Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

Find the angle between the lines $$\overrightarrow { r } =3i+2j-4k+\lambda \left( i+2j+2k \right) $$ and $$\overrightarrow { r } =\left( 5j-2k \right) +\mu \left( 3i+2j+6k \right) $$

A
$$\displaystyle \theta =\cos ^{ -1 }{ \left( \frac { 19 }{ 21 } \right) } $$

B
$$\displaystyle \theta =\sin ^{ -1 }{ \left( \frac { 19 }{ 21 } \right) } $$

C
$$\displaystyle \theta =\cos ^{ -1 }{ \left( \frac { 20 }{ 21 } \right) } $$

D
None of these

Solution

Let $$\theta$$ be the angle between the given lines.

The given lines are parallel to the vector.

$$\overrightarrow { { b }_{ 1 } } =1+2j+2k$$ and $$\overrightarrow { { b }_{ 2 } } =3i+2j+6k$$ respectively.

So, the angle $$\theta$$ between them is given by

$$\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } } }{ \left| \overrightarrow { { b }_{ 1 } } \right| \left| \overrightarrow { { b }_{ 1 } } \right| } =\dfrac { \left( i+2j+2k \right) \left( 3i+2j+6k \right) }{ \left| i+2j+2k \right| \left| 3i+2j+6k \right| } $$

$$\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 1+4+4 } \sqrt { 9+4+36 } } =\dfrac { 19 }{ 21 } $$

$$\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) } $$
Question 2
1 / -0

The d.c's of the normal to the plane $$2x-y+2z+5=0$$ are

A
$$(3, -2, 6)$$

B
$$\left (\displaystyle \dfrac{2}{7},\dfrac{3}{7},\dfrac{-6}{7}\right)$$

C
$$\left (\displaystyle \dfrac{3}{7}\dfrac{-2}{7},\dfrac{6}{7}\right)$$

D
$$\left (\displaystyle \dfrac{2}{3}\dfrac{-1}{3},\dfrac{2}{3}\right)$$

Solution

Given equation of plane is $$2x-y+2z+5=0$$
The dr's of the normal to the plane are $$(2,-1,2)$$.
The corrresponding dc's will be given by $$\left (\dfrac{2}{3} , \dfrac{-1}{3}, \dfrac{2}{3} \right )$$.
Question 3
1 / -0

Find the angle between the two lines having direction ratio $$(1,1,2)$$ and $$\left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right) $$.

A
$$\displaystyle \dfrac { \pi }{ 3 } $$

B
$$\displaystyle \dfrac { \pi }{ 2 } $$

C
$$\displaystyle \dfrac { \pi }{ 6 } $$

D
None of these

Solution

Let $$\overrightarrow { { m }_{ 1 } } $$ and $$\overrightarrow { { m }_{ 2 } }$$ be vectors parallel to the two given lines.

Then, angle between the two given lines is same as the angle between $$\overrightarrow { { m }_{ 1 } } $$ and $$\overrightarrow { { m }_{ 2 } }$$.

$$\overrightarrow { { m }_{ 1 } } =$$ Vector parallel to the line with direction ratios $$(1,1,2)=i+j+2k$$ and

$$\overrightarrow { { m }_{ 2 } } =$$ Vector parallel to the line with direction ratio $$\left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right) =\left( \sqrt { 3 } -1 \right) i+\left( -\sqrt { 3 } -1 \right) j+4k$$

Let $$\theta$$ be the angle between the given lines.

Then $$\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { m }_{ 1 } } .\overrightarrow { { m }_{ 2 } } }{ \left| \overrightarrow { { m }_{ 1 } } \right| \left| \overrightarrow { { m }_{ 2 } } \right| } =\dfrac { \left( \sqrt { 3 } -1 \right) -\left( \sqrt { 3 } +1 \right) +8 }{ \sqrt { 1+1+4 } \sqrt { { \left( \sqrt { 3 } -1 \right) }^{ 2 }+{ \left( \sqrt { 3 } +1 \right) }^{ 2 }+16 } } $$

$$\Rightarrow \displaystyle \cos { \theta } =\dfrac { 6 }{ \sqrt { 6 } \sqrt { 24 } } =\dfrac { 1 }{ 2 } $$

$$\Rightarrow \theta =\dfrac { \pi }{ 3 } $$
Question 4
1 / -0

The plane which passes through the point $$(-1, 0, -6)$$ and perpendicular to the line whose direction ratios is $$(6, 20, -1)$$ also passes through the point:

A
$$(1, 1, -26)$$

B
$$(0, 0, 0)$$

C
$$(2, 1, -32)$$

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

Solution

The d.r's of the perpendicular line give the normal.
Hence, the plane will be of the form $$6x + 20y -z = d$$
Since, it passes through $$(-1,0,-6)$$, $$d = 0$$
Hence, the plane passes through the origin.
Question 5
1 / -0

Equation of the plane through the mid-point of the join of $$A(4,5,-10)$$ and $$B(-1,2,1)$$ and perpendicular to $$AB$$ is

A
$$\vec{r}.(5i+3j-11k)+\dfrac{135}{2}=0$$

B
$$\vec{r}.(5i+3j-11k)=\dfrac{135}{2}$$

C
$$\vec{r}.\left (\dfrac{3}{2}\hat{i}+\dfrac{7}{2}\hat{j}-\dfrac{9}{2}\hat{k}\right)=5\hat{i}+3\hat{j}-11\hat{k}$$

D
$$\vec{r}.(5i+3j-11k)+\dfrac{185}{2}=0$$

Solution

Since, the line $$AB$$ is perpendicular to the plane, it has to be the normal to the plane.
DR's of normal $$=(4+1,5-2,-10-1)=(5,3,-11)$$
Hence, the equation of the plane is of the form,
$$5x + 3y - 11z = d$$
The midpoint of $$AB$$ is $$\left ( \dfrac{3}{2} , \dfrac{7}{2} , \dfrac{-9}{2} \right)$$
The plane passes through this point, so the value of d is $$ \dfrac{135}{2} $$.
Hence, option B is correct.
Question 6
1 / -0

If $$(2, 3, -1)$$ is the foot of the perpendicular from $$(4, 2, 1)$$ to a plane, then the equation of that plane is $$ax+by+cz=d$$. Then $$a+d$$ is

A
$$3$$

B
$$1$$

C
$$-2$$

D
$$2$$

Solution

Since, the line joining the two points is perpendicular to the plane, it's d.r's will give the normal to the plane.
Direction ratio of the normal $$= (4-2,2-3,1+1)=(2,-1,2)$$
Hence, $$a = 2$$, $$b = -1$$, $$c = 2$$
Since, the plane passes through $$(2,3,-1)$$,
$$d=2(2)+3(-1)-1(2)$$
$$d = -1$$
Hence, the required value of the expression is $$a +d = 1$$.
Question 7
1 / -0

lf $$\theta $$ is the angle between two lines whose d.cs are $$l_{1},m_{\mathrm{1}},n_{\mathrm{1}}$$ and $$l_{2},m_{2},n_{2}$$, then

$$\displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{\mathrm{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}=$$

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$2$$

Solution

Let $$\vec{a_1}$$ and $$\vec{a_2} $$ be unit vectors along the lines.

$$\vec{a_1} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}$$

$$\vec{a_2} = l_2 \hat{i} + m_2 \hat{j} + n_2 \hat{k}$$
$$\Rightarrow l_1^2+m_1^2+n_1^2 = 1$$ and $$l_2^2+m_2^2+n_2^2 = 1$$

Consider the dot product of $$\vec{a_1} $$ and $$\vec{a_2} $$

$$\vec{a_1}.\vec{a_2} = (l_1l_2)+(m_1m_2)+(n_1n_2) $$

$$\Rightarrow \cos \theta = (l_1l_2)+(m_1m_2)+(n_1n_2) $$

$$S = \displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}$$

$$\Rightarrow S = \dfrac{ 2 + 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{ 2 - 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \sin^2{\dfrac{\theta}{2}}} $$

$$\Rightarrow S = \dfrac{2+2\cos\theta}{ 4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{2-2\cos\theta}{4 \sin^2{\dfrac{\theta}{2}}}$$
Now, $$1+\cos\theta = 2\cos^2\dfrac{\theta}2$$ and $$1-\cos\theta = 2\sin^2\dfrac{\theta}2$$
$$\Rightarrow S = \dfrac22 + \dfrac22$$

$$\Rightarrow S =2 $$
Question 8
1 / -0

The plane $$2x+3y+kz-7=0$$ is parallel to the line whose direction ratios are $$(2, -3, 1)$$, then $$k=$$

A
$$5$$

B
$$8$$

C
$$1$$

D
$$0$$

Solution

Since, the plane is parallel to the line, the normal to the plane will be perpendicular to the line as well.
Hence, their dot product will be equal to $$0$$.
$$(2,3,k).(2,-3,1) = 0$$
$$4 - 9 +k =0$$
$$k= 5$$
Question 9
1 / -0

If the foot of the perpendicular from $$(0,0,0)$$ to the plane is $$(1,2,2)$$, then the equation of the plane is

A
$$-x+2y+8z-9=0$$

B
$$x+2y+2z-9=0$$

C
$$x+y+z-5=0$$

D
$$x+2y-3z+1=0$$

Solution

The line joining the origin to the point on the plane is perpendicular to the plane.
The dr's of the normal are $$(1,2,2)$$.
The equation of the plane will be of the form,
$$x+2y+2z = d$$
Since, $$(1,2,2)$$ lies on the plane, $$d = 9$$
Hence, $$x+2y+2z = 9$$
Question 10
1 / -0

The d.r's of the line of intersection of the planes $$x+y+z-1$$ $$=0$$ and $$2x+3y+4z-7$$ $$=0$$ are

A
$$1, 2, -3$$

B
$$2, 1, -3$$

C
$$4, 2, -6$$

D
$$1, -2, 1$$

Solution

The normal vector to the plane $$ax+by+cz+d=0$$ is represented by $$ai+bj+ck$$.

Hence, the normal to the first plane is $$i+j+k$$ and normal to the second plane is given by $$2i+3j+4k$$.

The line of intersection plane is perpendicular to the both these normal vectors.

It is found by the cross product of these normal vectors.

$$(i+j+k)\times(2i+3j+4k)=i-2j+k$$.

Hence, the direction ratios given by $$(1,-2,1)$$

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Three Dimensional Geometry Test - 22

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Three Dimensional Geometry Test - 22

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Question 1

$$\displaystyle \theta =\cos ^{ -1 }{ \left( \frac { 19 }{ 21 } \right) } $$

$$\displaystyle \theta =\sin ^{ -1 }{ \left( \frac { 19 }{ 21 } \right) } $$

$$\displaystyle \theta =\cos ^{ -1 }{ \left( \frac { 20 }{ 21 } \right) } $$

None of these

Solution

Question 2

$$(3, -2, 6)$$

$$\left (\displaystyle \dfrac{2}{7},\dfrac{3}{7},\dfrac{-6}{7}\right)$$

$$\left (\displaystyle \dfrac{3}{7}\dfrac{-2}{7},\dfrac{6}{7}\right)$$

$$\left (\displaystyle \dfrac{2}{3}\dfrac{-1}{3},\dfrac{2}{3}\right)$$

Solution

Question 3

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

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

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

None of these

Solution

Question 4

$$(1, 1, -26)$$

$$(0, 0, 0)$$

$$(2, 1, -32)$$

$$(1, 1, 1)$$

Solution

Question 5

$$\vec{r}.(5i+3j-11k)+\dfrac{135}{2}=0$$

$$\vec{r}.(5i+3j-11k)=\dfrac{135}{2}$$

$$\vec{r}.\left (\dfrac{3}{2}\hat{i}+\dfrac{7}{2}\hat{j}-\dfrac{9}{2}\hat{k}\right)=5\hat{i}+3\hat{j}-11\hat{k}$$

$$\vec{r}.(5i+3j-11k)+\dfrac{185}{2}=0$$

Solution

Question 6

$$3$$

$$1$$

$$-2$$

$$2$$

Solution

Question 7

$$1$$

$$0$$

$$-1$$

$$2$$

Solution

Question 8

$$5$$

$$8$$

$$1$$

$$0$$

Solution

Question 9

$$-x+2y+8z-9=0$$

$$x+2y+2z-9=0$$

$$x+y+z-5=0$$

$$x+2y-3z+1=0$$

Solution

Question 10

$$1, 2, -3$$

$$2, 1, -3$$

$$4, 2, -6$$

$$1, -2, 1$$

Solution

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