Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The direction cosines of the perpendicular from the origin to the plane $$\displaystyle 3x - y + 4z = 5$$ are

A
$$\displaystyle 4, \: -1, \: 3$$

B
$$\displaystyle 3, \: -1, \: 4$$

C
$$\displaystyle \frac{3}{\sqrt{26}}, \: \frac{-1}{\sqrt{26}}, \: \frac{4}{\sqrt{26}}$$

D
$$\displaystyle \frac{4}{\sqrt{26}}, \: \frac{-1}{\sqrt{26}}, \: \frac{3}{\sqrt{26}}$$

Solution

The direction ratios of the perpendicular from the origin to the plane $$\displaystyle 3x - y + 4z = 5$$ are $$(3,-1,4)$$
Therefore, direction cosines are $$\displaystyle \left( \frac { 3 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 }+4^{ 2 } } } ,\frac { -1 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 }+4^{ 2 } } } ,\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 }+{ 4 }^{ 2 } } } \right) =\left( \frac { 3 }{ \sqrt { 26 } } ,\frac { -1 }{ \sqrt { 26 } } ,\frac { 4 }{ \sqrt { 26 } } \right) $$

Ans: C
Question 2
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The equation of the right bisector plane of the segment joining $$(2,3,4)$$ and $$(6,7,8)$$ is

A
$$x+y+z+15=0$$

B
$$x+y+z-15=0$$

C
$$x-y+z-15=0$$

D
None of these

Solution

Let $$\vec { OA } =2\vec i+3\vec j+4\vec k$$ & $$\vec { OB }=6\vec i+7\vec j+8\vec k $$
Then $$\vec { AB }=\vec { OB } -\vec { OA } =4(\vec i+\vec j+\vec k) $$ is normal of the plane and passes through midpoint(M) of $$AB$$.
Therefore, $$\vec { OM } =\dfrac { \vec { OA } +\vec { OB } }{ 2 } =4\vec i+5\vec j+6\vec k$$
Equation of desired plane is $$\left( \vec { r } -\vec { OM } \right) .\vec { AB } =0$$
$$\Rightarrow \left( x\vec i+y\vec j+z\vec k-4\vec i-5\vec j-6\vec k \right) .\left( 4\vec i+4\vec j+4\vec k \right) =0$$
$$\Rightarrow x+y+z=15$$

Ans: B
Question 3
1 / -0

The line $$\dfrac{x-x_{1}}{0}=\dfrac{y-y_{1}}{1}=\dfrac{z-z_{1}}{2}$$ is:

A
parallel to x-axis

B
perpendicular to x-axis

C
perpendicular to YOZ plane

D
parallel to y-axis

Solution

$$L=\dfrac { x-{ x }_{ 1 } }{ 0 } =\dfrac { y-{ y }_{ 1 } }{ 1 } =\dfrac { z-{ z }_{ 1 } }{ 2 } $$

dr's of line $$ L =(0,1,2)={ d }_{ 1 }$$
dr's of x-axis =$$(1,0,0)={ d }_{ 2 }$$

clearly the line is not parallel to x-axis,if it was parallel then dr's would have been same.
$$\bar { { d }_{ 1 } } \cdot \bar { { d }_{ 2 } } =3\left( i+2\hat { k } \right) \cdot \left( \hat { i } \right) $$
$$=0 $$
So, the line $${ L }_{ 1 }$$ is perpendicular to x.
Question 4
1 / -0

If the direction cosines of the line joining the origin and a point at unit distance from the origin are $$\displaystyle \frac{1}{\sqrt{3}}$$, $$-\displaystyle \frac{1}{2}$$, $$\lambda$$ then value of $$\lambda$$ is?

A
$$ \displaystyle \frac{1}{2}\sqrt{\frac{5}{3}}$$

B
$$\displaystyle \frac{2}{\sqrt{3}}$$

C
$$ \displaystyle \frac{-1}{2}\sqrt{\frac{5}{3}}$$

D
none of these

Solution

Direction cosines of the line are given as $$\dfrac{1}{\sqrt3},\dfrac{-1}{2},\lambda$$
We know that
$$\cos^2a+\cos^2b+\cos^2c=1$$
i.e $$\dfrac{1}{\sqrt3}^2+\dfrac{-1}{2}^2+{\lambda}^2=1$$
$$\dfrac{1}{3}+\dfrac{1}{4}+{\lambda}^2=1$$
$$\dfrac{7}{12}+{\lambda}^2=1$$
$${\lambda}^2=\dfrac{5}{12}$$
Therefore, $$\lambda=\dfrac{\sqrt5}{2\sqrt3}$$
Hence, option A is correct.
Question 5
1 / -0

Vector equation of the plane $$\vec{r}=\widehat{i}-\widehat{j}+\lambda (\widehat{i}+\widehat{j}+\widehat{k})+ \mu (\widehat{i}-2\widehat{j}+3\widehat{k})$$ in the scalar dot product form is

A
$$\vec{r}.(5\widehat{i}+2\widehat{j}-3\widehat{k})=7$$

B
$$\vec{r}.(5\widehat{i}-2\widehat{j}+3\widehat{k})=7$$

C
$$\vec{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=7$$

D
$$\vec{r}.(5\widehat{i}+2\widehat{j}+3\widehat{k})=17$$

Solution

We have been given a plane which passes through three points namely, $$P = (1, -1, 0), Q = (2, 0, 1), R = (2, -3, 3)$$ substituting $$0,1$$ for $$\lambda, \mu.$$

$$\overrightarrow{PQ} = \hat{i} - \hat{j} + \hat{k}$$ and $$\overrightarrow{PR} = \hat{i} -2 \hat{j} + 3\hat{k}$$

Normal vector $$= \overrightarrow{PQ} \times \overrightarrow{PR}$$

$$ = (\bar{i} - \bar{j} + \bar{k}) \times (\bar{i} -2 \bar{j} + 3\bar{k}) $$
$$= -2\bar{k} - 3\bar{j} - \bar{k} + 3\bar{i} + \bar{j} + 2\bar{i} $$
$$= 5\bar{i} - 2\bar{j} - 3\bar{k}$$

Now, $$5(x - 1) - 2(y + 1) - 3(z - 0) = 0$$ represents the equation of plane.
$$\Rightarrow 5x - 2y - 3z = 7$$ or $$\bar{r}.(5\bar{i} - 2\bar{j} - 3\bar{k}) = 0$$
Question 6
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If $$\theta $$ is an angle given by $$\cos \theta $$ $$=$$ $$\displaystyle \frac{\cos ^{2}\alpha+\cos ^{2}\beta +\cos ^{2}\gamma}{\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma }$$ where $$\alpha $$, $$\beta $$, $$\gamma $$ are the angles made by a line with the positive directions of the axes of reference then the measure of $$\theta $$ is

A
$$\displaystyle \frac{\pi }{4}$$

B
$$\displaystyle \frac{\pi }{6}$$

C
$$\displaystyle \frac{\pi }{2}$$

D
$$\displaystyle \frac{\pi }{3}$$

Solution

The cosines of the angles made by a directed line segment with the coordinate axes are called as the direction cosines of that line.

if $$α, β, γ $$ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles and the cosines of these angles are the direction cosines of the line.
cos α, cos β and cos γ are called as the direction cosines
hence from the above diagram it is known that $$\theta$$ is equal to $$60 c^0=\dfrac{\pi}{3}$$.
Question 7
1 / -0

If a line makes angles of $$60^{\circ}$$ and $$45^{\circ}$$ with the positive directions of the $$x$$-axis and $$y$$-axis respectively, then the acute angle between the line and the $$z$$-axis is

A
$$60^{\circ}$$

B
$$45^{\circ}$$

C
$$75^{\circ}$$

D
$$15^{\circ}$$

Solution

Let the angle with $$z$$ axis be $$\gamma$$
And as we know that
$${\cos}^{2}$$$$\alpha$$ $$+$$ $${\cos}^{2}$$$$\beta$$ $$+$$ $${\cos}^{2}$$$$\gamma$$ $${=}$$ $$1$$
$${\cos}^{2}$$$${60}$$ $$+$$ $${\cos}^{2}$$45 $$+$$ $${\cos}^{2}$$$$\gamma$$ $${=}$$ $$1$$
$$\dfrac{1}{4}$$ $$+$$ $$\dfrac{1}{2}$$ $$+$$ $${\cos}^{2}$$$$\gamma$$ $${=}$$ $$1$$
$${\cos}^{2}$$$$\gamma$$ $${=}$$ $$1-$$ $$\dfrac{1}{4}$$ $$-$$ $$\dfrac{1}{2}$$
$${\cos}^{2}$$$$\gamma$$ $${=}$$ $$\dfrac{1}{4}$$
Since the angle is acute,
$$\gamma$$ $${=}60
^o$$
Question 8
1 / -0

The equation of the plane passing through the line $$\displaystyle \frac{x - 1}{2} = \frac{y + 1} {-1} = \frac{z}{3}$$ and parallel to the direction whose direction numbers are $$3, 4, 2$$ is

A
$$ \displaystyle 14x - 5y - 11z = 19$$

B
$$\displaystyle 3x + 4y + 2z + 1 = 0$$

C
$$\displaystyle 2x - y + 3z = 3$$

D
none of these

Solution

Equation of plane passing through given line is given by,
$$a(x-1)+b(y+1)+cz=0$$
Also direction ratio of line are $$2,-1,3$$
$$\Rightarrow 2a-b+3c=0......(1)$$
and plane is parallel to direction with dr $$3,4,2$$
$$\Rightarrow 3a+4b+2c=0.....(2)$$
Solving (1) and (2), we get $$a = \dfrac{-14c}{11}, b= \dfrac{5c}{11}$$
Hence, required plane is
$$14x-5y-11z=19$$
Question 9
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If the points $$A(1,2,-1)$$, $$B(2,6,2)$$ and $$\displaystyle C\left ( \lambda,-2,-4 \right )$$ are collinear, then $$\displaystyle \lambda $$ is

A
$$0$$

B
$$2$$

C
$$-2$$

D
$$1$$

Solution

D.R. of AB are $$2−1,6−2,2−(−1)i.e.,1,4,3.$$
D.R. of AC are $$λ.−1,−2−2,−4−(−1) $$
$$i.e., λ−1,−4,−3 $$
Since A, B, C are colinear,
$$ \therefore AB||BC$$
$$\therefore \dfrac{λ−1}{1}=\dfrac{−4}{4}=\dfrac{−3}{3}⇒λ−1=−1⇒λ=0$$
Question 10
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$$ABC$$ is a triangle where $$A= \left ( 2,3,5 \right ),B= \left ( -1,3,2 \right ) $$ and $$C= \left (\lambda ,5,\mu \right )$$. If the median through A is equally inclined with the axes, then

A
$$\lambda = 14,\mu = 20 $$

B
$$\lambda = 9,\mu = 6 $$

C
$$\lambda = \displaystyle \frac{7}{2},\mu = 20 $$

D
$$\lambda = 10,\mu = 7 $$

Solution

$$A$$ is a median, then it meets $$BC$$ at their mid-point $$D$$,
Coordinates of $$D$$ are $$\left(\dfrac{\lambda-1}{2},4,\dfrac{2+\mu}{2}\right)$$
$$AD$$ is equally inclined on the axes i.e. makes equal angle with all the three axes, which implies that, if $$l,m,n$$ are the direction ratios then they should be equal.
$$\therefore \dfrac{\lambda-1}{2}=4=\dfrac{2+\mu}{2}$$
$$\Rightarrow \lambda = 9, \mu = 6$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 30

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Three Dimensional Geometry Test - 30

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SHARING IS CARING

Question 1

$$\displaystyle 4, \: -1, \: 3$$

$$\displaystyle 3, \: -1, \: 4$$

$$\displaystyle \frac{3}{\sqrt{26}}, \: \frac{-1}{\sqrt{26}}, \: \frac{4}{\sqrt{26}}$$

$$\displaystyle \frac{4}{\sqrt{26}}, \: \frac{-1}{\sqrt{26}}, \: \frac{3}{\sqrt{26}}$$

Solution

Question 2

$$x+y+z+15=0$$

$$x+y+z-15=0$$

$$x-y+z-15=0$$

None of these

Solution

Question 3

parallel to x-axis

perpendicular to x-axis

perpendicular to YOZ plane

parallel to y-axis

Solution

Question 4

$$ \displaystyle \frac{1}{2}\sqrt{\frac{5}{3}}$$

$$\displaystyle \frac{2}{\sqrt{3}}$$

$$ \displaystyle \frac{-1}{2}\sqrt{\frac{5}{3}}$$

none of these

Solution

Question 5

$$\vec{r}.(5\widehat{i}+2\widehat{j}-3\widehat{k})=7$$

$$\vec{r}.(5\widehat{i}-2\widehat{j}+3\widehat{k})=7$$

$$\vec{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=7$$

$$\vec{r}.(5\widehat{i}+2\widehat{j}+3\widehat{k})=17$$

Solution

Question 6

$$\displaystyle \frac{\pi }{4}$$

$$\displaystyle \frac{\pi }{6}$$

$$\displaystyle \frac{\pi }{2}$$

$$\displaystyle \frac{\pi }{3}$$

Solution

Question 7

$$60^{\circ}$$

$$45^{\circ}$$

$$75^{\circ}$$

$$15^{\circ}$$

Solution

Question 8

$$ \displaystyle 14x - 5y - 11z = 19$$

$$\displaystyle 3x + 4y + 2z + 1 = 0$$

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

none of these

Solution

Question 9

$$0$$

$$2$$

$$-2$$

$$1$$

Solution

Question 10

$$\lambda = 14,\mu = 20 $$

$$\lambda = 9,\mu = 6 $$

$$\lambda = \displaystyle \frac{7}{2},\mu = 20 $$

$$\lambda = 10,\mu = 7 $$

Solution

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