Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 35

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Question 1
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If the points $$(p,0)$$, $$(0,q)$$ and $$(1,1)$$ are collinear, then $$\dfrac { 1 }{ p }+\dfrac { 1 }{ q }$$ is equal to:

A
$$-1$$

B
$$1$$

C
$$2$$

D
$$0$$

Solution

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other

points.

Slope of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 }

\right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ $$ = $$ $$\dfrac { { y

}_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$

So, slope of the line joining $$ (p,0) , (0,q) = $$ Slope of the line joining

$$ (0,q) $$ and $$ (1,1) $$

$$ \dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 } $$

$$ - \dfrac { q }{ p } = 1 - q $$
Dividing both sides by $$q$$,
$$ - \dfrac { 1 }{ p } = \dfrac { 1 }{ q } - 1 $$
$$ => \dfrac { 1 }{ p } + \dfrac { 1 }{ q } = 1 $$
Question 2
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The direction ratios of two lines are $$1,-2,-2$$ and $$0,2,1$$. The direction cosines of the line perpendicular to the above lines are

A
$$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 } $$

B
$$\displaystyle \frac { -1 }{ 3 } ,\frac { 2 }{ 3 } ,\frac { 2 }{ 3 } $$

C
$$\displaystyle \frac { 1 }{ 4 } ,\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } $$

D
None of these

Solution

Let $$a,b,c$$ be the direction ratios of the line whose direction cosines are required.
Then as this line is perpendicular to the given lines so we have
$$a(1)+b(-2)+c(-2)=0$$ and $$a(0)+b(2)+c(1)=0$$
Solving these simultaneously, we get
$$\displaystyle \frac { a }{ \left( -2 \right) \left( 1 \right) -\left( -2 \right) \left( 2 \right) } =\frac { b }{ \left( -2 \right) \left( 0 \right) -\left( 1 \right) \left( 1 \right) } =\frac { c }{ \left( 1 \right) \left( 2 \right) -\left( 0 \right) \left( -2 \right) } $$
$$\displaystyle \Rightarrow \frac { a }{ 2 } =\frac { b }{ -1 } =\frac { c }{ 2 } \Rightarrow a:b:c=2:-1:2$$
Therefore the required direction cosines are
$$\displaystyle \frac { 2 }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } ,\frac { -1 }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } ,\frac { 2 }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } \Rightarrow \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 } $$
Question 3
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If the foot of the perpendicular from the origin to a plane is $$\left( a,b,c \right) ,$$ the equation of the plane is

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

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

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

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

Solution

Let $$P\left( a,b,c \right) $$ be the foot of the perpendicular from the origin to the plane, then direction ratios of $$OP$$ are $$a-0,b-0,c-0$$ i.e. $$a,b,c.$$
So the equation of the plane passing through $$P\left( a,b,c \right) $$, the direction ratios of the normal to which are $$a,b,c$$ is
$$a\left( x-a \right) +b\left( y-b \right) +c\left( z-c \right) =0$$
$$\Rightarrow ax+by+cz={a}^{2}+{b}^{2}+{c}^{2}$$
Question 4
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Are the points (1, 1), (2, 3) and (8, 11) collinear ?

A
collinear

B
Non collinear

C
coplaner

D
None of above

Solution

Area of triangle formed by these vertices is
$$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 8 & 11 & 1 \end{vmatrix}$$
Applying $${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$$
$$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 7 & 10 & 0 \end{vmatrix}=\frac { 1 }{ 2 } \left( 10-14 \right) =2$$
Hence points are non collinear
Question 5
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The angle between the straight lines whose direction cosines are given by $$2l+2m-n=0,mn+nl+lm=0$$, is

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

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

C
$$\displaystyle \frac { \pi }{ 4 } $$

D
None of these

Solution

Relations are $$2l-2m-n=0\Rightarrow n=2l+2m$$ and $$mn+nl+lm=0$$.
Eliminating $$n$$, we have $$(m+l)(2l+2m)+lm=0$$
$$\Rightarrow2{ l }^{ 2 }+5lm+2{ m }^{ 2 }=0$$
$$\Rightarrow \left( 2l+m \right) \left( l+2m \right) =0$$
When $$2l+m=0$$, we have from $$2l+2m-n=0,n=m.$$
Thus $$\displaystyle \frac { l }{ -1 } =\frac { m }{ 2 } =\frac { n }{ 2 } ,$$
$$\Rightarrow$$ d.c's of one line are proportional to $$[-1,2,2]$$.
Again, when $$l+2m=0$$, we have from $$2l+2m-n=0,n=l$$
$$\displaystyle \therefore \frac { l }{ 2 } =\frac { m }{ -1 } =\frac { n }{ 2 } ,$$
$$\Rightarrow$$ d.c's of the other line are proportional to $$[2,-1,2]$$.
Now, if $$\theta$$ be the angle between the two lines, then
$$\displaystyle \cos { \theta } =\frac { -1.2+2.-1+2.2 }{ \sqrt { \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 } \right) } \sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } \right) } } =0,$$
$$\Rightarrow \displaystyle \theta =\frac { \pi }{ 2 } $$ or the two lines are at right angles to each other.
Question 6
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The planes $$3x-y+z+1=0,5x+y+3z=0$$ intersect in the line $$PQ$$. The equation of the plane through the point $$(2,1,4)$$ and perpendicular to $$PQ$$ is

A
$$x+y-2z=5$$

B
$$x+y-2z=-5$$

C
$$x+y+2z=5$$

D
$$x+y+2z=-5$$

Solution

Let $$[l,m,n]$$ be the direction cosines of $$PQ$$, then $$3l-m+n=0$$ and $$5l+m+3n=0$$
$$\displaystyle \therefore \frac { l }{ -3-1 } =\frac { m }{ 5-9 } =\frac { n }{ 3+5 } \Rightarrow \frac { l }{ 1 } =\frac { m }{ 1 } =\frac { n }{ -2 } $$
Now, a plane $$\bot$$ to $$PQ$$ will have $$l,m,n$$ as the coefficient of $$x,y$$ and $$z$$.
Hence, the plane $$\bot$$ to $$PQ$$ is $$x+y-2z=\lambda$$
It passes through $$(2,1,4)$$
$$\therefore 2+1-2.4=\lambda \Rightarrow \lambda =-5$$
Hence, the required plane is $$x+y-2z=-5$$
Question 7
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Directions For Questions

Given four points $$A(2, 1, 0); B(1, 0, 1); C(3, 0, 1)$$ and $$D(0, 0, 2)$$. The point $$D$$ lies on a line $$L$$ orthogonal to the plane determined by the point $$A, B$$ and $$C$$

...view full instructions

Equation of the line $$L$$ is -

A
$$\vec{r} = 2\hat{k} + \lambda(\hat{i} + \hat{k})$$

B
$$\vec{r} = 2\hat{k} + \lambda(2\hat{j} + \hat{k})$$

C
$$\vec{r} = 2\hat{k} + \lambda(\hat{j} + \hat{k})$$

D
none of these

Solution

$$\textbf{Step-1: Find the normal vector of the plane using given information}$$

$$\text{Given four points A(2,1,0), B(1,0,1), C(3,0,1) and D(0,0,2)}$$

$$\text{From the given points, we have}$$

$$\vec{BA}=(2-1)\hat i+(1-0)\hat j+(0-1)\hat k$$, $$\vec{BC}=(3-1)\hat i+(0-0) \hat j+(1-1)\hat k$$

$$\vec{BA}=\hat i+\hat j- \hat k$$, $$\vec{BC}=2\hat i$$

$$\text{So, the normal vector of the plane containing}$$ $$ABC$$ $$\text{is}$$

$$\vec{BA}\times \vec{BC}=(2i)\times (\hat i+\hat j-\hat k)$$

$$=2(\hat j+\hat k)$$

$$\text{Since the line is orthogonal to the plane, hence its directional vector is}$$

$$\text{ parallel to the normal vector of the plane.}$$

$$\text{Therefore the direction vector of the line is}$$ $$(j+k)$$.

$$\text{Now, it passes through}$$ $$D=2k$$

$$\text{Therefore the required equation of the line is}$$

$$\vec{r}=2k+\lambda(j+k)$$

$$\textbf{Hence, option - C is the answer}$$
Question 8
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If direction ratios of the normal of the plane which contains the lines $$\displaystyle\frac{x-2}{3}=\displaystyle\frac{y-4}{2}=\displaystyle\frac{z-1}{1}\;\&\;\displaystyle\frac{x-6}{3}=\displaystyle\frac{y+2}{2}=\displaystyle\frac{z-2}{1}$$ are $$(a,\,1,\,-26)$$, then $$a$$ is equal to

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

$$\dfrac { x-2 }{ 3 } =\dfrac { y-4 }{ 2 } =\dfrac { z-1 }{ 1 }$$
$$ \dfrac { x-6 }{ 3 } =\dfrac { y+2 }{ 2 } =\dfrac { z-2 }{ 1 } $$
$$ (a,1,-26) \rightarrow $$ DR's of normal
$$\Rightarrow { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 }= 0$$
$$ \Rightarrow 3(a)+(1)+1(-26) = 0$$
$$ \Rightarrow 3a=24$$
$$ \Rightarrow$$ $$a = 8$$
Hence the answer is $$8.$$
Question 9
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The equation of the plane which contains the lines $$\vec{r}\, =\, \hat{i}\, +\, 2 \hat{j}\, -\, \hat{k}\, +\, \lambda\, (\hat{i}\, +\, 2\, \hat{j}\, -\, \hat{k})$$ and $$\vec{r}\, =\, \hat{i}\, +\, 2 \hat{j}\, -\, \hat{k}\, +\, \mu\, (\hat{i}\, +\, \hat{j}\, +\, 3 \hat{k})$$ must be

A
$$\vec{r}.\, (7 \hat{i}\, -\, 4\, \hat{j}\, -\, \hat{k})\, =\, 0$$

B
$$7(x\, -\, 1)\, -\, 4(y\, -\, 2)\, -\, (z\, +\, 1)\, =\, 0$$

C
$$\vec{r}.\, (\hat{i}\, +\, 2\, \hat{j}\, -\, \hat{k})\, =\, 0$$

D
$$\vec{r}.\, (\hat{i}\, +\, \hat{j}\, +\, 3\, \hat{k})\, =\, 0$$

Solution

Clearly lines are parallel to $$\hat{i}+2\hat{j}-\hat{k}$$ and $$\hat{i}+\hat{j}+3\hat{k}$$
Thus normal vector of required plane is,
$$\vec{n}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\1&2&-1\\1&1&3\end{vmatrix}=7\hat{i}-4\hat{j}-\hat{k}$$
Hence, required plane is
$$[\vec{r}-(\hat{i}+2\hat{j}-\hat{k})]\cdot \vec{n}=0$$
$$\Rightarrow [\vec{r}-(\hat{i}+2\hat{j}-\hat{k})]\cdot (7\hat{i}-4\hat{j}-\hat{k})=0$$
$$\Rightarrow 7(x-1)-4(y-2)-(z+1)=0$$
Question 10
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Directions For Questions

In the shown figure, cuboid $$(AB = 4, AE = 3, AD = 5)$$ an insect can only crawl along the wall and it cannot fly.

...view full instructions

The direction ratios of a line followed by the insect during its journey from A to G along the shortest path are

A
$$(3, 4, 5)$$

B
$$(0, 4, 5)$$

C
$$(7, 0, 5)$$

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

Solution

Paths teken can be three:
$$i)AF$$ & $$FG=5+5=10$$
$$ii) AE$$ & $$EG=3+\sqrt { 14 } $$
$$iii)AB$$ & $$BG=4+\sqrt { 34 } $$
shortest one is $$ii)$$
$$\therefore A\longrightarrow E\longrightarrow G$$
$$\therefore$$ direction ratio $$\therefore A\longrightarrow E=(0,0,3)\equiv (0,0,1)$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 35

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Three Dimensional Geometry Test - 35

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SHARING IS CARING

Question 1

$$-1$$

$$1$$

$$2$$

$$0$$

Solution

Question 2

$$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 } $$

$$\displaystyle \frac { -1 }{ 3 } ,\frac { 2 }{ 3 } ,\frac { 2 }{ 3 } $$

$$\displaystyle \frac { 1 }{ 4 } ,\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } $$

None of these

Solution

Question 3

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

$$ax+by+cz=3$$

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

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

Solution

Question 4

collinear

Non collinear

coplaner

None of above

Solution

Question 5

$$\displaystyle \frac { \pi }{ 2 } $$

$$\displaystyle \frac { \pi }{ 3 } $$

$$\displaystyle \frac { \pi }{ 4 } $$

None of these

Solution

Question 6

$$x+y-2z=5$$

$$x+y-2z=-5$$

$$x+y+2z=5$$

$$x+y+2z=-5$$

Solution

Question 7

$$\vec{r} = 2\hat{k} + \lambda(\hat{i} + \hat{k})$$

$$\vec{r} = 2\hat{k} + \lambda(2\hat{j} + \hat{k})$$

$$\vec{r} = 2\hat{k} + \lambda(\hat{j} + \hat{k})$$

none of these

Solution

Question 8

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 9

$$\vec{r}.\, (7 \hat{i}\, -\, 4\, \hat{j}\, -\, \hat{k})\, =\, 0$$

$$7(x\, -\, 1)\, -\, 4(y\, -\, 2)\, -\, (z\, +\, 1)\, =\, 0$$

$$\vec{r}.\, (\hat{i}\, +\, 2\, \hat{j}\, -\, \hat{k})\, =\, 0$$

$$\vec{r}.\, (\hat{i}\, +\, \hat{j}\, +\, 3\, \hat{k})\, =\, 0$$

Solution

Question 10

$$(3, 4, 5)$$

$$(0, 4, 5)$$

$$(7, 0, 5)$$

$$(0, 0, 1)$$

Solution

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