Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the plane passing through the point $$(1, 1, 1)$$ and perpendicular to the planes $$2x+y-2z=5$$ and $$3x-6y-2z=7$$ is?

A
$$14x+2y-15z=1$$

B
$$-14x+2y+15z=3$$

C
$$14x-2y+15z=27$$

D
$$14x+2y+15z=31$$

Solution

NOTE:Equation if the second plane should be $$3x-6y-2z=7$$
Hence option (D) is the correct anwer.
Question 2
1 / -0

The acute angle between two lines such that the direction cosines $$l,\ m,\ n$$ of each of them satisfy the equation $$l+m+n=0$$ and $$l^2+m^2-n^2=0$$ is

A
$$30^\circ$$

B
$$45^\circ$$

C
$$60^\circ$$

D
$$15^\circ$$

Solution

Lines are $$l+m+n=0\implies -l=(m+n)$$and $$l^2-m^2+n^2=0\implies l^2=m^2-n^2$$Solving them gives,$$(-(m+n))^2=m^2+n^2+2mn=m^2-n^2\implies 2n(n+m)=0$$Now, for $$n=0$$$$m=-l$$, so d.c's $$\left(\dfrac 1{\sqrt 2}, -\dfrac 1{\sqrt 2}, 0\right)$$And for $$n=-m$$$$l=0$$, so d.c's $$\left(0, \dfrac 1{\sqrt 2}, -\dfrac 1{\sqrt 2}\right)$$Angles between the lines $$|\cos \theta|=\left|\dfrac 12\right|\implies \theta =\dfrac \pi 3$$
Question 3
1 / -0

The direction cosines of the ray $$P(1,-2,4)$$ and $$Q(-1,1,-2)$$ are

A
$$\left(-2,-3,-6\right)$$

B
$$\left(2,-3,-6\right)$$

C
$$\left(\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}\right)$$

D
$$\left(-\dfrac{2}{7},\dfrac{3}{7},-\dfrac{6}{7}\right)$$

Solution

$$P(1, -2, 4)\ Q(-1, 1, -2)$$
$$PQ = \sqrt{(1-(-1))^2 + (-2 -1)^2 + (4 - (-2))^2}$$
        $$= \sqrt{4 + 9 + 36}$$
        $$= \sqrt{49}$$
        $$= 7$$
$$DC = (\dfrac{-1 - 1}{7}, \dfrac{1 - (-2)}{7}, \dfrac{-2 - 4}{7})$$
        $$ = (\dfrac{-2}{7}, \dfrac{3}{7}, \dfrac{-6}{7})$$
Question 4
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The measure of the angle between the lines, whose direction numbers are $$l,m,n$$ and $$m-n,n-l,l-m$$ is ______

A
$$\cfrac { \pi }{ 4 } $$

B
$$\cfrac { \pi }{ 6 } $$

C
$$\cfrac { \pi }{ 2 } $$

D
$$\cfrac { \pi }{ 3 } $$

Solution

Here, $$(l,m,n).(m-n,n-l,l-m)$$
$$=l(m-n)+m(n-l)+m(l-m)$$
$$=lm-nl+mn-ml+nl-nm$$
$$=0$$
So, Lines are mutuall perpendicular
So, Angle between them is $$\cfrac { \pi }{ 2 } $$
Question 5
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The acute angle between two lines such that the direction cosines l,m,n of each of them satisfy the equations $$l+m+n=0$$ and $${ l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0$$ is :-

A
$$30$$

B
$$45$$

C
$$60$$

D
$$15$$

Solution

Lines are $$l+m+n=0\Rightarrow -l=(m+n)$$
and $${ l }^{ 2 }-{ m }^{ 2 }+{ n }^{ 2 }=0\Rightarrow { l }^{ 2 }={ m }^{ 2 }-{ n }^{ 2 }$$
Solving them gives, $${ \left( -\left( { m }{ +n } \right) \right) }^{ 2 }={ m }^{ 2 }+{ n }^{ 2 }+2mn={ m }^{ 2 }-{ n }^{ 2 }\Rightarrow 2n(n+m)=0$$
Now, for $$n=0$$
$$m=-l$$, so d.c's $$\left( \cfrac { 1 }{ \sqrt { 2 } } ,-\cfrac { 1 }{ \sqrt { 2 } } ,0 \right) $$
And for $$n=-m$$
$$l=0$$ so d.c's $$\left( 0,\cfrac { 1 }{ \sqrt { 2 } } ,-\cfrac { 1 }{ \sqrt { 2 } } \right) $$
Angle between the lines $$\left| \cos { \theta } \right| =\left| \cfrac { 1 }{ 2 } \right| \Rightarrow \theta =\cfrac { \pi }{ 3 } =60$$
Question 6
1 / -0

Find in a symmetrical form, the equations of the line formed by the planes $$x+y+z+1=0, 4x+y-2z+2=0$$ and find its direction-cosines.

A
$$\dfrac{x-\frac{1}{3}}{1}=\dfrac{y+\frac{2}{3}}{-2}=\dfrac{z-0}{1}; -\dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},\dfrac{1}{\sqrt 6}$$

B
$$\dfrac{x-\frac{1}{3}}{-1}=\dfrac{y-\frac{2}{3}}{2}=\dfrac{z-0}{1}; \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},-\dfrac{1}{\sqrt 6}$$

C
$$\dfrac{x+\frac{1}{3}}{-1}=\dfrac{y+\frac{2}{3}}{2}=\dfrac{z+0}{-1}; -\dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},-\dfrac{1}{\sqrt 6}$$

D
$$\dfrac{x+\frac{1}{3}}{1}=\dfrac{y-\frac{2}{3}}{2}=\dfrac{z+0}{1}; \dfrac{1}{\sqrt 6}, -\dfrac{2}{\sqrt{6}},\dfrac{1}{\sqrt 6}$$

Solution

The lines should be $$x + y + z = 0 $$

and, $$4x + y - 2z = (-2)$$

Now, equation should lie on both planes.

So, the equation of line should be perpendicular to both planes.

Let the DRs of the line $$(a, b, c)$$

So, $$(a, b, c) \perp (1,1, 1)$$

and, $$(a, b, c) \perp (4, ,1, -2)$$

$$\therefore a + b + c = 0$$

   $$4a + b - 2c = 0$$

On solving,
             $$\dfrac{a}{-2-1} = \dfrac{-b}{-2-4} = \dfrac{c}{1-4}$$

                 $$ \Rightarrow a = -3, b = 6, c = -3$$

                 $$ \Rightarrow a = -1, b = 2, c = -1$$

Let the coordinates of the point lying in $$xy$$ plane $$(x, y, 0)$$

on subtracting $$x + y = -1   -(1)$$

and $$4x + y = -2$$

$$\Rightarrow -3x = 1$$

$$\Rightarrow x = \dfrac{-1}{3}$$

Putting vbalue of $$x$$ in $$(1)$$

$$\dfrac{-1}{3} + y = -1$$

$$y = -1 + \dfrac{-1}{3}$$

   $$=\dfrac{-2}{3}$$

Coordinate $$(\dfrac{-1}{3} , \dfrac{-2}{3}, 0)$$

$$\therefore Equation = \dfrac{x + \dfrac{1}{3}}{-1} = \dfrac{y + \dfrac{2}{3}}{2} =\dfrac{z - 0}{-1}$$

DC = $$\dfrac{-1}{\sqrt{(-1)^2 + (2)^2 +(1)^2}} ,\dfrac{2}{\sqrt{(-1)^2 + (2)^2 +(1)^2}} , \dfrac{-1}{\sqrt{(-1)^2 + (2)^2 +(1)^2}}$$

    $$= (\dfrac{-1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{-1}{\sqrt{6}})$$
Question 7
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The equation of plane passing through $$(4, 5, -1)$$ having normal $$3\hat{i}-\hat{j}+\hat{k}$$ is ___________.

A
$$4x-5y+z=6$$

B
$$3x-y+z=6$$

C
$$3x+y+z=6$$

D
$$4x+5y-z=6$$

Solution

$$A(\bar{a})=(4, 5, -1)$$ and $$\bar{n}=(3, -1, 1)$$
$$\therefore$$ Required equation of plane $$\bar{r}\cdot \bar{n}=\bar{a}\cdot \bar{n}$$
$$\therefore (x, y, z)(3, -1, 1)=(4, 5, -1)\cdot (3, -1, 1)$$
$$\therefore 3x-y+z=12-5-1$$
$$\therefore 3x-y+z=6$$
Question 8
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Vector equation of line $$\dfrac{3-x}{3}=\dfrac{2y-3}{5}=\dfrac{z}{2}$$ is __________ $$k\in R$$.

A
$$\bar{r}=(3, 5, 2)+k(3, 3, 0)$$

B
$$\bar{r}=\left(3, \dfrac{3}{2}, 0\right)+k(-6, 5, 4)$$

C
$$\bar{r}=(3, 3, 0)+k(3, 5, 2)$$

D
$$\bar{r}=(-6, 5, 4)+$$ $$k\left( 3, \dfrac{3}{2}, 0\right)$$

Solution

$$\dfrac{3-x}{3}=\dfrac{2y-3}{5}=\dfrac{z}{2}$$
$$\therefore\dfrac{x-3}{-3}=\dfrac{y-\dfrac{3}{2}}{\dfrac{5}{2}}=\dfrac{z-0}{2}$$
Here $$A(\bar{a})=\left(3, \dfrac{3}{2}, 0\right)$$ and $$(\bar{l})=\left(-3, \dfrac{5}{2}, 2\right)$$
Vector equation of line : $$\bar{r}=\bar{a}+k(\bar{l})$$
$$\therefore r=\left(3, \dfrac{3}{2}, 0\right)+k'\left(-3, \dfrac{5}{2}, 2\right)$$
$$=\left(3, \dfrac{3}{2}, 0\right)+\dfrac{k'}{2}(-6, 5, 4)$$
$$=\left(3, \dfrac{3}{2}, 0\right)+k(-6, 5, 4)$$
$$\therefore k=\dfrac{k'}{2}\in R$$.
Question 9
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The vector equation of the plane which is at distance of $$10$$ unit from the origin and perpendicular to the vector $$4i+4j-2k$$ is

A
$$r.(4i+4j-2k)=10$$

B
$$r.(4i+4j-2k)=20$$

C
$$r.(4i+4j-2k)=30$$

D
$$r.(4i+4j-2k)=60$$

Solution

Consider the problem
Let the Normal vector be $$N=4\hat i + 4\hat j - 2\hat k$$
Then unit vector in the direction of $$\vec N$$ is
$$=\frac{\vec N}{|\vec N|}$$
$$ = \frac{{4\hat i + 4\hat j - 2\hat k}}{{\sqrt {{4^2} + {4^2} + {{( - 2)}^2}} }}$$
$$ = \frac{1}{6}\left( {4\hat i + 4\hat j - 2\hat k} \right)$$
As we know that the equation of the plane with the position vector $$\vec r$$
$$\vec r. \hat n=d$$
this imlies,
$$\vec r. \frac{1}{6}\left( {4\hat i + 4\hat j - 2\hat k} \right)=10$$
Hence,
the required equation
$$r.\left( {4i + 4j - 2k} \right) = 60$$

Hence option $$D$$ is the correct answer.
Question 10
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A line making angles $$45^o$$ and $$60^o$$ with the positive direction of $$x-$$ axis and $$y-$$ axis respectively. Then the angle made by the line with positive direction of $$z-$$ axis is

A
$$60^o$$

B
$$120^o$$

C
$$60^o$$ or $$120^o$$

D
$$None\ of\ these$$

Solution

$$cos^2\alpha + cos^2\beta + cos^2\gamma = 1$$
$$\alpha = 45^{\circ}$$
$$\beta = 60^{\circ}$$
$$\therefore cos^245^{\circ} + cos^260^{\circ} + cos^2\gamma = 1 $$
$$\dfrac{1}{2} + \dfrac{1}{4} + cos^2\gamma = 1 $$
$$cos^2\gamma = 1 - \dfrac{3}{4}$$
$$cos\gamma = \pm\dfrac{1}{2}$$
$$cos\gamma = \dfrac{1}{2}, cos\gamma = -\dfrac{1}{2}$$
$$\gamma = 60^{\circ}, \gamma = 120^{\circ}$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 42

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Three Dimensional Geometry Test - 42

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SHARING IS CARING

Question 1

$$14x+2y-15z=1$$

$$-14x+2y+15z=3$$

$$14x-2y+15z=27$$

$$14x+2y+15z=31$$

Solution

Question 2

$$30^\circ$$

$$45^\circ$$

$$60^\circ$$

$$15^\circ$$

Solution

Question 3

$$\left(-2,-3,-6\right)$$

$$\left(2,-3,-6\right)$$

$$\left(\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}\right)$$

$$\left(-\dfrac{2}{7},\dfrac{3}{7},-\dfrac{6}{7}\right)$$

Solution

Question 4

$$\cfrac { \pi }{ 4 } $$

$$\cfrac { \pi }{ 6 } $$

$$\cfrac { \pi }{ 2 } $$

$$\cfrac { \pi }{ 3 } $$

Solution

Question 5

$$30$$

$$45$$

$$60$$

$$15$$

Solution

Question 6

$$\dfrac{x-\frac{1}{3}}{1}=\dfrac{y+\frac{2}{3}}{-2}=\dfrac{z-0}{1}; -\dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},\dfrac{1}{\sqrt 6}$$

$$\dfrac{x-\frac{1}{3}}{-1}=\dfrac{y-\frac{2}{3}}{2}=\dfrac{z-0}{1}; \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},-\dfrac{1}{\sqrt 6}$$

$$\dfrac{x+\frac{1}{3}}{-1}=\dfrac{y+\frac{2}{3}}{2}=\dfrac{z+0}{-1}; -\dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},-\dfrac{1}{\sqrt 6}$$

$$\dfrac{x+\frac{1}{3}}{1}=\dfrac{y-\frac{2}{3}}{2}=\dfrac{z+0}{1}; \dfrac{1}{\sqrt 6}, -\dfrac{2}{\sqrt{6}},\dfrac{1}{\sqrt 6}$$

Solution

Question 7

$$4x-5y+z=6$$

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

$$3x+y+z=6$$

$$4x+5y-z=6$$

Solution

Question 8

$$\bar{r}=(3, 5, 2)+k(3, 3, 0)$$

$$\bar{r}=\left(3, \dfrac{3}{2}, 0\right)+k(-6, 5, 4)$$

$$\bar{r}=(3, 3, 0)+k(3, 5, 2)$$

$$\bar{r}=(-6, 5, 4)+$$ $$k\left( 3, \dfrac{3}{2}, 0\right)$$

Solution

Question 9

$$r.(4i+4j-2k)=10$$

$$r.(4i+4j-2k)=20$$

$$r.(4i+4j-2k)=30$$

$$r.(4i+4j-2k)=60$$

Solution

Question 10

$$60^o$$

$$120^o$$

$$60^o$$ or $$120^o$$

$$None\ of\ these$$

Solution

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