Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 33

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Weekly Quiz Competition

Question 1
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The equation of the plane which bisect the join of point $$(7, 2, 3)$$ and $$(-1, -4, 3)$$ perpendicularly is

A
$$x + 2y + 3= 0$$

B
$$2x -y + 3 = 0$$

C
$$x + y +2z + 7= 0$$

D
$$4x + 3y =9 $$

Solution

Lets assume $$P = (7, 2, 3)$$ and $$Q = (-1, -4, 3)$$
So mid point of $$PQ$$ is$$ (3, -1, 3) = M$$ (say), since plane bisect $$PQ$$ so $$M$$ will lie in required plane.
Also direction ratios of $$PQ$$ are $$8, 6, 0$$.
Also given $$PQ$$ is perpendicular to the plane, thus equation of plane is given by,
$$8(x-3)+6(y+1)+0(z-3)=0$$
$$\Rightarrow 4x+3y-9=0$$
Question 2
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Directions For Questions

Let $$A$$ be the given point whose position vectors with reference to origin $$O$$ be $$ \overrightarrow{a}$$ and $$ \overrightarrow{ON}= \overrightarrow{n}$$ Let $$P$$ be any point such that $$\overline{OP}= \overrightarrow{r}$$ lies on the plane & passes through $$A$$ and orthogonal to $$ON$$. Then for any point $$P$$ lies on the plane then.$$\overrightarrow{AP} \perp \overrightarrow{n}$$
$$\displaystyle \therefore \overrightarrow{AP}.\ \overrightarrow{n}=0$$
$$\displaystyle \Rightarrow \left ( \overrightarrow{OP}-\overrightarrow{OA} \right ).\overrightarrow{n}=0$$
$$\displaystyle \Rightarrow \overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$$ $$($$Knows as scalar product form$$)$$
$$\displaystyle \Rightarrow \overrightarrow{r}\cdot \overrightarrow{n}=p,$$ where $$p$$ is the $$\perp$$er distance from origin to the plane.
On the bases of above information answer the following questions

...view full instructions

Vector equation of the plane passing through a point having position vector $$\displaystyle 2\hat{i}+3\hat{j}-4\hat{k}$$ and perpendicular to the vector $$\displaystyle 2\hat{i}-\hat{j}+2\hat{k}$$ is

A
$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )+7=0$$

B
$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=7$$

C
$$\displaystyle \overrightarrow{r}\cdot \left ( -3\hat{i}-2\hat{j}-3\hat{k} \right )=0$$

D
$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=9$$

Solution

Using $$\displaystyle \left ( \overrightarrow{r} - \overrightarrow{a} \right )\cdot \overrightarrow{n}=0$$
i.e.$$\displaystyle \overrightarrow{r} \cdot \overrightarrow{n}= \overrightarrow{a} \cdot \overrightarrow{n}$$
$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )= \left ( 2\hat{i}+3\hat{j}-4\hat{k} \right ) \cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )$$
$$\displaystyle \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=4-3-8 $$
$$\displaystyle \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )= -7 $$
Hence option (a) is correct.
Question 3
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A line makes the same angle $$\theta $$ with each of the $$x$$ and $$z$$ axis. 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
$$\displaystyle \dfrac{3}{5}$$

B
$$\displaystyle \frac{1}{5}$$

C
$$\displaystyle \frac{2}{3}$$

D
$$\displaystyle \frac{2}{5}$$

Solution

If a line makes the angle $$\alpha,\beta,\gamma$$ with x,y,z axix respectively then
$$l^{2}+m^{2}+n^{2}=1$$
$$\Rightarrow 2l^{2}+m^{2}=1 \: or \: 2n^{2}+m^{2}=1$$
$$\Rightarrow 2\cos^{2}\theta=1- \cos ^{2} \beta\left ( \alpha=\gamma=\theta \right )$$
$$2 \cos ^{2}\theta = \sin ^{2} \beta$$
$$\Rightarrow 2\cos ^{2} \theta = 3\sin ^{2} \theta$$
$$\Rightarrow 5\cos ^{2} \theta=3$$
Question 4
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The line passes through the points $$\left ( 5,1,a \right )$$ & $$\left ( 3,b,1 \right )$$ crosses the $$yz$$ plane at the point $$\displaystyle \left ( 0,\frac{17}{2},-\frac{13}{2} \right )$$ ,then

A
$$a= 4, b= 6$$

B
$$a= 6, b= 4$$

C
$$a= 8, b= 2$$

D
$$a= 2, b= 8$$

Solution

Equation of the line through the points $$( 5,1,a ) $$ & $$( 3,b,1)$$ is
$$\displaystyle \frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=\lambda $$
Now it passes through $$\displaystyle \left ( 0,\frac{17}{2},\frac{-13}2{} \right )$$
$$\displaystyle \therefore \frac{0-5}{-2}=\frac{17/2-1}{b-1}=\frac{-13/2-a}{1-a}=\lambda \:$$
$$ \Rightarrow \lambda =\dfrac{5}{2}$$
$$\displaystyle \therefore \frac{17/2-1}{b-1}=\frac{5}{2} $$
$$\Rightarrow b=4$$
and $$\displaystyle \frac{-13/2-a}{1-a}=\frac{5}{2} $$
$$\Rightarrow a=6$$
Question 5
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If the three points with position vectors $$\displaystyle \bar{a}-2\bar{b}+3\bar{c}, \ 2\bar{a}+\lambda \bar{b}-4\bar{c}, \ -7\bar{b}+10\bar{c} $$ are collinear, then $$\displaystyle \lambda= $$

A
$$1$$

B

C
$$3$$

D
none of these

Solution

The given vectors are collinear, so $$l(\bar{a} - 2\bar{b} + 3\bar{c}) + k(2\bar{a} + \lambda\bar{b} - 4\bar{c}) = (l + k)(-7\bar{b} + 10\bar{c})$$
Comparing the coefficients of $$\bar{a} \rightarrow l + 2k = 0 $$
$$\bar{b} \rightarrow -2l + \lambda k = -7l -7k$$
$$\bar{c} \rightarrow 3l - 4k = 10l + 10k$$
$$\Rightarrow l = -2k$$ and so $$\lambda = 3$$
Question 6
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Find the equation of the plane containing the vectors $$\displaystyle \bar{\alpha} $$ and $$\displaystyle\bar{ \beta} $$ and passing through the point $$\displaystyle \bar{a} $$

A
$$\displaystyle (\bar{r}-\bar{a})\cdot (\bar{\alpha} \times \bar{\beta}) =0$$

B
$$\displaystyle (\bar{r}+\bar{a})(\bar{\alpha }\times \bar{\beta}) =0$$

C
$$\displaystyle (\bar{r}-\bar{a}) (\bar {a}\cdot \bar{b}) =0$$

D
none of these

Solution

Normal vector to the plane is, $$\bar{\alpha}\times \bar{\beta}$$
Thus equation of plane passing through $$\vec{a}$$ is given by,
$$(\vec{r}-\vec{a})\cdot (\bar{\alpha}\times \bar{\beta})=0$$
Question 7
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The number of lines which are equally inclined to the axes is

A
2

B
4

C
6

D
8

Solution

If a line makes angles $$\alpha,\beta,\gamma$$ with the axes we have $$\alpha = \beta = \gamma $$
$$cos \alpha = cos \beta = cos \gamma$$, or $$l = m = n .$$
$$ l^2 + m^2 + n^2 = 1$$
$$l^2 + l^2 + l^2 = 1$$
$$3l^2=1$$
$$l^2=\dfrac{1}{3}$$
$$\therefore l=\pm\dfrac {1}{\sqrt3}$$
$$\therefore ( l, m, n)(-l , -m , -n )$$ represent the d.c's of the same line.
Since there are four different groups of signs,
so there can be four different lines which makes equal angles with axes.
Question 8
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Direction cosines of the vector $$\displaystyle \bar{v}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$$ are

A
$$\displaystyle < a_{1},a_{2},a_{3} > $$

B
$$\displaystyle <-a_{1},-a_{2},-a_{3}>$$

C
$$\displaystyle <\frac {a_{1}}{\left|\bar{v}\right|},\frac {a_{2}}{\left|\bar{v}\right|,}\frac {a_{3}}{\left|\bar{v}\right|},>$$

D
none of these

Solution

The directions cosines will be
$$\dfrac{a_{1}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}},\dfrac{a_{2}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}},\dfrac{a_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}}$$

$$\Rightarrow\dfrac{a_{1}}{|\overline{v}|},\dfrac{a_{2}}{|\overline{v}|},\dfrac{a_{3}}{|\overline{v}|}$$
Question 9
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The acute angle between the lines $$ x =-2 + 2t, y = 3 -4t, z = -4 + t$$ and $$ x= 2 -t, y= 3 + 2t, z= -4 + 3t $$ is

A
$$\displaystyle \sin^{-1}\frac{1}{\sqrt{3}}$$

B
$$\displaystyle \cos^{-1}\frac{1}{\sqrt{6}}$$

C
$$\displaystyle \cos^{-1}\frac{1}{\sqrt{5}}$$

D
$$\displaystyle \cos^{-1}2/3$$

Solution

Given lines are $$ x =-2 + 2t, y = 3 -4t, z = -4 + t$$ and $$ x= 2 -t, y= 3 + 2t, z= -4 + 3t $$
$$\Rightarrow \displaystyle\frac { x+2 }{ 2 } =\displaystyle\frac { y-3 }{ -4 } =\displaystyle\frac { z+4 }{ 1 } $$ and $$\displaystyle\frac { x-2 }{ -1 } =\displaystyle\frac { y-3 }{ 2 } =\displaystyle\frac { z+4 }{ 3 } $$
if $$a_{1},b_{1},c_{1}$$ and $$a_{2},b_{2},c_{2}$$ are direction ratios of two lines, then angle between them is
$$\cos\theta = \displaystyle\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$$
$$\Rightarrow \cos\theta = \displaystyle\frac{-1}{\sqrt{6}}$$
$$\Rightarrow \theta = \pi-\cos ^{ -1 }{ \left(\displaystyle \frac { 1 }{ \sqrt { 6 } } \right) } $$
Hence, acute angle between the lines is $$\cos ^{ -1 }{ \left(\displaystyle \frac { 1 }{ \sqrt { 6 } } \right) }$$
Question 10
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Directions For Questions

$$\displaystyle L_{1}: \frac{x\, +\, 1}{-3}= \frac{y\, -\, 3}{2}= \frac{y\, +\, 2}{1}$$

$$\displaystyle L_{2}, : \, \frac{x}{1}= \frac{y\, -\, 7}{-3}= \frac{z\, +\, 7}{2}$$

...view full instructions

Equation of a plane containing $$L_{1}$$ and $$L_{2}$$ is

A
$$x + y + z = 0$$

B
$$3x -2y -z =0$$

C
$$x -3y + 2z = 0$$

D
$$x + y + z = 42$$

Solution

Normal vector of the plane is given by,
$$\vec{n} = \begin{vmatrix}

\hat{i}&\hat{j}&\hat{k}\\-3&2&1\\1&-3&2

\end{vmatrix}=7\hat{i}+7\hat{j}+7\hat{k}=7(\hat{i}+\hat{j}+\hat{k})$$
Hence equation of plane is,
$$ (\vec{r} - (7\hat{j}-7\hat{k}))\cdot \vec{n}=0$$
$$\Rightarrow 7(x-0)+7(y-7)+7(z+7)=0$$
$$\Rightarrow x+y+z=0$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 33

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Three Dimensional Geometry Test - 33

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SHARING IS CARING

Question 1

$$x + 2y + 3= 0$$

$$2x -y + 3 = 0$$

$$x + y +2z + 7= 0$$

$$4x + 3y =9 $$

Solution

Question 2

$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )+7=0$$

$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=7$$

$$\displaystyle \overrightarrow{r}\cdot \left ( -3\hat{i}-2\hat{j}-3\hat{k} \right )=0$$

$$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=9$$

Solution

Question 3

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

$$\displaystyle \frac{1}{5}$$

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{2}{5}$$

Solution

Question 4

$$a= 4, b= 6$$

$$a= 6, b= 4$$

$$a= 8, b= 2$$

$$a= 2, b= 8$$

Solution

Question 5

$$1$$

2

$$3$$

none of these

Solution

Question 6

$$\displaystyle (\bar{r}-\bar{a})\cdot (\bar{\alpha} \times \bar{\beta}) =0$$

$$\displaystyle (\bar{r}+\bar{a})(\bar{\alpha }\times \bar{\beta}) =0$$

$$\displaystyle (\bar{r}-\bar{a}) (\bar {a}\cdot \bar{b}) =0$$

none of these

Solution

Question 7

2

4

6

8

Solution

Question 8

$$\displaystyle < a_{1},a_{2},a_{3} > $$

$$\displaystyle <-a_{1},-a_{2},-a_{3}>$$

$$\displaystyle <\frac {a_{1}}{\left|\bar{v}\right|},\frac {a_{2}}{\left|\bar{v}\right|,}\frac {a_{3}}{\left|\bar{v}\right|},>$$

none of these

Solution

Question 9

$$\displaystyle \sin^{-1}\frac{1}{\sqrt{3}}$$

$$\displaystyle \cos^{-1}\frac{1}{\sqrt{6}}$$

$$\displaystyle \cos^{-1}\frac{1}{\sqrt{5}}$$

$$\displaystyle \cos^{-1}2/3$$

Solution

Question 10

$$x + y + z = 0$$

$$3x -2y -z =0$$

$$x -3y + 2z = 0$$

$$x + y + z = 42$$

Solution

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