Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 52

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Question 1
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A plane passes through the point $$(0, -1, 0)$$ and $$(0, 0, 1)$$ and makes an angle of $$\dfrac{\pi}{4}$$ with the plane $$y-z=0$$ then the point which satisfies the desired plane is?

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

B
$$(\sqrt{2}, 1, 2)$$

C
$$(\sqrt{2}, 1, 4)$$

D
$$(\sqrt{2}, 2, 4)$$

Solution

$$ax+by+cz=1$$, $$-b=1$$, $$c=1$$
$$ax-y+z=1$$
$$\dfrac{1}{\sqrt{2}}=\cos\dfrac{\pi}{4}=\dfrac{-1-1}{\sqrt{2}\sqrt{a^2+2}}$$
$$\Rightarrow a^2+2=4\Rightarrow a=-\sqrt{2}$$
$$\Rightarrow$$ Plane is $$-\sqrt{2}x-y+z=1$$
Clearly $$(\sqrt{2}, 1, 4)$$ satisfy the plane.
Question 2
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The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is?

A
$$0^o$$

B
$$45^o$$

C
$$90^o$$

D
$$30^o$$

Solution

$$\dfrac{x}{1/2}=\dfrac{y}{1/3}=\dfrac{x}{-1}$$; $$\dfrac{x}{1/6}=\dfrac{y}{-1}=\dfrac{x}{-1/4}$$

Here: $$\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{1-4+3}{12}=0$$

Lines are perpendicular.
Question 3
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The angle between two adjacent sides $$\vec { a } $$ and $$\vec { b } $$ of parallelogram is $$\cfrac{\pi}{6}$$. If $$\vec { a } =\left( 2,-2,1 \right) $$ and $$\left| \vec { b } \right| =2\left| \vec { a } \right| $$, then area of this parallelogram is ______

A
$$9$$

B
$$18$$

C
$$\cfrac{9}{2}$$

D
$$\cfrac{3}{4}$$

Solution

$$\vec { a } =2\hat { i } -2\hat { j } +\hat { k } ,\left| \vec { a } \right| =3$$
$$\sin { { 30 }^{ o } } =\cfrac { x }{ \left| \vec { b } \right| } \Rightarrow x=3\quad \quad $$
area $$=3\times 3=9$$
Question 4
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The equation of the plane passing through the point $$(-1, 2, 1)$$ and perpendicular to the line joining the points $$(-3, 1, 2)$$ and $$(2, 3, 4)$$ is ________.

A
$$\bar{r}\cdot (5\hat{i}+2\hat{j}+2\hat{k})=1$$

B
$$\bar{r}\cdot (5\hat{i}+2\hat{j}+2\hat{k})=-1$$

C
$$\bar{r}\cdot (5\hat{i}-2\hat{j}+2\hat{k})=-5$$

D
$$\bar{r}\cdot (5\hat{i}-2\hat{j}-2\hat{k})=1$$

Solution

The plane is perpendicular to the line joining the points $$(-3, 1, 2)$$ and $$(2, 3, 4).$$ So, its normal vector is $$(5, 2, 2).$$
Therefore, the equation of the plane is $$\bar{r} . (5\hat{i} + 2\hat{j} + 2\hat{k}) = a$$ for some constant $$a.$$
It is given that the plane passes through the point $$(-1, 2, 1).$$ So, we substitute $$\bar{r} = (-1, 2, 1)$$ in the above equation and get $$a = -5 + 4 + 2 = 1.$$
So the equation of the plane is $$\bar{r}. (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1.$$
Question 5
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The direction ratios of the line perpendicular to the lines

$$\dfrac {x - 7}{2} = \dfrac {y + 17}{-3} = \dfrac {z - 6}{1}$$ and, $$\dfrac {x + 5}{1} = \dfrac {y + 3}{2} = \dfrac {z - 4}{-2}$$ are proportional to

A
$$4, 5, 7$$

B
$$4, -5, 7$$

C
$$4, -5, -7$$

D
$$-4, 5, 7$$

Solution

We have,$$\dfrac{x-7}{2}=\dfrac{y+17}{-3}=\dfrac{z-6}{1}$$
and
$$\dfrac{x+5}{1}=\dfrac{y+3}{2}=\dfrac{z-4}{-2}$$

The direction ratios of the given lines are proportional to $$2,-3,1$$ and $$1,2,-2$$

The vectors parallel to the given vectors are

$$\vec{{b}_{1}}=2\hat{i}-3\hat{j}+\hat{k}$$ and $$\vec{{b}_{2}}=\hat{i}+2\hat{j}-2\hat{k}$$

Vector perpendicular to the given two lines is

$$\vec{b}=\vec{{b}_{1}}\times\vec{{b}_{2}}$$
$$=\left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2\end{matrix} \right|$$

$$=\left(6-2\right)\hat{i}-\left(-4-1\right)\hat{j}+\left(4+3\right)\hat{k}$$
$$=4\hat{i}+5\hat{j}+7\hat{k}$$

Hence, the direction ratios of the line perpendicular to the given two lines are proportional to $$4,5,7$$
Question 6
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The projections of a line segment on $$X, Y$$ and $$Z$$ axes are $$12, 4$$ and $$3$$ respectively. The length and direction cosines of the line segment are

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

B
$$19; \dfrac {12}{19}, \dfrac {4}{19}, \dfrac {3}{19}$$

C
$$11; \dfrac {12}{11}, \dfrac {4}{11}, \dfrac {3}{11}$$

D
None of these

Solution

If a line makes angles $$\alpha,\,\beta$$ and $$\gamma$$ with the axes, then

$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$$ ---- ( 1 )

Let $$r$$ be the length of the line segment, then,
$$r\cos\alpha=12,\,r\cos\beta=4,\,r\cos\gamma=3$$ ---- ( 2 )
$$\Rightarrow$$ $$(r\cos\alpha)^2+(r\cos\beta)^2+(r\cos\gamma)^2=(12)^2+(4)^2+(3)^2$$
$$\Rightarrow$$ $$r^2\cos^2\alpha+r^2\cos^2\beta+r^2\cos^2\gamma=144+16+9$$
$$\Rightarrow$$ $$r^2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)=169$$
$$\Rightarrow$$ $$r^2(1)=169$$ [ From ( 1 ) ]
$$\Rightarrow$$ $$r=\sqrt{169}$$
$$\Rightarrow$$ $$r=\pm13$$

Since, the length cannot be negative
$$\Rightarrow$$ $$r=13$$

Substituting $$r=13$$ in ( 2 )
we get,

$$\cos\alpha=\dfrac{12}{13},$$ $$\cos\beta=\dfrac{4}{13},$$ $$\cos\gamma=\dfrac{3}{13}$$
$$\therefore$$ The direction cosines of the line are $$\dfrac{12}{13},\,\dfrac{4}{13},\,\dfrac{3}{13}$$
Question 7
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A line passes through the point $$A(-2,4,-5)$$ and is parallel to the line $$\cfrac{x+3}{3}=\cfrac{y-4}{5}=\cfrac{z+8}{6}$$. The vector equation of the line is

A
$$\vec { r } =\left( -3\hat { i } +4\hat { j } -8\hat { k } \right) +\lambda \left( -2\hat { i } +4\hat { j } -5\hat { k } \right) $$

B
$$\vec { r } =\left( -2\hat { i } +4\hat { j } -5\hat { k } \right) +\lambda \left( -3\hat { i } +5\hat { j } +6\hat { k } \right) $$

C
$$\vec { r } =\left( 3\hat { i } +5\hat { j } +6\hat { k } \right) +\lambda \left( -2\hat { i } +4\hat { j } -5\hat { k } \right) $$

D
$$\vec { r } =\left( -2\hat { i } +4\hat { j } -5\hat { k } \right) +\lambda \left( 3\hat { i } +5\hat { j } +6\hat { k } \right) $$

Solution
Question 8
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A line passes through the point $$A(5,-2,4)$$ and it is parallel to the vector $$\left(2 \hat { i } -\hat { j } +3\hat { k } \right) $$. The vector equation of the line is

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

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

C
$$\vec { r } .\left( 5\hat { i } -2\hat { j } +4\hat { k } \right) =\sqrt{14}$$

D
none of these

Solution
Question 9
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The Cartesian equations of a line are $$\cfrac{x-1}{2}=\cfrac{y+2}{3}=\cfrac{z-5}{-1}$$. Its vector equation is

A
$$\vec { r } =\left( -\hat { i } +2\hat { j } -5\hat { k } \right) +\lambda \left( 2\hat { i } +3\hat { j } -\hat { k } \right) $$

B
$$\vec { r } =\left(2 \hat { i } +3\hat { j } -\hat { k } \right) +\lambda \left( \hat { i } -2\hat { j } +5\hat { k } \right) $$

C
$$\vec { r } =\left( \hat { i } -2\hat { j } +5\hat { k } \right) +\lambda \left( 2\hat { i } +3\hat { j } -\hat { k } \right) $$

D
none of these

Solution
Question 10
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The angle between the lines $$\cfrac{x}{2}=\cfrac{y}{2}=\cfrac{z}{1}$$ and $$\cfrac{x-5}{4}=\cfrac{y-2}{1}=\cfrac{z-3}{8}$$ is

A
$$\cos ^{ -1 }{ \left( \cfrac { 3 }{ 4 } \right) } $$

B
$$\cos ^{ -1 }{ \left( \cfrac { 5 }{ 6 } \right) } $$

C
$$\cos ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } $$

D
$$\cfrac { \pi }{ 3 } $$

Solution