Three Dimensional Geometry Test

Result

Three Dimensional Geometry Test - 54

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$P_1 : \overrightarrow{r} \cdot \overrightarrow{n_1} - d_1 = 0, P_2 : \overrightarrow{r} \cdot \overrightarrow{n_2} - d_2 = 0$$ and $$P_3 : \overrightarrow{r} \cdot \overrightarrow{n_3} - d_3 = 0$$ are three planes and $$\overrightarrow{n_1}, \overrightarrow{n_2}$$ and $$\overrightarrow{n_3}$$ are three non-copllanar vectors, then three lines $$P_1 = 0, P_2 = 0; P_2 = 0, P_3 = 0$$ and $$P_3 = 0, P_1 = 0$$ are

A
parallel lines

B
coplanar lines

C
coincident lines

D
concurrent lines

Solution

d. $$P_1 = P_2 = 0, P_2 = P_3 = 0$$ and $$P_3 = P_1 = 0$$ are lines of intersection of the three planes $$P_1, P_2$$ and $$P_3$$. As $$\overrightarrow{n_1}, \overrightarrow{n_2}$$ and $$\overrightarrow{n_3}$$ are non-coplanar, planes $$P_1, P_2$$ and $$P_3$$ will intersect at unique point. So the given lines will pass through a fixed point.
Question 2
1 / -0

Directions For Questions

A ray of light comes along the line $$L = 0$$ and strikes the plane mirror kept along the plane $$P = 0$$ at $$B$$. $$A(2, 1, 6)$$ is a point on the line $$L = 0$$ whose image about $$P = 0$$ is $$A'$$. It is given that $$L = 0$$ is $$\frac{x - 2}{3} = \frac{y - 1}{4} = \frac{z - 6}{5}$$ and $$P = 0$$ is $$x + y - 2x = 3$$.

...view full instructions

If $$L_1 = 0$$ is the reflected ray, then its equation is

A
$$\frac{x + 10}{4} = \frac{y - 5}{4} = \frac{z + 2}{3}$$

B
$$\frac{x + 10}{3} = \frac{y + 15}{5} = \frac{z + 14}{5}$$

C
$$\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$$

D
none of these

Solution

The equation of the reflected ray $$L_1 = 0$$ is the line joining $$Q(x_2, y_2, z_2)$$ and $$B(-10, -15, -14)$$.
$$\frac{x + 10}{16} = \frac{y + 15}{20} = \frac{z + 14}{12}$$
or $$\frac{x + 10}{3} = \frac{y + 15}{5} = \frac{z + 14}{5}$$
Question 3
1 / -0

If $$\cos {\alpha},\ \cos {\beta},\ \cos {\gamma}$$ are direction
cosines of line, then the value of $$\sin^{2}\alpha + \sin^{2}\beta +
\sin^{2}\gamma$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 4
1 / -0

If $$\cos {\alpha},\ \cos {\beta},\ \cos {\gamma}$$ are direction
cosines of line, then the value of $$\sin^{2}\alpha + \sin^{2}\beta +
\sin^{2}\gamma$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 5
1 / -0

If $$\alpha, \ \beta,\ \gamma$$ are direction angles of a line and $$\alpha = 60^{o},\ \beta=45^{o},\ \gamma =$$ ____.

A
$$30^o$$ or $$90^o$$

B
$$45^o$$ or $$60^o$$

C
$$90^o$$ or $$30^o$$

D
$$60^o$$ or $$120^o$$

Solution
Question 6
1 / -0

The angle between the lines 2x = 3 y = - z and 6 x = -y = -4 z is

A
$$45^{\circ}$$

B
$$30^{\circ}$$

C
$$0^{\circ}$$

D
$$90^{\circ}$$
Question 7
1 / -0

The direction ratios of the line which is perpendicular to the two lines $$\dfrac{x-7}{2}=\dfrac{y+17}{-3}=\dfrac{z-6}{1}and\dfrac{x+5}{1}=\dfrac{y+3}{2}=\dfrac{z-6}{-2}$$ are

A
$$ 4 , 5 , 7 $$

B
$$4 , -5 , 7$$

C
$$4 , -5 , -7$$

D
$$-4 , 5 , 8 $$

Solution
Question 8
1 / -0

Te direction ratios of the line $$3x + 1 = 6 y - 2 = 1 -z $$ are

A
$$ 2 , 1 , 6 $$

B
$$2 , 1 , -6 $$

C
$$2, -1 , 6 $$

D
$$-2 , 1, 6$$

Solution
Question 9
1 / -0

Position vectors of two points are
$$P(2\hat i+\hat j+3\hat k)$$ and $$Q(-4\hat i-2\hat j+\hat k)$$
Equation of plane passing through $$Q$$ and perpendicular of $$PQ$$ is

A
$$\vec r.(6\hat i+3\hat j+2\hat k)=28$$

B
$$\vec r.(6\hat i+3\hat j+2\hat k)=32$$

C
$$\vec r.(6\hat i+3\hat j+2\hat k)+28=0$$

D
$$\vec r.(6\hat i+3\hat j+2\hat k)+32=0$$

Solution

Let position vector of point $$P$$.
$$\vec a=2\hat i+\hat j+3\hat k$$
and position vector of point $$Q$$.
$$\vec b=-4\hat i-2\hat j+\hat k$$
then $$\hat PQ=$$position vector of $$Q-$$ position of vector of $$P$$
$$\vec n=-4\hat i-2\hat j+\hat k-(2\hat j+\hat k-/(2\hat i+\hat j+3\hat k)$$
$$\vec n=-6\hat i-3\hat j-2\hat k$$$
$$\therefore $$ Equation of plane passing through point $$Q(\vec b)$$ perpendicular to $$PQ$$ is
$$(\vec r-\vec b).\vec n=0$$
$$\vec r.\vec n=\vec b.\vec n$$
$$=-4\times -6+(-2)\times (-3)+1\times -2$$
$$=24+6-2=28$$
$$\Rightarrow \hat r(-6\hat i-3\hat j-2\hat k)=28$$
$$\Rightarrow -\hat r(6\hat i+3\hat j-2\hat k)=28$$
$$\Rightarrow \hat r(6\hat i+3\hat j-2\hat k)+28=0$$
Question 10
1 / -0

which of the following group is not direction cosines of a line:

A
$$1,1,1$$

B
$$0,0,-1$$

C
$$-1,0,0$$

D
$$0,-1,0$$

Solution

Direction cosines of a line are proportional to direction ratio's.
Let $$a,b$$ and $$c$$ are direction ratio's, then according to equation
$$1\alpha \dfrac{1}{a}$$
$$\rightarrow =\dfrac{l}{a}$$ where $$k$$ is any constant
Similarly
$$k=\dfrac{m}{b}$$
$$k=\dfrac{n}{c}$$
$$k=\dfrac{l}{a}=\dfrac{m}{b}=\dfrac{n}{c}=\dfrac{\sqrt{l^2+m^2+n^2}}{\sqrt{a^2+b^2+c^2}}$$
$$k=\dfrac{1}{\sqrt{a^2+b^2+c^2}}$$
$$\because l^2+m^2+n^2=1$$
But $$\sqrt{a^2+b^2+c^2}\ne 1$$
$$l=\dfrac{a}{\sqrt{a^2+b^2+c^2}}\ne 1$$
$$m=\dfrac{a}{\sqrt{a^2+b^2+c^2}}\ne 1$$
$$n=\dfrac{a}{\sqrt{a^2+b^2+c^2}}\ne 1$$
$$\therefore $$ Group$$(1,1,1)$$ is not dc's.