Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

The vector equation $$r=i-2j-k+t(6j-k)$$ represents a straight line passing through the points:

A
$$(0, 6, -1)$$ and $$(1, -2, -1)$$

B
$$(0, 6, -1)$$ and $$(-1, -4, -2)$$

C
$$(1, -2, -1)$$ and $$(1, 4, -2)$$

D
$$(1, -2, -1)$$ and $$(0, -6, 1)$$

Solution

Cartesian representation of the given line is,

$$\dfrac{x-1}{0}=\dfrac{y+2}{6}=\dfrac{z+1}{-1}=t$$
So any point on the given line is of the form $$(1,6t-2, -t-1)$$
where $$t$$ can be any real numbers

so for $$t=0$$ and $$1$$ the corresponding points are $$(1,-2,-1)$$ and $$(1,4,-2)$$

You can check other options does not satisfy above point for any $$t$$.
Question 2
1 / -0

A line makes the same angle $$\theta$$ with each of the $$X$$ and $$Z$$-axes. If the angle $$\beta$$, which it makes with $$Y$$-axis, is such that $$\sin ^{ 2 }{ \beta } =3\sin ^{ 2 }{ \theta } $$, then $$\cos ^{ 2 }{ \theta } $$ equals

A
$$\dfrac {2}{5}$$

B
$$\dfrac {1}{5}`$$

C
$$\dfrac {3}{5}$$

D
$$\dfrac {2}{3}$$

Solution

Let $$l,m$$ and $$n$$ be the direction cosines

Then, $$l=\cos { \theta } ,m=\cos { \beta } ,n=\cos { \theta } $$

We have $$\quad { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$

$$\Rightarrow \cos ^{ 2 }{ \theta } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \theta } =1\quad $$

$$\Rightarrow 2\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \beta } =0$$

$$\Rightarrow 2\cos ^{ 2 }{ \theta } -3\sin ^{ 2 }{ \beta } =0\left[ \because \sin ^{ 2 }{ \beta } =3\sin ^{ 2 }{ \theta } \right] $$

$$\Rightarrow \tan ^{ 2 }{ \theta } =\dfrac {2}{3}$$

Therefore, $$\cos ^{ 2 }{ \theta } =\cfrac { 1 }{ 1+\tan ^{ 2 }{ \theta } } =\cfrac { 3 }{ 5 } $$
Question 3
1 / -0

Equation to a line parallel to the vector $$2\hat{i}-\hat{j}{+}\hat{k}$$ and passing through the point $$\hat{i}+\hat{j}{+\hat{k}}$$

A
$$(x-1)\hat{i}+(y-1)\hat{j}+(z -1)\hat{k}=\lambda(2\hat{i}-\hat{j}+\hat{k})$$

B
$$(x-2)\hat{i}+(y-1)\hat{j}+(z+1)\hat{k}=\lambda(\hat{i}+\hat{j}+\hat{k})$$

C
$$2x\hat{i}-y\hat{j}+z\hat{k}=\lambda(\hat{i}+\hat{j}+\hat{k})$$

D
$$(x-1)\hat{i}+(y+1)\hat{j}+(z+1)\hat{k}=\lambda(\hat{i}+\hat{j}+\hat{k})$$

Solution

Equation parallel to a vector $$\vec{a}$$ and passing through point of position vector $$\vec{b}$$ is given by $$\vec{r}=\vec{a}+\lambda \vec{b}$$
$$\vec{r}= (\hat{i}+\hat{j}+\hat{k})+\lambda (2\hat{i}-\hat{j}+\hat{k})$$
$$x\hat{i}+y\hat{j}+z\hat{k}= \hat{i}+\hat{j}+\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$$
$$(x-1)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k}= \lambda (2\hat{i}-\hat{j}+\hat{k})$$
Question 4
1 / -0

The points with position vectors $$ 60i + 3j, 40i -8j$$ and $$ ai -52j $$ are collinear if

A
$$a = -40$$

B
$$a = 40$$

C
$$a = 20$$

D
None of these

Solution

Denoting $$a,b,c$$ by the given vectors respectively
These vectors will be collinear if there is some constant $$k$$ such that $$c-a=K\left( b-a \right) $$
$$\Rightarrow a-60=-20K$$ and $$-55=-11K$$
$$\Rightarrow a=-100+60=-40$$
Question 5
1 / -0

If the direction cosines of a line are $$\left(\displaystyle \dfrac{1}{c},\dfrac{1}{c},\dfrac{1}{c}\right)$$ then $$c=$$______

A
$$3$$

B
$$1/\sqrt{3}$$

C
$$\pm\sqrt{3}$$

D
$$\displaystyle \pm\dfrac{1}{\sqrt{3}}$$

Solution

If direction cosines are $$\left ( \dfrac{1}{c}, \dfrac{1}{c}, \dfrac{1}{c} \right)$$
$$\Rightarrow \left (\dfrac {1}{c}\right)^2+\left (\dfrac {1}{c}\right)^2 +\left (\dfrac {1}{c}\right)^2 = 1 $$
$$\Rightarrow c^{2} = 3$$
$$\Rightarrow c=\pm\sqrt 3$$
Question 6
1 / -0

$$l = m =n = 1$$ represents the direction cosines of

A
$$x-$$axis

B
$$y-$$axis

C
$$z-$$axis

D
none of these

Solution

Suppose, $$l,m,n$$ are direction cosines
$$\implies l^2+m^2+n^2=1$$
But $$l=m=n=1$$
$$\implies 3m^2=1$$
$$\implies l=m=n=\dfrac{1}{\sqrt{3}}$$ which are not direction cosines of either of the three co-ordinate axes.
Hence, option D is correct.
Question 7
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The direction ratios of the diagonal of the cube joining the origin to the opposite corner are (when the $$3$$ concurrent edges of the cube are coordinate axes)

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

B
$$1,1,1$$

C
$$2,-2,1$$

D
$$1,2,3$$

Solution

Since, a cube is a symmetric figure, the vertex we are talking about will be at the diagonally opposite end of the origin. i.e. it will be equally inclined to the three axes.
Let the side of the cube be $$a$$, then the corner opposite to origin will have coordinates $$(a,a,a)$$.
Direction ratios of a line joining two points $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2)$$ is given by $$(x_2-x_1,y_2-y_1,z_2-z_1)$$
Then, direction ratios of two point $$(0,0,0)$$ and $$(a,a,a)$$ will be $$(a-0,a-0,a-0)=(a,a,a)=a(1,1,1)$$
Hence, the direction ratios are $$1,1,1$$.

Question 8

1 / -0

List I	List II
1) d.c's of $$x -$$ axis	a) $$(1,1,1)$$
2) d.c's of $$y -$$ axis	b)$$\left(\displaystyle \frac{]}{\sqrt{3}}\frac{]}{\sqrt{3}},\frac{]}{\sqrt{3}}\right)$$
3) d.c's of $$z -$$ axis	c) $$(1,0,0)$$
4) d.c's of a line makes equal angles with axes	d) $$(0,1,0)$$
	e) $$(0,0,1)$$

The correct order for 1, 2, 3, 4 is

$$c,d,e,b$$

$$a,b,c,e$$

$$c,d,a,b$$

$$b,c,a,e$$

Solution

On the $$x-$$ axis, the other $$2$$ coordinates $$y,z$$ will be $$0$$.

So, the point is $$(1,0,0)$$

Direction cosines of $$x-$$ axis i.e $$(1,0,0)$$ is $$(1,0,0)$$

$$\therefore$$ Direction cosines of $$x-$$ axis are $$(1,0,0)$$

Similarly, direction cosines of $$y$$ and $$z$$ axes are $$(0,1,0)$$ and $$(0,0,1)$$ respectively.

$$\implies 1\rightarrow c, 2\rightarrow d, 3\rightarrow e$$
For the fourth part where we need to find the dc's of a line that makes equal angles with the axes.
The components of such a line will always be equal.
Moreover, the sum of the squares of the dc's of a line should be equal.

If the point is $$(a,a,a)$$, then $$a^2+a^2+a^2=3a^2$$

So, the direction cosines are $$\left(\dfrac{a}{\sqrt{3a^2}},\dfrac{a}{\sqrt{3a^2}},\dfrac{a}{\sqrt{3a^2}}\right)=\left(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}\right)$$

$$\implies 4\rightarrow b$$

Hence, option A is correct.

Question 9
1 / -0

If $$P(x, y, z)$$ moves such that $$x=0, z=0$$, then the locus of $$P$$ is the line whose d.cs are

A
$$y$$-axis

B
$$1, 0, 0$$

C
$$0, 1, 0$$

D
$$0, 0, 0$$

Solution

When $$P$$ moves then $$x=0,z=0$$ but $$y$$ is not given.
Let $$y=y$$
Then the coordinates of the point will be $$(0,y,0)$$
Now, direction cosines with respect to $$(0,y,0)$$ is given by
$$\cos \alpha = \dfrac{0}{\sqrt{0^2+y^2+0^2}}=\dfrac{0}{y}=0$$
$$\cos \beta = \dfrac{y}{\sqrt{0^2+y^2+0^2}}=\dfrac{y}{y} = 1$$
$$\cos \gamma = \dfrac{0}{\sqrt{0^2+y^2+0^2}}=\dfrac{0}{y}=0$$
Hence, the direction cosines are $$0,1,0$$.
Question 10
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If the points $$A(\overline{a}), B(\overline{b}), C(\overline{c})$$ satisfy the relation $$3\mathrm{a}-8\mathrm{b}+5\mathrm{c}=0$$ then the points are

A
Vertices of an equilateral triangle

B
Collinear

C
Vertices of a right angled triangle

D
Vertices of a isosceles triangle

Solution

The points $$A(\vec{a}), B(\vec{b}), C(\vec{c})$$ satisfy the relation $$3a-8b+5c=0$$
$$\implies 3\vec{a}-8\vec{b}+5\vec{c}=0$$
$$\implies 3\vec{a}+5\vec{c}=8\vec{b}$$
$$\implies \dfrac{3\vec{a}+5\vec{c}}{8}=\vec{b}$$
$$\implies \dfrac{3\vec{a}+5\vec{c}}{3+5}=\vec{b}$$
Similarly, $$\vec{a}=\dfrac{8\vec{b}-5\vec{c}}{8-5}$$ and $$\vec{c}=\dfrac{8\vec{b}-3\vec{a}}{8-3}$$
This is a section formula.
Hence $$\vec{a},\vec{b},\vec{c}$$ are collinear.

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Three Dimensional Geometry Test - 14

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Question 1

$$(0, 6, -1)$$ and $$(1, -2, -1)$$

$$(0, 6, -1)$$ and $$(-1, -4, -2)$$

$$(1, -2, -1)$$ and $$(1, 4, -2)$$

$$(1, -2, -1)$$ and $$(0, -6, 1)$$

Solution

Question 2

$$\dfrac {2}{5}$$

$$\dfrac {1}{5}`$$

$$\dfrac {3}{5}$$

$$\dfrac {2}{3}$$

Solution

Question 3

$$(x-1)\hat{i}+(y-1)\hat{j}+(z -1)\hat{k}=\lambda(2\hat{i}-\hat{j}+\hat{k})$$

$$(x-2)\hat{i}+(y-1)\hat{j}+(z+1)\hat{k}=\lambda(\hat{i}+\hat{j}+\hat{k})$$

$$2x\hat{i}-y\hat{j}+z\hat{k}=\lambda(\hat{i}+\hat{j}+\hat{k})$$

$$(x-1)\hat{i}+(y+1)\hat{j}+(z+1)\hat{k}=\lambda(\hat{i}+\hat{j}+\hat{k})$$

Solution

Question 4

$$a = -40$$

$$a = 40$$

$$a = 20$$

None of these

Solution

Question 5

$$3$$

$$1/\sqrt{3}$$

$$\pm\sqrt{3}$$

$$\displaystyle \pm\dfrac{1}{\sqrt{3}}$$

Solution

Question 6

$$x-$$axis

$$y-$$axis

$$z-$$axis

none of these

Solution

Question 7

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

$$1,1,1$$

$$2,-2,1$$

$$1,2,3$$

Solution

Question 8

$$c,d,e,b$$

$$a,b,c,e$$

$$c,d,a,b$$

$$b,c,a,e$$

Solution

Question 9

$$y$$-axis

$$1, 0, 0$$

$$0, 1, 0$$

$$0, 0, 0$$

Solution

Question 10

Vertices of an equilateral triangle

Collinear

Vertices of a right angled triangle

Vertices of a isosceles triangle

Solution

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