Introduction to Three Dimensional Geometry Test

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Introduction to Three Dimensional Geometry Test - 19

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Question 1
1 / -0

A point on the line $$\displaystyle \frac{{x + 2}}{1} = \frac{{y - 3}}{{ - 4}} = \frac{{z - 1}}{{2\sqrt 2 }}$$ at a distance 6 from the point (2, 3, 1) is

A
$$(4-21, 1+12\sqrt{2})$$

B
$$\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$$

C
$$\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$$

D
$$\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$$

Solution

$$(\lambda -2, -4\lambda+3, 2\sqrt 2\lambda+1)\leftrightarrow (-2, 3, 1)$$
$$(\lambda-2+2)^2+(-4\lambda+3-3)^2+2\sqrt 2\lambda+1-1)^2=36$$
$$\lambda^2+16\lambda^2+8\lambda^2=36$$
$$\lambda=\pm \frac {6}{5}$$
$$\therefore point=(\frac {-4}{5}, \frac {-9}{5}, 1)$$
Question 2
1 / -0

If $$A (1, 2, 3), B ( 2, 3, 1), C (3, 1, 2)$$ then the length of the altitude through $$C$$ is

A
$$3$$

B
$$3\sqrt{3}$$

C
$$3\sqrt{2}$$

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

Solution

So $$\bigtriangleup \ ABC$$ is equilateral
So length of median $$=$$ length of attitude
$$D$$ is mid point of $$AB$$
P.V of $$D=(\dfrac{3}{2},\dfrac{5}{2},2)$$
$$CD=\sqrt{\dfrac{9}{4}+\dfrac{9}{4}}$$
$$=\dfrac{3}{\sqrt{2}}$$
Question 3
1 / -0

A plane intersects the co ordinate axes at $$A, B, C$$. If $$O= (0, 0, 0)$$ and $$(1, 1, 1)$$ is the centroid of the tetrahedron $$O ABC$$, then the sum of the reciprocals of the intercepts of the plane

A
$$12$$

B
$$\displaystyle \dfrac{4}{3}$$

C
$$1$$

D
$$\displaystyle \dfrac{3}{4}$$

Solution

Let the point of intersections be,
$$A(a,0,0) , B(0,b,0)$$ and $$C(0,0,c)$$
The coordinates of the centroid are $$\left ( \dfrac{a}{4} , \dfrac{b}{4} , \dfrac{c}{4} \right )$$
Comparing it with the coordinates given, we get
$$a = 4 , b = 4 , c = 4$$
Hence, $$\left ( \dfrac{1}{a} , \dfrac{1}{b} , \dfrac{1}{c} \right) = \dfrac{3}{4} $$
Question 4
1 / -0

The plane $$ax+by+cz+(-3)=0$$ meet the co-ordinate axes in A,B,C. Then centroid
of the triangle is

A
$$(3a,3b,3c)$$

B
$$(\displaystyle \frac{3}{a}\frac{3}{b},\frac{3}{c})$$

C
$$\left ( \dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3} \right )$$

D
$$(\displaystyle \dfrac{1}{a}, \dfrac{1}{b},\dfrac{1}{c})$$

Solution

For finding the coordinates of the point where the plane $$ax + by + cz - 3 = 0$$ cuts the x axis, we equate y and z to zero.
The point becomes $$\left (\cfrac{3}{a}, 0, 0 \right )$$
Similarly, the point on y axis becomes $$\left (0, \cfrac{3}{b}, 0 \right )$$
And that on z axis becomes $$\left (0, 0, \cfrac{3}{c} \right )$$
The centroid of the triangle formed by these points would be
$$\left ( \cfrac{\cfrac{3}{a} + 0 + 0}{3}, \cfrac{0 + \cfrac{3}{b} + 0}{3}, \cfrac{0 + 0 + \cfrac{3}{c}}{3} \right )$$
$$= \left ( \cfrac{1}{a}, \cfrac{1}{b}, \cfrac{1}{c} \right )$$
Question 5
1 / -0

The name of the figure formed by the points $$(-1, -3, 4), (5, -1,1), (7, -4, 7)$$ and $$(1, -6, 10)$$ is a

A
square

B
rhombus

C
parallelogram

D
rectangle

Solution

Keeping the above points as vertices and using the distance formula,
$$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$$
We get that the sides of the parallelogram formed by the above lines are equal and all the sides are of $$7$$ units.
Hence the parallelogram will be either a rhombus or a square. For the parallelogram to be square, all the adjacent sides of the parallelogram should make an angle of $$\dfrac{\pi}{2}$$.
Consider the vector equation of $$AB$$ as $$(5-(-1))i'+(-1-(-3))j'+(1-4)k'$$
$$6i'+2j'-3k'$$
Consider the vector equation of $$BC$$ as $$(7-5)i'+(-4-(-1))j'+(7-1)k'$$
$$2i'-3j'+6k'$$
Taking dot product, we get $$6(2)+2(-3)-3(6)$$
$$=12-6-18$$
$$=-12$$
Thus the parallelogram is a rhombus.
Question 6
1 / -0

The end points of a body diagonal of a rectangular parallelepiped whose faces are parallel to the coordinate planes are $$(2, 3, 5)\ and\ (5, 7, 10)$$. The lengths of its sides are

A
$$5, 7, 3$$

B
$$6, 5, 3$$

C
$$3, 6, 2$$

D
$$3, 4, 5$$

Solution

$$\bar { AB } =3\hat { i } +4\hat { j } +5\hat { k } $$
$$\overrightarrow { AB } $$ is the resultant of $$3\hat { i } $$ (side parallel to x axis), $$4\hat { j } $$ (side parallel to y axis),$$5\hat { k } $$ (side parallel to x axis)
So $$(3,4,5)$$ are lengths of sides.
Question 7
1 / -0

A tetrahedron is a three dimensional figure bounded by non coplanar triangular planes. So, a tetrahedron has four non-coplanar points as its vertices. Suppose a tetrehedron has points A,B,C,D as its vertices which have coordinates $$(x1, y1, z_{1}) (x_{2}, y_{2}, z_{2})$$ , $$(x_{3}, y_{3}, z_{3})$$ and $$(x_{4}, y_{4}, z_{4})$$, respectively in a rectangular three dimensional space. Then, the coordinates of its centroid are $$[\dfrac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\dfrac{y_{1}+y_{2}+y_{3}+y_{4}}{4},\dfrac{z_{1}+z_{2}+z_{3}+z_{4}}{4}]$$.
Let a tetrahedron have three of its vertices represented by the points $$(0,0,0) ,(6,5,1)$$ and $$(4,1,3)$$ and its centroid lies at the point $$(1,2,5)$$. Now, answer the following question. The coordinate of the fourth vertex of the tetrahedron is:

A
$$(-6, 2,16)$$

B
$$(1, -2,13)$$

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

D
$$(1, -1,1)$$

Solution

Given three vertices of tetrahedron are $$(0,0,0), (6,5,1),(4,1,3)$$
Now equation of plane passing through $$(0,0,0)$$ is given by,
$$ax+by+cz=0$$
Also this line passing through other two points
$$\Longrightarrow 6a+5b+c=0, 4a+b+3c=0$$
Solving these, we get $$a=-c,b=c$$
Hence required plane will be $$-x+y+z=0$$
From the above equationn the coordinate of the fourth vertex of the tetrahedron is $$(-6,2,16)$$
Question 8
1 / -0

If the extremities of a diagonal of a square are $$(1, -2, 3)$$ and $$(4, 2, 3)$$ then the area of the square is

A
$$25$$

B
$$50$$

C
$$\displaystyle \frac{25}{2}$$

D
$$\sqrt{50}$$

Solution

If $$a$$ is the length of a side of square then the length of diagonal is given by $$\sqrt{2}a$$. Distance between two given points is $$\sqrt{(1-4)^2+(-2-2)^2+(3-3)^2}=5=\sqrt{2}a$$. Hence the area is given by $$a^2=\dfrac{25}{2}$$.
Question 9
1 / -0

Directions For Questions

Let $$A(1, 2, 3), B(0, 0, 1)$$ and $$C(-1, 1,1)$$ are the vertices of $$\Delta$$ $$ABC$$.

...view full instructions

The equation of median through C to side AB is

A
$$r = - \hat i + \hat j + \hat k + p (3\hat i - 2\hat k)$$

B
$$r = - \hat i + \hat j + \hat k + p (3\hat i + 2\hat k)$$

C
$$r = - \hat i + \hat j + \hat k + p (-3\hat i + 2\hat k)$$

D
$$r = -\hat i + \hat j + \hat k + p (3\hat i + 2 \hat j )$$

Solution

Given, coordinates of A = (1 , 2 , 3 )
coordinates of B = (0 , 0 , 1 )
coordinates of C = (-1 , 1 , 1 )
Now CD is the median through C on AB.
Now we have to find the coordinate of D.
Hence, D = $$\left( \dfrac{1+0}{2} , \dfrac{2+0}{2} , \dfrac{3+1}{2} \right) $$ [Since D is the midpoint of AB ]
$$ \therefore D = \left( \dfrac{1}{2} , 1 , 2 \right) \, \, [\therefore \, \textrm{midpoint} \, = \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} , \dfrac{z_1 + z_2}{2} ] $$
$$ \therefore $$ The direction radios of CD $$ = \left( \dfrac{1}{2} + 1 , 1 - 1 , 2-1 \right) \\ = \left( \dfrac{3}{2} , 0 , 1 \right) = (3 , 0 , 2) $$
And this line passes through C
$$\therefore $$ Eqn of line CD , $$ \dfrac{x+1}{3/2} = \dfrac{y-1}{0} = \dfrac{z-1}{1} $$
$$ \therefore \vec{r} = (-\hat{i} + \hat{j} + \hat{k}) + P \, (3 \hat{i} + 2 \hat{k}) $$
Question 10
1 / -0

From which of the following the distance of the point $$(1, 2, 3)$$ is $$\sqrt{10}$$?

A
Origin

B
$$x-$$axis

C
$$y-$$axis

D
$$z-$$axis

Solution

The point is $$P(1,2,3)$$ so distance of point from $$y$$ axis is
$${=}$$ $$\sqrt{{(1)}^{2} + {(3)}^{2} } $$
$$=\sqrt{10}$$

Hence, option C is correct.

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Introduction to Three Dimensional Geometry Test - 19

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Introduction to Three Dimensional Geometry Test - 19

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Question 1

$$(4-21, 1+12\sqrt{2})$$

$$\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$$

$$\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$$

$$\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$$

Solution

Question 2

$$3$$

$$3\sqrt{3}$$

$$3\sqrt{2}$$

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

Solution

Question 3

$$12$$

$$\displaystyle \dfrac{4}{3}$$

$$1$$

$$\displaystyle \dfrac{3}{4}$$

Solution

Question 4

$$(3a,3b,3c)$$

$$(\displaystyle \frac{3}{a}\frac{3}{b},\frac{3}{c})$$

$$\left ( \dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3} \right )$$

$$(\displaystyle \dfrac{1}{a}, \dfrac{1}{b},\dfrac{1}{c})$$

Solution

Question 5

square

rhombus

parallelogram

rectangle

Solution

Question 6

$$5, 7, 3$$

$$6, 5, 3$$

$$3, 6, 2$$

$$3, 4, 5$$

Solution

Question 7

$$(-6, 2,16)$$

$$(1, -2,13)$$

$$(-2,4,-2)$$

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

Solution

Question 8

$$25$$

$$50$$

$$\displaystyle \frac{25}{2}$$

$$\sqrt{50}$$

Solution

Question 9

$$r = - \hat i + \hat j + \hat k + p (3\hat i - 2\hat k)$$

$$r = - \hat i + \hat j + \hat k + p (3\hat i + 2\hat k)$$

$$r = - \hat i + \hat j + \hat k + p (-3\hat i + 2\hat k)$$

$$r = -\hat i + \hat j + \hat k + p (3\hat i + 2 \hat j )$$

Solution

Question 10

Origin

$$x-$$axis

$$y-$$axis

$$z-$$axis

Solution

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