Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 51

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Question 1
1 / -0

If $$A(3\hat { i } +2\hat { j } +3\hat { k } ),B(-\hat { i } -\hat { j } +8\hat { k } ),C(-4\hat { i } +4\hat { j } +6\hat { k } )$$ are the vertices of a triangle then the equation of the line passing through the circumcentre and parallel to $$\vec { A B }$$ is

A
$$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

B
$$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

C
$$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } -\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

D
$$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } -\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$
Question 2
1 / -0

The cartesian from of equation a line passing through the point position vector $$2\hat{i}-\hat{j}+2\hat{k}$$ and is in the direction of $$-2\hat{i}+\hat{j}+\hat{k}$$, is

A
$$\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}$$

B
$$\dfrac{x+4}{-2}=\dfrac{y-1}{1}=\dfrac{z+2}{1}$$

C
$$\dfrac{x+2}{4}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}$$

D
$$None \ of \ these$$

Solution

Equation of a line passing through a point with position vector $$\vec{a}$$ and parallel to $$\vec{b}$$ is

$$\vec{r}=\vec{a}+\lambda\vec{b}$$

Here $$\vec{a}= 2\hat{i}−\hat{j}+2\hat{k}$$ and $$\vec{b}=−2\hat{i}+\hat{j}+\hat{k}$$

So,$$\vec{r}=2\hat{i}−\hat{j}+2\hat{k}+\lambda\left(−2\hat{i}+\hat{j}+\hat{k}\right)$$

Equation of the line in vector form is $$2\hat{i}−\hat{j}+2\hat{k}+\lambda\left(−2\hat{i}+\hat{j}+\hat{k}\right)$$

Since the line passes through a point with position vector $$2\hat{i}−\hat{j}+2\hat{k}$$

$$\therefore {x}_{1}=2,\,{y}_{1}=-1,\,{z}_{1}=2$$

Also, line is in the direction of $$−2\hat{i}+\hat{j}+\hat{k}$$

Direction ratios :$$a=-2,\,b=1,\,c=1$$

Equation of line in cartesian form is

$$\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}$$
Question 3
1 / -0

If $$\cos { \alpha ,\quad \cos { \beta ,\quad \cos { \gamma } } }$$ are the direction cosine of a line, then find the value of $${ cos }^{ 2 }\alpha +\left( \cos { \beta +\sin { \gamma } } \right)$$$$\left( \cos { \beta - \sin { \gamma } } \right)$$

A
$$2$$

B
$$0$$

C
$$-1$$

D
$$1$$
Question 4
1 / -0

$$\dfrac { x - 2 } { 1 } = \dfrac { y - 3 } { 1 } = \dfrac { z - 4 } { - 1 }$$ & $$\dfrac { x - 1 } { k } = \dfrac { y - 4 } { 2 } = \dfrac { z - 5 } { 2 }$$ are coplanar then k=?

A
any value

B
exactly one value

C
exactly $$2$$ values

D
exactly $$3$$ values

Solution

If this two lines are co-planar
$$\begin{vmatrix} 1& 1&-1 \\ k & 2 & 2\\ 2-1&3-4&4-5\end{vmatrix}=0$$

$$\begin{vmatrix} 1&1&1 \\k&2 &2\\ 1&-1&-1\end{vmatrix}=0$$

$$1(-2+2)-1(-k-2)-1(-k-2)=0$$

$$k + 2+ k + 2 = 0$$

$$k = -2$$
Question 5
1 / -0

A normal to the plane $$ x=2 $$ is...

A
$$
(0,1,1)
$$

B
$$
(2,0,2)
$$

C
$$
(1,0,0)
$$

D
$$
(0,1,0)
$$

Solution

The plane $$x=2$$ is perpendicular to $$ x axis$$
So the angle is $$\dfrac{\pi}{2},\cos \dfrac{\pi}{2}=0$$
The plane $$x=2$$ is parallel to both $$y axis$$ and $$ z axis $$
So the angle is $$0,\cos 0=1$$
So the Drs are $$(0,1,1)$$
Question 6
1 / -0

The straight lines $$ \dfrac {x-1}{1} = \dfrac {y-2}{2} = \dfrac {z-3}{3} $$ and $$ \dfrac {x-1}1{} = \dfrac {y-2}{2} = \dfrac {z-3}3{} $$ are :

A
Parallel lines

B
Intersecting at $$60^o $$

C
Skew lines

D
Intersecting at right angle

Solution

The given straight line $$\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z - 3}}{3}\,and,\,\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z - 3}}{3}$$ are intersecting at right angle because by solving it with matrix method we get $$0$$.
Hence, the option $$D$$ is the correct answer.
Question 7
1 / -0

Direction ratio of line given by $$\dfrac { x-1 }{ 3 } =\dfrac { 6-2y }{ 10 } =\dfrac { 1-z }{ -7 } $$ are:

A
$$<3,10,-7>$$

B
$$<3,-5,7>$$

C
$$<3,5,7>$$

D
$$<3,5,-7>$$

Solution

$$\cfrac{x-1}{3} = \cfrac{6 - 2y}{10} = \cfrac{1 - z}{-7}$$
$$\cfrac{x-1}{3} = \cfrac{2y - 6}{-10} = \cfrac{z - 1}{- \left( -7 \right)}$$
$$\cfrac{x - 1}{3} = \cfrac{y - 3}{\left( \cfrac{-10}{2} \right)} = \cfrac{z - 1}{7}$$
$$\cfrac{x-1}{3} = \cfrac{y - 3}{-5} = \cfrac{z-1}{7}$$
Therefore,
Direction ratios are $$3, -5, 7$$
Question 8
1 / -0

A line with direction cosines proportional to 2 , 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by ______________.

A
(3a, 3a, 3a), (a,a,a)

B
(3a,2a, 3a), (a,a,a)

C
(3a, 2a, 3a), (a, a, 2a)

D
(2a, 3a, 3a), (2a, a, a)

Solution
Question 9
1 / -0

the points $$(\alpha ,\beta )(\gamma ,\delta ),(\alpha ,\delta )and (\gamma ,\beta )$$ where $$\alpha ,\beta ,\gamma ,\delta $$ are different real numbers, are

A
Collinear

B
vertices of square

C
vertices of rhomubs

D
concyclic

Solution
Question 10
1 / -0

The angle between the lines whose direction cosines are given by $$2l-m+2n=0$$, $$lm+mn+nl=0$$ is

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

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

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

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

Solution

$$\Rightarrow$$ $$mn+nl+lm=0$$ ----- ( 1 )
$$\Rightarrow$$ $$2l-m+2n=0$$
$$\Rightarrow$$ $$m=2l+2n$$ ----- ( 2 )
Substituting ( 2 ) in ( 1 ), we get,
$$\Rightarrow$$ $$(2l+2n)n+nl+l(2l+2n)=0$$
$$\Rightarrow$$ $$2ln+2n^2+nl+2l^2+2ln=0$$
$$\Rightarrow$$ $$2n^2+5ln+2l^2=0$$
$$\Rightarrow$$ $$2n^2+4ln+ln+2l^2=0$$
$$\Rightarrow$$ $$2n(n+2l)+l(n+2l)=0$$
$$\Rightarrow$$ $$(2n+1)(n+2l)=0$$
$$\Rightarrow$$ $$2n+1=0$$ or $$n+2l=0$$
$$\Rightarrow$$ $$l=-2n$$ or $$2l=-n$$ ---- ( 3 )
Substituting the values of ( 3 ) in ( 2 ) we get,
For the $$1^{st}$$ line :
$$\Rightarrow$$ $$m=2(-2n)+2n$$
$$\Rightarrow$$ $$m=-4n+2n$$
$$\Rightarrow$$ $$m=-2n$$
The direction ratios for the $$1^{st}$$ line is $$(-2n,-2n,n).$$
For the $$2^{nd}$$ line:
$$\Rightarrow$$ $$m=-n+2n$$
$$\Rightarrow$$ $$m=n$$
The direction ratios for the $$2^{nd}$$ line is $$\left(\dfrac{-n}{2},n,n\right)$$.
We know that the angle between the lines with direction ratios proportional to $$(a_1,b_1,c_1)$$ and $$(a_2,b_2,c_2)$$ is given by:
$$\Rightarrow$$ $$\theta=\cos^{-1}\left(\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right)$$
Here, $$a_1=-2n,\,a_2=\dfrac{-n}{2},\,b_1=-2n,\,b_2=n,\,c_1=n$$ and $$c_2=n$$

$$\Rightarrow$$ $$\theta=\cos^{-1}\left(\dfrac{\left(-2n\times \dfrac{-n}{2}\right)+(-2n\times n)+(n\times n)}{\sqrt{(-2n)^2+(-2n)^2+n^2}\sqrt{\left(\dfrac{n}{2}\right)^2+n^2+n^2}}\right)$$

$$\Rightarrow$$ $$\theta=\cos^{-1}\left(\dfrac{n^2-2n^2+n^2}{\sqrt{4n^2+4n^2+n^2}\sqrt{\dfrac{n^2}{4}+n^2+n^2}}\right)$$

$$\Rightarrow$$ $$\theta=\cos^{-1}\left(\dfrac{0}{\sqrt{9n^2}\sqrt{\dfrac{9n^2}{4}}}\right)$$

$$\Rightarrow$$ $$\theta=\cos^{-1}(0)$$
$$\therefore$$ $$\theta=\dfrac{\pi}{2}$$
$$\therefore$$ The angle between two lines is $$\dfrac{\pi}{2}$$ or $$90^o.$$

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Three Dimensional Geometry Test - 51

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Three Dimensional Geometry Test - 51

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SHARING IS CARING

Question 1

$$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

$$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

$$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } -\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

$$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } -\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$

Question 2

$$\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}$$

$$\dfrac{x+4}{-2}=\dfrac{y-1}{1}=\dfrac{z+2}{1}$$

$$\dfrac{x+2}{4}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}$$

$$None \ of \ these$$

Solution

Question 3

$$2$$

$$0$$

$$-1$$

$$1$$

Question 4

any value

exactly one value

exactly $$2$$ values

exactly $$3$$ values

Solution

Question 5

$$ (0,1,1) $$

$$ (2,0,2) $$

$$ (1,0,0) $$

$$ (0,1,0) $$

Solution

Question 6

Parallel lines

Intersecting at $$60^o $$

Skew lines

Intersecting at right angle

Solution

Question 7

$$<3,10,-7>$$

$$<3,-5,7>$$

$$<3,5,7>$$

$$<3,5,-7>$$

Solution

Question 8

(3a, 3a, 3a), (a,a,a)

(3a,2a, 3a), (a,a,a)

(3a, 2a, 3a), (a, a, 2a)

(2a, 3a, 3a), (2a, a, a)

Solution

Question 9

Collinear

vertices of square

vertices of rhomubs

concyclic

Solution

Question 10

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

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

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

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

Solution

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$$
(0,1,1)
$$

$$
(2,0,2)
$$

$$
(1,0,0)
$$

$$
(0,1,0)
$$