Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 56

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Question 1
1 / -0

The vector equation of the plane through the point $$\vec i+2\vec j-\vec k$$ and $$\bot$$ to the line of intersection of the plane $$\overrightarrow { r } .\left( 3\vec i-\vec j+\vec k \right) =1$$ and $$\overrightarrow { r } .\left(\vec i+4\vec j-2\vec k \right) =2$$ is

A
$$\overrightarrow { r } .\left( 2\vec i+7\vec j-13\vec k \right) =1$$

B
$$\overrightarrow { r } .\left( 2\vec i-7\vec j-13k \right) =1$$

C
$$\overrightarrow { r } .\left( 2\vec i+7\vec j+13\vec k \right) =0$$

D
None of these

Solution

The line of inetrsection of the planes $$\overrightarrow { r } .\left( 3\vec i-\vec j+\vec k \right) =1$$ and $$\overrightarrow { r } .\left( \vec i+4\vec j-2\vec k \right) =2$$ is common to both the planes.
Therefore, it is $$\bot$$ to normals to the two planes i.e. $$\overrightarrow { { n }_{ 1 } } =3\vec i-\vec j+\vec k$$ and $$\overrightarrow { { n }_{ 2 } } =\vec i+4\vec j-2\vec k$$.
Hence, it is parallel to the vector $$\overrightarrow { { n }_{ 1 } } \times \overrightarrow { { n }_{ 2 } } =-2\vec i+7\vec j+13\vec k$$.
Thus, we have to find the equation of the plane passing through $$\overrightarrow { a } =\vec i+2\vec j-\vec k$$ and normal to the vector $$\overrightarrow { n } =\overrightarrow { { n }_{ 1 } } \times \overrightarrow { { n }_{ 2 } } $$.
The equation of the required plane is
$$\left( \overrightarrow { r } -\overrightarrow { a } \right) .\overrightarrow { n } \Rightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { a } .\overrightarrow { n } \\ \Rightarrow \overrightarrow { r } .\left( -2\vec i+7\vec j+13\vec k \right) =\left( \vec i+2\vec j+\vec k \right) .\left( -2\vec i+7\vec j+13\vec k \right) \\ \Rightarrow \overrightarrow { r } .\left( 2\vec i-7\vec j-13\vec k \right) =1$$
Question 2
1 / -0

The direction cosine of a line equally inclined to the axes are

A
$$\displaystyle \frac { 1 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 1 }{ 3 } $$

B
$$\displaystyle -\frac { 1 }{ 3 } ,-\frac { 1 }{ 3 } ,-\frac { 1 }{ 3 } $$

C
$$\displaystyle \frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } $$

D
none of these

Solution

If $$a$$ line makes angles $$\alpha ,\beta ,\gamma $$ with the axes, we have $$\alpha =\beta =\gamma$$
$$\displaystyle \therefore \cos { \alpha } =\cos { \beta } =\cos { \gamma } \Rightarrow l=m=n$$
$$\displaystyle \because { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
$$\displaystyle \Rightarrow { l }^{ 2 }+{ l }^{ 2 }+{ l }^{ 2 }\quad $$
$$\Rightarrow { 3l }^{ 2 }=1$$
$$\displaystyle \Rightarrow { l }^{ 2 }=\frac { 1 }{ 3 } \quad $$
$$\Rightarrow l=\pm \dfrac { 1 }{ \sqrt { 3 } } $$
Question 3
1 / -0

Equation of the line which passes through the point with p.v. (2, 1, 0) and perpendicular to the plane containing the vectors $$\widehat{i}+\widehat{j}\:and\: \widehat{j}+\widehat{k}$$ 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

if line perpendicular to plane then eq
$$\vec{r}=position\ vector\ \vec{a}+t(normal\ vector\ \vec{n})$$
normal vector of plane
Given $$\vec{a}=\hat{i}+\hat{j}$$
$$\vec{b}=\hat{j}+\hat{k}$$
$$\vec{n}=\vec{a}\times\vec{b}$$
$$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\0&1&1\end{vmatrix}$$
$$\vec{a}\times\vec{b}=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)$$
$$\vec{a}\times\vec{b}=\hat{i}-\hat{j}+\hat{k}$$
eq become
$$\vec{r}=(2,1,0)+t(1,-1,1)$$

Question 4

1 / -0

Statement-1 : If a line makes acute angles $$\alpha, \beta, \gamma, \delta$$ with diagonals of a cube, then $$ \displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta=\frac{4}{3}$$
Statement 2 : If a line makes equal angle (acute) with the axes, then its direction cosine are $$ \displaystyle \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{2}}$$ and $$\dfrac{1}{\sqrt{3}}$$

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

Solution

Let l,m,n be the direction ratio of a line making angles ,,, with four diagonals of a cube then as four diagonals have direction ratio,$$( a, a, a)$$;$$(a, a, -a)$$;$$( a, -a, a,)$$ ;$$(-a, a, a) $$(where $$a$$ is the side of the cube)

Now, $$\displaystyle \cos\alpha =\left| \dfrac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left|\dfrac{l+m+n}{\sqrt3 \sqrt{l^2+m^2+n^2}} \right| $$

$$\displaystyle \cos\beta =\left| \dfrac{al+am-an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left| \dfrac{l+m-n}{\sqrt{3} \sqrt{l^2+m^2+n^2}} \right| $$

$$\displaystyle \cos\gamma = \left|\dfrac{al-am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left|\dfrac{l-m+n}{\sqrt3 \sqrt{l^2+m^2+n^2}} \right|$$

$$\displaystyle \cos\delta = \left|\dfrac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|= \left| \dfrac{-l+m+n}{\sqrt{3} \sqrt{l^2+m^2+n^2}} \right|$$

and $$\cos^2\alpha +\cos^2\beta+\cos^2\gamma+\cos^2\delta$$

$$\displaystyle =\dfrac{1}{3}\times \dfrac{(4l^2+4m^2+4n^2)}{l^2+m^2+n^2}=\dfrac{4}{3}$$

So statement 1 is true and statement-2 is not true .

Hence, option 'C' is correct.

Question 5
1 / -0

The direction cosines of a line whose equations are $$\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-2}{-3}$$

A
$$\dfrac{1}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{2}{\sqrt{14}}$$

B
$$\dfrac{2}{\sqrt{29}},\dfrac{4}{\sqrt{29}},\dfrac{-3}{\sqrt{29}}$$

C
$$\dfrac{1}{\sqrt{29}},\dfrac{-3}{\sqrt{29}},\dfrac{2}{\sqrt{29}}$$

D
$$2,4,-3$$

Solution

Step 1:
The direction ratios of the line joining these points are
$$(x_2−x_1),(y_2−y_1),(z_2−z_1)$$
$$(i.e) 2,4,-3$$
$$\therefore$$ Direction cosines are
$$\dfrac{2}{\sqrt {2^2+4^2+(-3)^2}},\dfrac{4}{\sqrt {2^2+4^2+(-3)^2}},\dfrac{-3}{\sqrt {2^2+4^2+(-3)^2}}$$
$$\Longrightarrow \dfrac{2}{\sqrt {29}},\dfrac{4}{\sqrt {29}},\dfrac{-3}{\sqrt {29}}$$
Question 6
1 / -0

If the foots of the perpendicular from the origin to a plane is $$(a,b,c)$$, the equation of the plane is

A
$$\displaystyle \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =3$$

B
$$ax+by+cz=3$$

C
$$ax+by+cz={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$$

D
$$ax+by+cz=a+b+c$$

Solution

Let the foot of the perpendicular be $$P(a,b,c)$$.
Then, direction ratios of $$OP$$ are $$a-0,b-0,c-0$$ i.e. $$a,b,c$$
So, the equation of the plane passing through $$P(a,b,c)$$, the direction ratios of the normal to which are $$a,b,c$$ is
$$a\left( x-a \right) +b\left( y-b \right) +c\left( z-c \right) =0\\ \Rightarrow ax+by+cz={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }.$$
Question 7
1 / -0

If equation of the plane through the straight line $$\displaystyle \dfrac{x -1}{2}=\dfrac{y +2}{-3}=\dfrac{z}{5}$$ and perpendicular to the plane $$x - y + z + 2 = 0 \: $$is$$ \:ax- by + cz + 4 = 0,$$ then find the value of $$ a^2 + b^2 + c$$

A
$$12$$

B
$$14$$

C
$$16$$

D
$$18$$

Solution

Let equation of a plane containing the line be $$\ell(x- 1) + m(y + 2) + nz = 0$$
then $$2\ell-3m + 5n = 0 \:$$ and $$ \:\ell - m + n = 0$$
$$\therefore \dfrac{\ell}{2}=\dfrac{m}{3}=\dfrac{n}{1}$$
$$\therefore$$ the plane is$$ \:2(x- 1) + 3 (y + 2) + z = 0$$
i.e. $$2x + 3y + z + 4 = 0$$
$$ \therefore a = 2, b = - 3, c = 1$$
$$\therefore a^2 + b^2 + c = 14$$
Question 8
1 / -0

The line $$\displaystyle \frac{x - 1}{2} = \frac{y}{-1} = \frac{z + 2}{2}$$ cuts the plane $$\displaystyle x + y + z = 1$$ at $$\displaystyle P$$. If the foot of the perpendicular from $$\displaystyle P$$ to a point $$Q\displaystyle \left ( 3, \: -4, \: 1 \right )$$ on the plane $$S$$ then the equation of the plane $$S$$ is

A
$$\displaystyle 3x - 2y - z = 0$$

B
$$\displaystyle 2x - y + 2z = 12$$

C
$$\displaystyle 2x - 10y + 5z = 51$$

D
none of these

Solution

We have plne,
$$x+y+z=1.......(1)$$
And line,
$$\dfrac{x-1}{2}=\dfrac{y}{-1}=\dfrac{z+2}{2}=\lambda......(2)$$
Assume line (2) cuts plane (1) at point P,
$$P=(2\lambda+1,-\lambda,2\lambda-2)$$
Since line (2) cuts the plane (1) , Therefore Point P lies on plane (1).
$$2\lambda+1-\lambda+2\lambda-2=1$$
$$3\lambda=2$$
$$\lambda=\dfrac{2}{3}$$

Therefore,
$$P=(\dfrac{7}{3},\dfrac{-2}{3},\dfrac{-2}{3})$$

direction ratios of normal of plane are,as we have two points, point P and Q(3,-4,1),
$$(3-\dfrac{7}{3},-4+\dfrac{2}{3},1-\dfrac{-2}{3})$$
$$(\dfrac{2}{3},\dfrac{-10}{3},\dfrac{5}{3})$$

Required Equation of plane,
$$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$$

$$\dfrac{2}{3}(x-3)+\dfrac{-10}{3}(y+4)+\dfrac{5}{3}(z-1)=0$$
$$2x-10y+5z-51=0$$
$$2x-10y+5z=51$$
Therefore option (C) is correct.
Question 9
1 / -0

if a line makes angles $$\alpha,\beta,\gamma,\delta$$ with four diagonals a cube then value of $$sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma+sin^{2}\delta$$ equals

A
$$2$$

B
$$\displaystyle \frac{4}{3}$$

C
$$\displaystyle \frac{8}{3}$$

D
$$1$$

Solution

Diagonals are $$OE=(1,1,1)$$
$$\vec { GB } =(1,1,0)-(0,0,1)=(1,1,-1)$$
$$\vec { FC } =(1,-1,1)$$
$$\vec { AD } =(-1,1,1)$$
Let the line have direction ratio's as $$l,m,n$$
Angle between $$OE$$ and $$L$$
$$\cos { \alpha } =\dfrac { l+m+n }{ \sqrt { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } \times 2\sqrt { 3 } } $$
$$\cos { \beta } =\dfrac { l+m-n }{ \sqrt { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } \times \sqrt { 3 } } $$
$$\cos { \gamma } =\dfrac { l-m+n }{ \sqrt { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } \times \sqrt { 3 } } $$
$$\cos { \delta } =\dfrac { -l+m+n }{ \sqrt { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } \times \sqrt { 3 } } $$
$$\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } +\sin ^{ 2 }{ \delta } =4-(\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } +\cos ^{ 2 }{ \delta } )$$
$$\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } +\sin ^{ 2 }{ \delta } =4-(4\{ \dfrac { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } }{ { 3(l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }) } \} )$$
$$=\dfrac { 8 }{ 3 } $$
Question 10
1 / -0

Determine the equation of the plane on which the co - ordinates of the foot of perpendicular drawn from origin O is the point $$\displaystyle P\left ( \alpha ,\beta ,\gamma \right ).$$

A
$$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha ^{3}+\beta ^{3}+\gamma ^{3}$$

B
$$\displaystyle \alpha x+\beta y+\gamma z=\alpha ^{2}+\beta ^{2}+\gamma ^{2}$$

C
$$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha +\beta +\gamma $$

D
none of these

Solution

Given that $$OP$$ is normal to plane
So the Drs of normal to the plane is $$\alpha,\beta,\gamma$$
Therefore the equation of plane is $$\alpha x+\beta y+\gamma z=k$$
The point $$(\alpha,\beta,\gamma)$$ lies on the plane
So we have $${ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }=k$$
Therefore the equation of plane is $$\alpha x+\beta y+\gamma z={ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }$$
Therefore the correct option is $$B$$

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Three Dimensional Geometry Test - 56

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Three Dimensional Geometry Test - 56

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SHARING IS CARING

Question 1

$$\overrightarrow { r } .\left( 2\vec i+7\vec j-13\vec k \right) =1$$

$$\overrightarrow { r } .\left( 2\vec i-7\vec j-13k \right) =1$$

$$\overrightarrow { r } .\left( 2\vec i+7\vec j+13\vec k \right) =0$$

None of these

Solution

Question 2

$$\displaystyle \frac { 1 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 1 }{ 3 } $$

$$\displaystyle -\frac { 1 }{ 3 } ,-\frac { 1 }{ 3 } ,-\frac { 1 }{ 3 } $$

$$\displaystyle \frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } $$

none of these

Solution

Question 3

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

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

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

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

Solution

Question 4

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

Solution

Question 5

$$\dfrac{1}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{2}{\sqrt{14}}$$

$$\dfrac{2}{\sqrt{29}},\dfrac{4}{\sqrt{29}},\dfrac{-3}{\sqrt{29}}$$

$$\dfrac{1}{\sqrt{29}},\dfrac{-3}{\sqrt{29}},\dfrac{2}{\sqrt{29}}$$

$$2,4,-3$$

Solution

Question 6

$$\displaystyle \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =3$$

$$ax+by+cz=3$$

$$ax+by+cz={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$$

$$ax+by+cz=a+b+c$$

Solution

Question 7

$$12$$

$$14$$

$$16$$

$$18$$

Solution

Question 8

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

$$\displaystyle 2x - y + 2z = 12$$

$$\displaystyle 2x - 10y + 5z = 51$$

none of these

Solution

Question 9

$$2$$

$$\displaystyle \frac{4}{3}$$

$$\displaystyle \frac{8}{3}$$

$$1$$

Solution

Question 10

$$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha ^{3}+\beta ^{3}+\gamma ^{3}$$

$$\displaystyle \alpha x+\beta y+\gamma z=\alpha ^{2}+\beta ^{2}+\gamma ^{2}$$

$$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha +\beta +\gamma $$

none of these

Solution

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