Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 29

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Weekly Quiz Competition

Question 1
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Equation of the line which passes through the point with position vector $$(2, 1, 0)$$ and perpendicular to the plane containing the vectors $$i + j$$ and $$j + k$$ is

A
$$\vec{r}= (2, 1, 0) + t(1, -1, 1)$$

B
$$\vec{r} = (2, 1, 0) + t(-1, 1, 1)$$

C
$$\vec{r}= (2, 1, 0) + t(1, 1, -1)$$

D
$$\vec{r} = (2, 1, 0) + t(1, 1, 1)$$

Solution

$$\vec{r}=position\ vector\ \vec{a}+t(normal\ vector\ \vec{n})$$
normal vector of plane
Given $$\vec{a}=\hat{i}+\hat{j}$$
$$\vec{b}=\hat{j}+\hat{k}$$
$$\vec{n}=\vec{a}\times\vec{b}$$
$$\Rightarrow\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\0&1&1\end{vmatrix}$$
$$\Rightarrow\vec{a}\times\vec{b}=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)$$
$$\Rightarrow\vec{a}\times\vec{b}=\hat{i}-\hat{j}+\hat{k}$$
eq become
$$\Rightarrow\vec{r}=(2,1,0)+t(1,-1,1)$$
Question 2
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The projections of a directed line segment on the coordinate axes $$12, 4, 3$$. The direction cosines of the line are

A
$$\dfrac{12}{13}, -\dfrac{4}{13}, \dfrac{3}{13}$$

B
$$-\dfrac{12}{13}, -\dfrac{4}{13}, \dfrac{3}{13}$$

C
$$\dfrac{12}{13}, \dfrac{4}{13}, \dfrac{3}{13}$$

D
none of these

Solution

Assuming position vector of line segment ends point are $$\vec{A}$$ and $$\vec{B}$$
Now given projection of line segment on the axes are $$(12, 4, 3)$$
So $$\vec{AB} = 12 \hat{i} + 4\hat{j} + 3 \hat{k}$$
Hence, direction cosine of line segment are $$\dfrac{12}{\sqrt{12^2+4^2+3^2}}, \dfrac{4}{\sqrt{12^2+4^2+3^2}}, \dfrac{3}{\sqrt{12^2+4^2+3^2}}$$ $$\Rightarrow \dfrac{12}{13}, \dfrac{4}{13}, \dfrac{3}{13}$$
Question 3
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Equation of plane passing through the points $$(2, 2, 1)$$, $$(9, 3, 6)$$ and perpendicular to the plane $$2x+ 6y + 6z-1= 0$$ is

A
$$3x+ 4y+ 5z= 9$$

B
$$3x+ 4y- 5z+ 9= 0$$

C
$$3x+ 4y- 5z- 9=0$$

D
None of these

Solution

Let $$ax+by+cz=1$$ be a plane
Since, it is perpendicular to $$2x+6y+6z=1$$
Therefore, $$2a+6b+6c=1$$ ....(1)
and passes through $$(2,2,1)$$ & $$(9,3,6)$$
Therefore, $$2a+2b+c=1$$ ....(2)
and $$9a+3b+6c=1$$ ....(3)
Solving $$(1),(2)$$ and $$(3)$$ simultaneoulsly, we get
$$a=\dfrac {3}{9}$$, $$b=\dfrac {4}{9}$$, $$c=-\dfrac {5}{9}$$
Therfore, desired plane is $$3x+4y-5z-9=0$$

Ans: C
Question 4
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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 altitude through $$B$$ to side $$AC$$ is

A
$$r= k+t(7 i - 10 + 2k)$$

B
$$r = k + t (- 9 i + 6 j - 2k)$$

C
$$r = k + t (7i - 10 j - 2k)$$

D
$$r = k + t (7i + 10 j + 2k)$$

Solution

Equation of line $$AC$$ is given by,
$$\displaystyle \dfrac{x-1}{1+1}=\dfrac{y-2}{2-1}=\dfrac{z-3}{3-1}$$

$$\displaystyle \dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{2}=t$$ (say)

So any point on line $$AC$$ is $$P(2t+1, t+2, 2t+3)$$ such that $$BP\perp AC $$

$$\Rightarrow (2t+1-0)(2)+(t+2-0)(1)+(2t+3-1)(2)=0$$

$$\Rightarrow 4t+2+t+2+4t+4=0\Rightarrow t=-\dfrac{8}{9}$$

$$\therefore\displaystyle P =\left(-\dfrac{7}{9},\dfrac{10}{9},\dfrac{11}{9} \right)$$

Thus the equation altitude through $$B$$ to side $$AC$$, i.e. the equation of line $$BP$$ is,

$$\displaystyle \dfrac{x-0}{-7/9}=\dfrac{y-0}{10/9}=\dfrac{z-1}{11/9-1}$$

$$\displaystyle \dfrac{x}{-7}=\dfrac{y}{10}=\dfrac{z-1}{2}$$

In vector form, it is given by, $$r=k+t(-7i+10j+2k)$$
Question 5
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Given $$A(1,-1,0)$$; $$B(3,1,2)$$;$$C(2,-2,4)$$ and $$D(-1,1,-1)$$ which of the following points neither lie on $$AB$$ nor on $$CD$$

A
$$(2,2,4)$$

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

C
$$(2,0,1)$$

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

Solution

Given:- $$A(1,-1,0)$$; $$B(3,1,2)$$;$$C(2,-2,4)$$ and $$D(-1,1,-1)$$
The equation of line $$AB$$ is given by $$r=i-j+t(2i+2j+2k)$$ or $$\dfrac{x-1}{2} = \dfrac{y+1}{2} = \dfrac{z}{2} $$
The points $$(2,0,1)$$ and $$(0,-2,1)$$ lies on $$AB$$.
The equation of line $$CD$$ is given by $$r=2i-2j+4k+t(3i-3j+5k)$$ or $$ \dfrac{x-2}{3} = \dfrac{y+2}{-3} = \dfrac{z-4}{5}$$
$$(2,-2,4)$$ lie on $$CD$$.
$$(2,2,4)$$ does not lie on any of the lines AB or CD.
Hence, option A is correct.
Question 6
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The vector equation of the line $$\displaystyle \frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1$$ is $$\displaystyle \overrightarrow { r } =\left( 2\hat { i } +\frac { 5 }{ 2 } \hat { j } -\hat { k } \right) +\lambda \left( 2\hat { i } -\frac { 3 }{ 2 } \hat { j } +x\hat { k } \right) $$, where $$x$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

The given line is $$\displaystyle\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1$$ $$\Rightarrow\displaystyle\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } =\frac { z+1 }{ 0 } $$
This shows that the given line passes through point $$\left( 2,\displaystyle\frac { 5 }{ 2 } ,-1 \right) $$ and has direction ratios $$\left( 2, \displaystyle\frac { -3 }{ 2 } ,0 \right) $$
So the direction cosines are
$$\displaystyle\frac { 2 }{ \sqrt { { 2 }^{ 2 }+{ \left( -\dfrac { 3 }{ 2 } \right) }^{ 2 } } } ,\frac { -\dfrac { 3 }{ 2 } }{ \sqrt { { 2 }^{ 2 }+{ \left( -\dfrac { 3 }{ 2 } \right) }^{ 2 } } } ,\frac { 0 }{ \sqrt { { 2 }^{ 2 }+{ \left( -\dfrac { 3 }{ 2 } \right) }^{ 2 } } } $$ or $$\left( \displaystyle\frac { 4 }{ 5 } ,-\frac { 3 }{ 5 } ,0 \right) $$
Thus the given line passes through the point having position vector $$\vec { a } =2\hat { i } +\displaystyle\frac { 5 }{ 2 } \hat { j } -\hat { k } $$ and is parallel to the vector $$\vec { b } =2\hat { i } -\displaystyle\frac { 3 }{ 2 } \hat { j } +0\hat { k } $$
So, its vector equations is $$\vec { r } =2\hat { i } +\displaystyle\frac { 5 }{ 2 } \hat { j } -\hat { k } +\lambda \left( 2\hat { i } -\frac { 3 }{ 2 } \hat { j } +0\hat { k } \right) $$
Hence, $$x=0$$.
Question 7
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If the points $$(0, 1, -2), (3$$, $$\lambda$$,$$ 1)$$ and ($$\mu$$, $$7, 4$$) are collinear, the point on the same line is

A
$$(5, 6, 3)$$

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

C
$$(-5, -6, -3)$$

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

Solution

The direction vector using the first two points we get $$3i+(\lambda-1)j+3k$$.
Using the first and third point $$\mu i+6j+6k$$.
Since they are up to multiplication by a constant we get $$\lambda=4$$ and $$\mu=6$$.
Hence direction vector is $$i+j+k$$.
Any point on the line is given by $$(0,1,-2)+t(1,1,1)$$. Plugging $$t=5$$ we get the point $$(5,6,3)$$.
Question 8
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A line makes angles $$\alpha$$, $$\beta $$, $$\gamma $$ with the positive directions of the axes of reference. The value of $$\cos 2\alpha +\cos 2\beta +\cos
2\gamma$$ is

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$0$$

Solution

The value of $$\cos 2\alpha+\cos 2\beta+\cos 2\gamma$$
$$=(1+\cos 2\alpha+1+\cos 2\beta+1+\cos 2\gamma)-3$$
$$=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3$$
$$=2-3$$
$$=-1$$
Question 9
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The direction cosines of the line joining the points $$(2,3,-1)$$ and $$(3,-2,1) $$ are

A
$$-1,5,-2$$

B
$$\dfrac{1}{\sqrt{30}}$$,$$-\sqrt{\dfrac{5}{6}}$$,$$\sqrt{\dfrac{2}{15}}$$

C
$$\dfrac{-1}{30 }$$,$$\dfrac{1}{6 }$$,$$-\dfrac{1}{15 }$$

D
none of these

Solution

Step 1:
The direction ratios of the line joining these points are
$$x_2−x_1,y_2−y_1,z_2−z_1$$
$$(i.e) (3−2),(-2−3),(1−(-1))$$
$$\Longrightarrow$$ we get $$(1,-5,2)$$
Step 2:
Its direction cosines are :$$ \dfrac {1}{\sqrt {30}},-\sqrt{\cfrac 56},\sqrt{ \cfrac {2}{15}}$$
Question 10
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Given $$A(1,-1,0)$$; $$B(3,1,2)$$; $$C(2,-2,4)$$ and $$D(-1,1,-1)$$ which of the following points neither lie on $$AB$$ nor on $$CD$$?

A
$$(2,2,4)$$

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

C
$$(2,0,1)$$

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

Solution

$$A(1,-1,0) , B(3,1,2), C(2,-2,4), D(-1,1,-1)\\ \vec { AB } = <3-1, 1-(-1), 2-0>$$
$$= <2,2,2>$$
$$\therefore \quad Equation\quad of\quad line\quad AB:\\ \dfrac { x-1 }{ 2 } =\dfrac { y-(-1) }{ 2 } =\dfrac { z-0 }{ 2 } \quad \Longrightarrow \quad \dfrac { x-1 }{ 2 } =\dfrac { y+1 }{ 2 } =\dfrac { z-0 }{ 2 }$$
$$\vec { CD } = <-1-2, 1-(-2), -1-4>$$
$$= <-3,3,-5>$$
$$\therefore \quad Equation\quad of\quad line\quad CD:\\ \dfrac { x-2 }{ -3 } =\dfrac { y-(-2) }{ 3 } =\dfrac { z-4 }{ -5 } \quad \Longrightarrow \quad \dfrac { x-2 }{ -3 } =\dfrac { y+2 }{ 3 } =\dfrac { z-4 }{ -5 }$$
By putting the values of points given in the choices in the equation of lines $$AB$$ and $$CD$$, $$(2,2,4)$$ is the point which neither lie on $$AB$$ nor on $$CD$$.

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 29

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Three Dimensional Geometry Test - 29

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SHARING IS CARING

Question 1

$$\vec{r}= (2, 1, 0) + t(1, -1, 1)$$

$$\vec{r} = (2, 1, 0) + t(-1, 1, 1)$$

$$\vec{r}= (2, 1, 0) + t(1, 1, -1)$$

$$\vec{r} = (2, 1, 0) + t(1, 1, 1)$$

Solution

Question 2

$$\dfrac{12}{13}, -\dfrac{4}{13}, \dfrac{3}{13}$$

$$-\dfrac{12}{13}, -\dfrac{4}{13}, \dfrac{3}{13}$$

$$\dfrac{12}{13}, \dfrac{4}{13}, \dfrac{3}{13}$$

none of these

Solution

Question 3

$$3x+ 4y+ 5z= 9$$

$$3x+ 4y- 5z+ 9= 0$$

$$3x+ 4y- 5z- 9=0$$

None of these

Solution

Question 4

$$r= k+t(7 i - 10 + 2k)$$

$$r = k + t (- 9 i + 6 j - 2k)$$

$$r = k + t (7i - 10 j - 2k)$$

$$r = k + t (7i + 10 j + 2k)$$

Solution

Question 5

$$(2,2,4)$$

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

$$(2,0,1)$$

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

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 7

$$(5, 6, 3)$$

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

$$(-5, -6, -3)$$

$$(0, 0, 0)$$

Solution

Question 8

$$1$$

$$2$$

$$-1$$

$$0$$

Solution

Question 9

$$-1,5,-2$$

$$\dfrac{1}{\sqrt{30}}$$,$$-\sqrt{\dfrac{5}{6}}$$,$$\sqrt{\dfrac{2}{15}}$$

$$\dfrac{-1}{30 }$$,$$\dfrac{1}{6 }$$,$$-\dfrac{1}{15 }$$

none of these

Solution

Question 10

$$(2,2,4)$$

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

$$(2,0,1)$$

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

Solution

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