Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

The angle between the lines $$\cfrac{x-2}{2}=\cfrac{y-1}{7}=\cfrac{z+3}{-3}$$ and $$\cfrac{x+2}{-1}=\cfrac{y-4}{2}=\cfrac{x-5}{4}$$ is

A
$$\cfrac { \pi }{ 6 } $$

B
$$\cfrac { \pi }{ 3 } $$

C
$$\cfrac { \pi }{ 2 } $$

D
$$\cos ^{ -1 }{ \left( \cfrac { 3 }{ 8 } \right) } $$

Solution
Question 2
1 / -0

The Cartesian equation of a line are $$\cfrac{x-2}{2}=\cfrac{y+1}{3}=\cfrac{z-3}{-2}$$. What is its vector equation?

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

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

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

D
none of these

Solution
Question 3
1 / -0

If the points $$A(-1,3,2),B(-4,2,-2)$$ and $$C(5,5,\lambda)$$ are collinear then the value of $$\lambda$$ is

A
$$5$$

B
$$7$$

C
$$8$$

D
$$10$$

Solution
Question 4
1 / -0

The direction cosines of the perpendicular from the origin to the plane $$\vec{r}\cdot (6\hat{i}-3\hat{j}+2\hat{k})+1=0$$ are?

A
$$\dfrac{6}{7}, \dfrac{3}{7}, \dfrac{-2}{7}$$

B
$$\dfrac{6}{7}, \dfrac{-3}{7}, \dfrac{2}{7}$$

C
$$\dfrac{-6}{7}, \dfrac{3}{7}, \dfrac{2}{7}$$

D
None of these

Solution
Question 5
1 / -0

The direction consines of the line drawn from $$P\left ( -5,3,1 \right )\,to\,Q\left ( 1,5,-2 \right )$$ is

A
$$\left ( 6,2,-3 \right )$$

B
$$\left ( 2,-4,1 \right )$$

C
$$\left ( -4,8,-1 \right )$$

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

Solution
Question 6
1 / -0

The equation of the plane which passes through the x-axis and perpendicular to the line $$\dfrac {(x - 1)}{cos\theta} = \dfrac {(y + 2)}{sin\theta} = \dfrac {(z - 3)}{0}$$ is

A
$$x\, tan\theta + y\,sec\theta = 0$$

B
$$x\, sec\theta + y\,tan\theta = 0$$

C
$$x\, cos\theta + y\,sin\theta = 0$$

D
$$x\, sin\theta - y\,cos\theta = 0$$

Solution
Question 7
1 / -0

The direction cosines of the normal to the plane $$5y+4=0$$ are?

A
$$0, \dfrac{-4}{5}, 0$$

B
$$0, 1, 0$$

C
$$0, -1, 0$$

D
None of these

Solution

$$5y+4=0$$

$$\Rightarrow -5y=4$$

$$\Rightarrow -y=\dfrac{4}{5}$$

$$\Rightarrow 0x-1\cdot y+0\cdot z=\dfrac{4}{5}$$

D.r.'s of the normal to this plane are $$0, -1, 0$$ and $$\sqrt{0^2+(-1)^2+0^2}=\sqrt{1}=1$$

$$\therefore$$ d.c.'s of the normal to the given plane are $$0, -1, 0$$.
Question 8
1 / -0

If O is the origin and $$P(1, 2, -3)$$ is a given point, then the equation of the plane through P and perpendicular to OP is?

A
$$x+2y-3z=14$$

B
$$x-2y+3z=12$$

C
$$x-2y-3z=14$$

D
None of these

Solution

Let the required equation of the plane through $$P(1, 2, -3)$$ be
$$a(x-1)+b(y-2)+c(z+3)=0$$ ........(i)

D.r.'s of OP are $$(1-0), (2-0), (-3, 0)$$ i.e., $$1, 2, -3$$

$$\therefore a=1, b=2, c=-3$$

Hence, the required equation of the plane is

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

$$\Rightarrow x+2y-3z=14$$.
Question 9
1 / -0

If the directions cosines of a line are $$ k, k, k, $$ then

A
$$ k>0 $$

B
$$ 0< k< 1 $$

C
$$ k=1 $$

D
$$ k=\dfrac{1}{\sqrt{3}} $$ or $$ -\dfrac{1}{\sqrt{3}} $$

Solution

since, direction cosines of a line are $$ k, k $$ and $$ k $$
$$ \therefore l=k, m=k $$ and $$ n=k $$
We know that, $$ l^{2}+m^{2}+n^{2}=1 $$
$$ \Rightarrow k^{2}+k^{2}+k^{2}=1 $$
$$ \Rightarrow k^{2}=\dfrac{1}{3} $$
$$ \therefore k=\pm \dfrac{1}{\sqrt{3}} $$
Question 10
1 / -0

What is the equation of the plane which passes through the z-axis and is perpendicular to the line
$$\dfrac{x - a} {\cos \theta} = \dfrac{y + 2} {\sin \theta} = \dfrac{z - 3} {0} ?$$

A
$$x + y \tan \theta = 0$$

B
$$y + x \tan \theta = 0$$

C
$$x \cos \theta - y \sin \theta = 0$$

D
$$x \sin \theta - y \cos \theta = 0$$

Solution

The plane is perpendicular to the line $$\dfrac{x - a} {\cos \theta} = \dfrac{y + 2} {\sin \theta} = \dfrac{z - 3} {0} $$.
Hence, the direction ratios of the normal of the plane are $$\cos \theta, \sin \theta$$ and $$0$$. (i)
Now, the required plane passes through the z-axis. Hence the point $$\left (0, 0, 0 \right )$$ lies on the plane.
From Eqs. (i) and (ii), we get equation of the plane as
$$\cos \theta \left (x - 0 \right ) + \sin \theta \left (y - 0 \right ) + 0 \left (z - 0\right ) = 0$$
$$\cos \theta \space x + \sin \theta \space y = 0$$
$$x + y \tan \theta = 0$$