Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the plane passing through $$(a,b,c)$$ and parallel to the plane $$r.(\hat{i}+\hat{j}+\hat{k})=2$$ is,

A
$$x+y+z=1$$

B
$$ax+by+cz=1$$

C
$$x+y+z=a+b+c$$

D
None of these

Solution

Let any point on the plane be $$(x,y,z)$$ then,
$$1(x-a)+1(y-b)+1(z-c)=0\implies x+y+z=a+b+c$$
Question 2
1 / -0

Angle between the lines $$3x=6y=2z$$ and $$3x+2y+z-5=0=x+y-2z-3$$ is?

A
$$\dfrac{\pi}{6}$$

B
$$\dfrac{\pi}{3}$$

C
$$\dfrac{\pi}{4}$$

D
$$\dfrac{\pi}{2}$$

Solution

The given two planes have a line of intersection,
whose equation can be found by,
$$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i } } & { \widehat { j } } & { \widehat { k } } \\ 3 & 2 & 1 \\ 1 & 1 & { -2 } \end{array} } \right| \\ =\widehat { i\, } (-4-1)-\widehat { j } (-6-1)+\widehat { k } (3-2) \\ =-5\widehat { i } +7\widehat { j } +\widehat { k } \end{array}$$

Given line is,
$$\begin{array}{l} 3x=6y=2z \\ \frac { x }{ 2 } =y=\frac { z }{ 3 } \end{array}$$
so, its vector equation is $$2\widehat i + \widehat j + 3\widehat k$$.

Angle between line is cos$$\theta $$=$$\frac{{( - 5)(2) + 7(1) + 1(3)}}{{\sqrt {25 + 49 + 1} \,\,\,\sqrt {4 + 1 + 9} }}$$

= $$\frac{0}{{\sqrt {25 + 49 + 1} \,\,\,\sqrt {4 + 1 + 9} }}$$
cos$$\theta $$ = 0
Therefor $$\theta $$ = $$\frac{\pi }{2}$$
So that the correct Option is D
Question 3
1 / -0

The points $$i + j + k, \, i + 2j, \, 2i+2j+k,\, 2i+3j+2k$$ are

A
collinear

B
coplanar but not collinear

C
non-coplanar

D
none

Solution

$$\begin{matrix} A& B& C& D\\i+j+k, &i+2j, &2i+2j+k,&2i+3j+2k \end{matrix}$$
$$\overline{AC} = (2-1)i + (2-1)j + k-k$$
$$=i+j$$
$$\overline{AB} = o + j - \overline{k} = j - \overline{k}$$
$$\overline{AD} = i + 2j + k$$
$$\begin{vmatrix} 1&1&0 \\0&2 &1\end{vmatrix} = 1(1+2)-1(0+1)$$
$$=3-1 = 2 \neq 0$$
Non coplanar.
Question 4
1 / -0

If the dr's the line are $$(1+\lambda, 1-\lambda, 2)$$ and it makes an angle $${60}^{o}$$ with the Y-axis then $$\lambda$$ is

A
$$1\pm \sqrt{3}$$

B
$$4\pm \sqrt{5}$$

C
$$2\pm 2\sqrt{3}$$

D
$$2\pm \sqrt{5}$$

Solution

Given that,
$$\begin{array}{l} \frac { { \left| { 1-\lambda } \right| } }{ { \sqrt { { { \left( { 1+\lambda } \right) }^{ 2 } }+{ { \left( { 1-\lambda } \right) }^{ 2 } }+4 } } } =\frac { 1 }{ 2 } \\ \frac { { \left| { 1-\lambda } \right| } }{ { \sqrt { 6+2{ \lambda ^{ 2 } } } } } =\frac { 1 }{ 2 } \\ 4{ \left( { \left| { 1-\lambda } \right| } \right) ^{ 2 } }=6+2{ \lambda ^{ 2 } } \\ 2{ \lambda ^{ 2 } }-2-8\lambda =0 \\ { \lambda ^{ 2 } }-4\lambda -1=0 \\ \lambda =\frac { { 4\pm \sqrt { 16+4 } } }{ 2 } =\frac { { 4\pm 2\sqrt { 5 } } }{ 2 } \\ =2\pm \sqrt { 5 } \end{array}$$
Then,
We get $$2 \pm \sqrt 5 $$
Option $$D$$ is correct answer.
Question 5
1 / -0

The line joining the points $$(-2, 1, -8)$$ and $$(a, b, c)$$ is parallel to the line whose direction ratios are $$6, 2, 3$$. The value of $$a, b, c$$ are

A
$$(4, 3, -5)$$

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

C
$$(10, 5, -2)$$

D
$$(-5, 3, 4)$$

Solution

If lines are parallel
then their direction
ratios must be equal
$$\Rightarrow a+2=6$$
$$a=4$$
$$b-1=2$$
$$b=3$$
$$c+8=3$$
$$c=-5$$
$$\therefore (a,b,c)$$ is $$(4,3,-5)$$
Question 6
1 / -0

If a line has the direction ratio $$18, 12, 4 $$, then its direction cosines are:

A
$$\dfrac{9}{11}$$, $$\dfrac{6}{11}$$, $$\dfrac{2}{11}$$

B
$$\dfrac{9}{13}$$, $$\dfrac{6}{13}$$, $$\dfrac{2}{13}$$

C
$$\dfrac{9}{7}$$, $$\dfrac{6}{7}$$, $$\dfrac{2}{7}$$

D
None of these

Solution

Dr's of the line are : 18, 12, 4
Dc's = $$\dfrac{18}{\sqrt{18^2+12^2+4^2}}, \dfrac{12}{\sqrt{18^2+12^2+4^2}}, \dfrac{4}{\sqrt{18^2+12^2+4^2}}$$

$$=\dfrac{18}{22}, \dfrac{12}{22}, \dfrac{4}{22}$$

$$=\dfrac{9}{11}, \dfrac{6}{11}, \dfrac{2}{11}$$
Question 7
1 / -0

The angle between the lines $$\dfrac {x - 1}{1} = \dfrac {y - 1}{1} = \dfrac {z - 1}{2}$$ and, $$\dfrac {x - 1}{-\sqrt {3} - 1} = \dfrac {y - 1}{\sqrt {3} - 1} = \dfrac {z - 1}{4}$$ is

A
$$\cos^{-1}\left (\dfrac {1}{65}\right )$$

B
$$\dfrac {\pi}{6}$$

C
$$\dfrac {\pi}{3}$$

D
$$\dfrac {\pi}{4}$$

Solution

Given the lines are $$\dfrac {x - 1}{1} = \dfrac {y - 1}{1} = \dfrac {z - 1}{2}$$ and,

$$\dfrac {x - 1}{-\sqrt {3} - 1} = \dfrac {y - 1}{\sqrt {3} - 1} = \dfrac {z - 1}{4}$$.

The direction ratios of the lines are $$1,1,2$$ and $$-\sqrt{3}-1,\sqrt{3}-1,4$$.
Then the angle between the given two lines is

$$\cos^{-1}\left(\dfrac{1.(-\sqrt{3}-1)+1.(\sqrt{3}-1)+2.4}{\sqrt{1^2+1^2+2^2}.\sqrt{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2+4^2}}\right)$$

$$=\cos^{-1}\left(\dfrac{6}{\sqrt{6}.\sqrt{24}}\right)$$
$$=\cos^{-1}\left(\dfrac{6}{12}\right)$$

$$=\dfrac{\pi}{3}$$.
Question 8
1 / -0

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

A
$${\vec{r}}=(1+2{t})\hat{i}+(1+{t})\hat{j}{+}(1+2t)\hat{k}$$

B
$${\vec{r}}=(1+2{t})\hat{i}+(1+{t})\hat{j}{+}(2{t}-1)\hat{k}$$

C
$${\vec{r}}=(2{t}-1)\hat{i}+(1+{t})\hat{j}{+}(2{t}+1)\hat{k}$$

D
$${\vec{r}}=(1-2{t})\hat{i}+(1+{t})\hat{i}{+}(2{t}+1)\hat{k}$$

Solution

The required equation of the line in Cartesian format will be
$$\dfrac{x-1}{2}=\dfrac{y-1}{1}=\dfrac{z+1}{2}=t$$
Hence, any point on the line can be written as
$$x=2t+1$$, $$y=t+1$$ and $$z=2t-1$$
Hence in vector format,
$$\vec{r}=xi+yj+zk$$
$$=(2t+1)i+(t+1)j+(2t-1)k$$
Question 9
1 / -0

If the points with position vectors $$60\hat{i}+3\hat{j},40\hat{i}-8\hat{j}$$ and $$a\hat{i}-52\hat{j}$$ are collinear then $$a$$ is equal to

A
$$-40$$

B
$$-20$$

C
$$20$$

D
$$40$$

Solution
Question 10
1 / -0

Equation of a plane passing through the points $$A=\hat{i}+\hat{j}{+}\hat{k},\ B=\hat{j}{+\hat{k}}, C=\hat{k}$$ is

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

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

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

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

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Three Dimensional Geometry Test - 19

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Three Dimensional Geometry Test - 19

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Question 1

$$x+y+z=1$$

$$ax+by+cz=1$$

$$x+y+z=a+b+c$$

None of these

Solution

Question 2

$$\dfrac{\pi}{6}$$

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{4}$$

$$\dfrac{\pi}{2}$$

Solution

Question 3

collinear

coplanar but not collinear

non-coplanar

none

Solution

Question 4

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

$$4\pm \sqrt{5}$$

$$2\pm 2\sqrt{3}$$

$$2\pm \sqrt{5}$$

Solution

Question 5

$$(4, 3, -5)$$

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

$$(10, 5, -2)$$

$$(-5, 3, 4)$$

Solution

Question 6

$$\dfrac{9}{11}$$, $$\dfrac{6}{11}$$, $$\dfrac{2}{11}$$

$$\dfrac{9}{13}$$, $$\dfrac{6}{13}$$, $$\dfrac{2}{13}$$

$$\dfrac{9}{7}$$, $$\dfrac{6}{7}$$, $$\dfrac{2}{7}$$

None of these

Solution

Question 7

$$\cos^{-1}\left (\dfrac {1}{65}\right )$$

$$\dfrac {\pi}{6}$$

$$\dfrac {\pi}{3}$$

$$\dfrac {\pi}{4}$$

Solution

Question 8

$${\vec{r}}=(1+2{t})\hat{i}+(1+{t})\hat{j}{+}(1+2t)\hat{k}$$

$${\vec{r}}=(1+2{t})\hat{i}+(1+{t})\hat{j}{+}(2{t}-1)\hat{k}$$

$${\vec{r}}=(2{t}-1)\hat{i}+(1+{t})\hat{j}{+}(2{t}+1)\hat{k}$$

$${\vec{r}}=(1-2{t})\hat{i}+(1+{t})\hat{i}{+}(2{t}+1)\hat{k}$$

Solution

Question 9

$$-40$$

$$-20$$

$$20$$

$$40$$

Solution

Question 10

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

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

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

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

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