Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

If the points $$(0, 1, -2), (3, \lambda, -1)$$ and $$(\mu, -3, -4)$$ are collinear, the point on the same line is

A
$$(12, 9, 2)$$

B
$$(1, -1, -2)$$

C
$$(5, -3, 4)$$

D
$$(0, 0, 0)$$

Solution

We know condition of collinearity,
$$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$$

We have points $$(0,1,-2),(3,\lambda,-1) and (\mu,-3,-4)$$
$$\dfrac{x-0}{3-0}=\dfrac{y-1}{\lambda-1}=\dfrac{z+2}{-1+2}.........(1)$$

Since $$(\mu,-3,-4)$$ is collinear , so it satisfy equation (1)
taking y and z coordinates,
$$\dfrac{-3-1}{\lambda-1}=\dfrac{-4+2}{-1+2}$$
$$\lambda=3$$

Now equation (1) becomes
$$\dfrac{x-0}{3-0}=\dfrac{y-1}{3-1}=\dfrac{z+2}{-1+2}.........(2)$$

Option (A) (12,9,2) satisfy equation (2).

Therefore option (A) is correct.
Question 2
1 / -0

If O is the origin and A is the point $$(a, b, c)$$, then the equation of the plane through A and at right angles to OA is

A
$$a (x - a) - b (y - b) - c (z - c) = 0$$

B
$$a(x + a) + b (y + b) + c (z + c) = 0$$

C
$$a(x - a) + b (y - b) + c (z - c) = 0$$

D
None of the above

Solution

A plane with normal $$ai+bj+ck$$ is given by $$ax+by+cz+d=0$$.
Since it passes through $$(a,b,c)$$ we have $$d=-(a^2+b^2+c^2)$$.
Hence, the equation of plane is $$ax+by+cz-a^2-b^2-c^2=0$$.
By rearranging we get, $$a(x-a)+b(y-b)+c(z-c)=0$$
Question 3
1 / -0

The values of $$a$$ for which point $$(8, -7, a), (5, 2, 4)$$ and $$(6, -1, 2)$$ are collinear.

A
$$-4$$

B
$$-2$$

C
$$0$$

D
$$2$$

Solution

Let the points be $$A(8,-7,a), B(5,2,4), C(6,-1,2)$$
Direction ratio of $$BC =(1,-3,-2)$$
If $$A,B,C$$ are collinear points then direction ratio of $$BC$$ and $$AB$$ must be proportional.
Direction ratio of $$AB=(-3,9.4-a)$$
$$\therefore -3=\lambda1$$
$$\therefore 9=\lambda(-3)$$
$$\therefore 4-a=\lambda(-2)$$ and $$\lambda=-3$$
$$\therefore \dfrac{4-a}{-3}=-2$$

$$\therefore a=-2$$
Question 4
1 / -0

The equation of the perpendicular from the point $$(\alpha, \beta, \gamma)$$ to the plane $$ax + by + cz + d = 0$$ is

A
$$a(x-\alpha) + b (y - \beta) + c ( z - \gamma) = 0$$

B
$$\displaystyle \frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - y}{c}$$

C
$$a (x - \alpha) + b(y - \beta) + c (z - \gamma) = abc$$

D
None of these

Solution

Equation of given plane is $$ax+by+cz+d=0$$
Equation of the perpendicular plane is $$ax+by+cz+d=k$$
This planes passes through $$P(\alpha,\beta,\gamma)=\dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{\gamma}{c}+d=k$$
$$\therefore $$ Equation of the required plane is
$$ax+by+cz+d=\dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{\gamma}{c}+d=k$$
$$i.e. \dfrac{x-\alpha}{a}+\dfrac{y-\beta}{b}+\dfrac{z-\gamma}{c}$$
Question 5
1 / -0

The straight line $$\displaystyle \frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}$$ is

A
parallel to x-axis

B
parallel to y-axis

C
parallel to z-axis

D
perpendicular to z-axis

Solution

The given line is $$\dfrac{x-3}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{0}$$ which passes through the points $$(3,2,1)$$ and having the direction cosine's as $$a_1=(3,1,0)$$
Direction cosine's of z axis are $$a_{2}=(0,0,1)$$
Now, $$a_{1}\cdot a_{2}=(3,1,0)\cdot(0,0,1)=3\times 0+1\times 0+0\times 1=0$$
$$\therefore a_1\cdot a_2=0$$
Thus, $$z-$$axis and given line are perpendicular.
Hence, option D is correct.
Question 6
1 / -0

A straight line $$L$$ on the $$xy$$-plane bisects the angle between $$OX$$ and $$OY$$. What are the direction cosines of$$ L$$?

A
$$ (1/\sqrt 2, 1/\sqrt 2, 0) $$

B
$$( 1/2, \sqrt 3/2, 0 )$$

C
$$(0, 0, 1) $$

D
$$(2/3, 2/3, 1/3) $$

Solution

L makes an angle $$\dfrac{\pi}{4}$$ with X and Y axis and $$\dfrac{\pi}{2}$$
Therefore, d.cs are $$\displaystyle \left( \cos { \frac { \pi }{ 4 } } ,\cos { \frac { \pi }{ 4 } } ,\cos { \frac { \pi }{ 2 } } \right) =\left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } ,0 \right) $$
Question 7
1 / -0

The equation of the plane through $$(1, 2, 3)$$ and parallel to the plane $$2x + 3y - 4z = 0$$ is

A
$$2x + 3y + 4z = 4$$

B
$$2x + 3y + 4z + 4 = 0$$

C
$$2x - 3y + 4z + 4 = 0$$

D
$$2x + 3y - 4z + 4 = 0$$

Solution

Equation of plane parallel to $$2x + 3y - 4z = 0$$ is given by,
$$2x + 3y - 4z +\lambda= 0$$
Also it passes through $$(1,2,3)$$
$$\Rightarrow 2(1)+3(2)-4(3)+\lambda = 0\Rightarrow \lambda = 4$$
Hence, required plane is
$$2x + 3y - 4z +4= 0$$
Question 8
1 / -0

If $$M$$ denotes the mid-point of the line segment joining $$ A( 4\widehat{i} + 5\widehat{j}- 10 \widehat{k})$$ and $$ B(-\widehat{i} + 2 \widehat{j} + \widehat{k})$$, then the equation of the plane through $$M$$ and perpendicular to $$AB$$, is

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

B
$$\vec{r}.\left ( \dfrac{3}{2} \widehat{i}+\dfrac{7}{2} \widehat{j} - \dfrac{9}{2} \widehat{k}\right) + \dfrac{135}{2}= 0 $$

C
$$\vec{r}. ( 4 \widehat{i}+ 5\widehat{j} - 10 \widehat{k}) + 4= 0 $$

D
$$\vec{r}. ( - \widehat{i}+ 2 \widehat{j} + \widehat{k}) + 4= 0 $$

Solution

$$\vec { OM } =\dfrac { \vec { OA } +\vec { OB } }{ 2 } =\dfrac { 3i+7j-9k }{ 2 } $$
Then the equation of the plane through $$M$$ and perpendicular to $$AB$$ is
$$\left( \vec { r } -\vec { OM } \right) .\left( \vec { OB } -\vec { OA } \right) =0$$
$$\Rightarrow \left( \vec { r } -\dfrac { 3\vec i+7\vec j-9\vec k }{ 2 } \right) .\left( -\vec i+2\vec j+\vec k-\left( 4\vec i+5\vec j-10\vec k \right) \right) =0$$
$$\Rightarrow \vec { r } .(-5\vec i-3\vec j+11\vec k)+\dfrac { 135 }{ 2 } =0$$

Ans: A
Question 9
1 / -0

The point collinear with $$(4, 2, 0)$$ and $$(6, 4, 6)$$ among the following is

A
$$(0,4,6)$$

B
$$(8,6,8)$$

C
$$(1, -4, -6)$$

D
None of these

Solution

Cartesian form of a line passing through $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2) $$ is:
$$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}=\lambda$$

Cartesian form of the line passing through (4,2,0) and (6,4,6) is:
$$\dfrac{x-4}{6-4}=\dfrac{y-2}{4-2}=\dfrac{z-0}{6-0}=\lambda$$

$$\dfrac{x-4}{2}=\dfrac{y-2}{2}=\dfrac{z-0}{6}=\lambda$$

Therefore general point on the line is $$(2\lambda+4,2\lambda+2,6\lambda)$$
Now if we compare this coordinate with given options,we get different values of $$\lambda$$.

that means none of the points given in options lie on the line(Non-Colinear).

Therefore, (D) option is correct.
Question 10
1 / -0

If the points $$a(1, 2, -1), B(2, 6, 2)$$ and $$c(\lambda, -2, -4)$$ are collinear then $$\lambda$$ 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 collinear $$\therefore AB||BC $$
$$\therefore \dfrac {\lambda-1}{1} =\dfrac{-4}{4}=\dfrac{-3}{3}$$
$$\Longrightarrow \lambda-1=-1$$
$$\therefore \lambda=0$$

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Three Dimensional Geometry Test - 28

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Three Dimensional Geometry Test - 28

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Question 1

$$(12, 9, 2)$$

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

$$(5, -3, 4)$$

$$(0, 0, 0)$$

Solution

Question 2

$$a (x - a) - b (y - b) - c (z - c) = 0$$

$$a(x + a) + b (y + b) + c (z + c) = 0$$

$$a(x - a) + b (y - b) + c (z - c) = 0$$

None of the above

Solution

Question 3

$$-4$$

$$-2$$

$$0$$

$$2$$

Solution

Question 4

$$a(x-\alpha) + b (y - \beta) + c ( z - \gamma) = 0$$

$$\displaystyle \frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - y}{c}$$

$$a (x - \alpha) + b(y - \beta) + c (z - \gamma) = abc$$

None of these

Solution

Question 5

parallel to x-axis

parallel to y-axis

parallel to z-axis

perpendicular to z-axis

Solution

Question 6

$$ (1/\sqrt 2, 1/\sqrt 2, 0) $$

$$( 1/2, \sqrt 3/2, 0 )$$

$$(0, 0, 1) $$

$$(2/3, 2/3, 1/3) $$

Solution

Question 7

$$2x + 3y + 4z = 4$$

$$2x + 3y + 4z + 4 = 0$$

$$2x - 3y + 4z + 4 = 0$$

$$2x + 3y - 4z + 4 = 0$$

Solution

Question 8

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

$$\vec{r}.\left ( \dfrac{3}{2} \widehat{i}+\dfrac{7}{2} \widehat{j} - \dfrac{9}{2} \widehat{k}\right) + \dfrac{135}{2}= 0 $$

$$\vec{r}. ( 4 \widehat{i}+ 5\widehat{j} - 10 \widehat{k}) + 4= 0 $$

$$\vec{r}. ( - \widehat{i}+ 2 \widehat{j} + \widehat{k}) + 4= 0 $$

Solution

Question 9

$$(0,4,6)$$

$$(8,6,8)$$

$$(1, -4, -6)$$

None of these

Solution

Question 10

$$0$$

$$2$$

$$-2$$

$$1$$

Solution

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