - 2j + 2k$$

D
$$i + 2j + 2k$$

Solution

Unit vector with direction cosine $$\dfrac {2}{3}, -\dfrac {1}{3}, \dfrac {2}{3}$$ is given by
$$\hat{a}=\dfrac{2}{3}\hat i-\dfrac{1}{3}\hat j+\dfrac{2}{3}\hat k$$

Thus vector with magnitude $$3$$ and direction cosine as above is given by
$$3\hat a =2\hat i-\hat j+2\hat k$$
Question 2
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If a plane passing through the point $$(2, 2, 1)$$ and is perpendicular to the planes $$3x + 2y + 4z + 1 = 0$$ and $$2x + y + 3z + 2 = 0$$. Then, the equation of the plane is

A
$$2x - y - z - 1 = 0$$

B
$$2x + 3y + z - 1 = 0$$

C
$$2x + y + z + 3 = 0$$

D
$$x - y + z - 1 = 0$$

Solution

Equation of plane passing through $$(2, 2, 1)$$ is

$$a(x - 2) + b(y - 2) + c(z - 1) = 0 .... (i)$$

Since, above plane is perpendicular to

$$3x + 2y + 4z + 1 = 0$$

and $$2x + y + 3z + 2 = 0$$

$$\therefore 3a + 2b + 4c = 0 .... (ii)$$

and $$2a + b + 3c = 0 .... (iii)$$

$$[\because$$ for perpendicular, $$a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0]$$

On multiplying eq. (iii) by $$2$$, we get

$$4a + 2b + 6c = 0 .... (iv)$$

On subtracting eq, (iv) from eq. (ii), we get

$$\implies c = \dfrac {-a}{2}$$

On putting $$c = \dfrac {-a}{2}$$ in eq. (iii), we get $$b = \dfrac {-a}{2}$$

On putting $$b = \dfrac {-a}{2}$$ and $$c = \dfrac {-a}{2}$$ in eq. (i),

we get $$a(x - 2) - \dfrac {a}{2}(y - 2) - \dfrac {a}{2} (z - 1) = 0$$

$$\implies \dfrac {a}{2} [2(x - 2) - (y - 2) - (z - 1)] = 0$$

$$\implies 2x - 4 - y + 2 - z + 1 = 0$$

$$\implies 2x - y - z - 1 = 0$$ is the required equation of plane.
Question 3
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If a line segment $$OP$$ makes angles of $$\dfrac {\pi}{4}$$ and $$\dfrac {\pi}{3}$$ with X-axis and Y-axis, respectively. Then, the direction cosines are

A
$$\dfrac {1}{\sqrt {2}}, \dfrac {\sqrt {3}}{2}, \dfrac {1}{\sqrt {2}}$$

B
$$\dfrac {1}{\sqrt {2}}, \dfrac {1}{2}, \dfrac {1}{2}$$

C
$$1, \sqrt {3}, 1$$

D
$$1, \dfrac {1}{\sqrt {3}}, 1$$

Solution

Let $$\alpha, \beta$$ and $$\gamma$$ be the angles made by the line segment $$OP$$ with X-axis, Y-axis and Z-axis respectively.
Given: $$\alpha = \dfrac {\pi}{4}$$ and $$\beta = \dfrac {\pi}{3}$$
We know that, $$\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$$
$$\therefore \cos^{2}\dfrac {\pi}{4} + \cos^{2}\dfrac {\pi}{3} +\cos^{2}\gamma = 1$$
$$\Rightarrow \left (\dfrac {1}{\sqrt {2}}\right )^{2} + \left (\dfrac {1}{2}\right )^{2} + \cos^{2}\gamma = 1$$
$$\Rightarrow \dfrac {1}{2} + \dfrac {1}{4} + \cos^{2}\gamma = 1$$
$$\Rightarrow \cos^{2}\gamma = \dfrac {1}{4}$$
$$\Rightarrow \cos \gamma = \dfrac {1}{2}$$
$$\therefore \gamma = \dfrac {\pi}{3}$$
Hence, direction cosines are $$\cos \alpha, \cos \beta, \cos \gamma$$
i.e. $$\dfrac {1}{\sqrt {2}}, \dfrac {1}{2}, \dfrac {1}{2}$$.
Question 4
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If a line makes $${ 45 }^{ o }$$, $${ 60 }^{ o }$$ with positive direction of axes $$x$$ and $$y$$ then the angles it makes with the $$z$$-axis is

A
$${ 30 }^{ o }$$

B
$${ 90 }^{ o }$$

C
$${ 45 }^{ o }$$

D
$${ 60 }^{ o }$$

Solution

Fact: If $$\alpha, \beta, \gamma$$ are the angle made by any line with $$x,y$$ and $$z$$-axis, then
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$$

Here given $$\alpha=45^o, \beta = 60^o$$
$$\Rightarrow \cos^245^o+\cos^260^o+\cos^2\gamma=1$$
$$\Rightarrow \dfrac{1}{2}+\dfrac{1}{4}+\cos^2\gamma=1$$
$$\Rightarrow \cos^2\gamma=\dfrac{1}{4}$$
$$\Rightarrow \cos\gamma=\dfrac{1}{2}$$
$$\therefore \gamma =60^o$$
Question 5
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The vector equation of a plane passing through a point whose. P.V. is $$\overrightarrow a$$ and perpendicular to a vector $$\overrightarrow n$$, is

A
$$\overrightarrow r . \overrightarrow n = \overrightarrow a . \overrightarrow n$$

B
$$\overrightarrow r \times \overrightarrow n = \overrightarrow a \times \overrightarrow n$$

C
$$\overrightarrow r + \overrightarrow n = \overrightarrow a + \overrightarrow n$$

D
$$\overrightarrow r - \overrightarrow n = \overrightarrow a - \overrightarrow n$$

Solution

Let $$\vec r$$ be any point in the plane
So the vectors $$\vec r-\vec a$$ and $$\vec n$$ will be perpendicular to each other
$$\Rightarrow (\vec r-\vec a)\cdot \vec n=0$$
$$\Rightarrow \vec r\cdot \vec n-\vec a\cdot \vec n=0$$
$$\Rightarrow \vec r\cdot \vec n=\vec a\cdot \vec n$$
Question 6
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What are the direction cosines of a line which is equally inclined to the positive directions of the axes?

A
$$\left( \cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right) $$

B
$$\left( -\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right) $$

C
$$\left( -\cfrac { 1 }{ \sqrt { 3 } } ,-\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right) $$

D
$$\left( \cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right) $$

Solution

We know sum of the squares of the direction cosines is one.
$$i.e. cos^2\alpha+cos^2\beta+cos^2\gamma=1$$
but its given that $$ \alpha=\beta=\gamma$$
$$\therefore cos^2\alpha+cos^2\alpha+cos^2\alpha=1$$
$$3cos^2\alpha=1$$
$$\therefore cos^2\alpha=\dfrac{1}{3}$$
$$\therefore$$ Positive directions of the axes are $$(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3})$$
Question 7
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What are the direction ratios of normal to the plane $$2x - y + 2z + 1 = 0$$ ?

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

B
$$(1, \frac{1}{2}, 1)$$

C
$$(1, -2, 1)$$

D
None of the above

Solution

Direction ratios of normal to the plane $$ax+by+cz+d=0$$, are $$a,b,c$$.
So, here in the question the given plane is $$2x - y + 2z + 1 = 0$$
Thus, the direction ratios are $$2,-1,2$$
Hence, A is correct.
Question 8
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What is the sum of the squares of direction cosines of $$x$$-axis?

A
$$0$$

B
$$\dfrac{1}{3}$$

C
$$1$$

D
$$3$$

Solution

The direction cosines of the vector a are the cosines of angles that the vector forms with the coordinate axes.
The direction cosines uniquely set the direction of vector.
$$\therefore $$ The sum of the squares of the direction cosines is equal to one.
$$i.e. cos^2\alpha+cos^2\beta+cos^2\gamma=1$$
Option C is correct answer
Question 9
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Let a vector $$\vec{r}$$ make angles $$60^o, 30^o$$ with it and y-axes respectively.Find the angle $$\vec{r}$$ make with z-axis.

A
$$30^o$$

B
$$60^o$$

C
$$90^o$$

D
$$120^o$$

Solution

Sum of squares of direction cosines $$ =1$$

$${ l }^{ 2 }+{ m }^{ 2 }+n^{ 2 }=1$$

Here, $$l=\cos { 60 }^{ 0 }, m= \cos { 30 }^{ 0 }(given), n=?$$

$${ n }^{ 2 }=1-{ l }^{ 2 }-{ m }^{ 2 }\\ { n }^{ 2 }=1-\left(\dfrac{1}{2}\right)^{ 2 }-\left(\cfrac { \sqrt { 3 } }{ 2 } \right)^{ 2 }\\ \quad \quad =0$$

Hence, $$n=0=\cos { \theta }$$

$$\Rightarrow \theta ={ 90 }^{ 0 }$$
Question 10
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Directions For Questions

[passage-header]undefined[/passage-header]For the next two (02) items that follow :
Let a vector $$\vec{r}$$ make angles $$60^o, 30^o$$ with it and y-axes respectively.[passage-footer]undefined[/passage-footer]

...view full instructions

What are the direction cosines of $$\vec{r}$$ ?

A
$$\left (\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0 \right )$$

B
$$\left (\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right )$$

C
$$\left (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right )$$

D
$$\left (-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right )$$

Solution

Sum of squares of direction cosines $$ =1$$
$${ l }^{ 2 }+{ m }^{ 2 }+n^{ 2 }=1$$
Here, $$l=\cos { 60 }^{ 0 }, m= \cos { 30 }^{ 0 }, n=?$$
$${ n }^{ 2 }=1-{ l }^{ 2 }-{ m }^{ 2 }\\ { n }^{ 2 }=1-\left(\dfrac{1}{2}\right)^{ 2 }-\left(\cfrac { \sqrt { 3 } }{ 2 } \right)^{ 2 }\\ \quad \quad =0$$
Hence, $$n=0=\cos { \theta }$$
$$\Rightarrow \theta ={ 90 }^{ 0 }$$
Direction cosines $$(l,m,n) \equiv \left(\cfrac { 1 }{ 2 } ,\cfrac { \sqrt { 3 } }{ 2 } ,0\right)$$

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Three Dimensional Geometry Test - 38

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Three Dimensional Geometry Test - 38

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SHARING IS CARING

Question 1

$$2i + j + 2k$$

$$2i - j + 2k$$

$$i - 2j + 2k$$

$$i + 2j + 2k$$

Solution

Question 2

$$2x - y - z - 1 = 0$$

$$2x + 3y + z - 1 = 0$$

$$2x + y + z + 3 = 0$$

$$x - y + z - 1 = 0$$

Solution

Question 3

$$\dfrac {1}{\sqrt {2}}, \dfrac {\sqrt {3}}{2}, \dfrac {1}{\sqrt {2}}$$

$$\dfrac {1}{\sqrt {2}}, \dfrac {1}{2}, \dfrac {1}{2}$$

$$1, \sqrt {3}, 1$$

$$1, \dfrac {1}{\sqrt {3}}, 1$$

Solution

Question 4

$${ 30 }^{ o }$$

$${ 90 }^{ o }$$

$${ 45 }^{ o }$$

$${ 60 }^{ o }$$

Solution

Question 5

$$\overrightarrow r . \overrightarrow n = \overrightarrow a . \overrightarrow n$$

$$\overrightarrow r \times \overrightarrow n = \overrightarrow a \times \overrightarrow n$$

$$\overrightarrow r + \overrightarrow n = \overrightarrow a + \overrightarrow n$$

$$\overrightarrow r - \overrightarrow n = \overrightarrow a - \overrightarrow n$$

Solution

Question 6

$$\left( \cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right) $$

$$\left( -\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right) $$

$$\left( -\cfrac { 1 }{ \sqrt { 3 } } ,-\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right) $$

$$\left( \cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right) $$

Solution

Question 7

$$(2, -1, 2)$$

$$(1, \frac{1}{2}, 1)$$

$$(1, -2, 1)$$

None of the above

Solution

Question 8

$$0$$

$$\dfrac{1}{3}$$

$$1$$

$$3$$

Solution

Question 9

$$30^o$$

$$60^o$$

$$90^o$$

$$120^o$$

Solution

Question 10

$$\left (\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0 \right )$$

$$\left (\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right )$$

$$\left (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right )$$

$$\left (-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right )$$

Solution

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