Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The three points $$ABC$$ have position vectors $$(1,x,3),(3,4,7)$$ and $$(y,-2,-5)$$ are collinear then $$(x,y)=$$

A
$$(2,-3)$$

B
$$(-2,3)$$

C
$$(-2,-3)$$

D
$$(2,3)$$

Solution

$$(1,x,3)=\lambda(3,4,7) + \mu (y,-2,-5)$$
$$1=3\lambda +\mu y$$
$$x= 4\lambda +(-2\mu)$$
$$3 = 7\lambda -5 \mu$$
$$2-x= (-3-y)\mu$$
So only $$x=2$$, $$y=-3$$
Question 2
1 / -0

The points with position vectors $$\vec{a}+\vec{b},\vec{a}-\vec{b}$$ and $$\vec{a}+\lambda\vec{b}$$ are collinear for

A
Only integrals values of $$\lambda$$

B
No value of $$\lambda$$

C
All real values of $$\lambda$$

D
Only rational values of $$\lambda$$
Question 3
1 / -0

If $$A = (1, 2, 3), B = (2, 10, 1)$$, $$Q$$ are collinear points and $$Q_x=-1$$, then $$Q_z=$$

A
$$-3$$

B
$$7$$

C
$$-14$$

D
$$-7$$

Solution

The direction ratios of the line $$AB$$ are $$ (2-1) : (10-2) : (1-3) = 1 : 8 : -2 $$.

Hence, any point $$Q$$ lying on the line $$AB$$ can be written in the form of $$(1+k, 2+8k, 3-2k)$$, where $$k$$ is a real number.

$$Q_x = -1 $$

Hence, $$1+k = -1$$

This gives, $$k = -2$$.

$$ {Q}^{}_{z} = 3- (2\times (-2)) = 7$$

Hence, option B is correct.
Question 4
1 / -0

If $$PQR$$ are the three points with respective position vectors $$\hat{i}+\hat{j},\ \hat{i}-\hat{j}$$ and $$a\hat{i}+b\hat{j}+c\hat{k}$$, then the points $$PQR$$ are collinear if

A
$$a=b=c=1$$

B
$$a=b=c=0$$

C
$$a=1,\ b,c$$ in $$R$$

D
$$a=1, c=0,\ b\in R$$

Solution

Given, $$P(\hat{i}+\hat{j}),Q(\hat{i}-\hat{j})$$ and $$R(a\hat{i}+b\hat{j}+c\hat{k})$$ are collinear
Refer image 1.
Consider $$PQ=\vec{a}$$ and $$QR=\vec{b}$$
Since , PQR are collinear, angle between $$\vec{a}$$ and $$\vec{b}$$ should be equal to $$0^0$$
$$\Rightarrow \vec{a}\times \vec{b}=ab\sin\theta=a b\sin O=0$$

So, $$\boxed{\vec{a}\times \vec{b}=0}$$............(i)

$$\vec{a}=(\hat{i}-\hat{j})-(\hat{i}+\hat{j})$$
$$=\hat{i}-\hat{j}-\hat{i}-\hat{j}$$
$$=-2\hat{j}$$

$$\vec{b}=(a\hat{i}+b\hat{j}+c\hat{k})-(\hat{i}-\hat{j})$$
$$=a\hat{i}+b\hat{j}+c\hat{k}-\hat{i}+\hat{j}$$
$$=(a-1)\hat{i}+(b+1)\hat{j}+c\hat{k}$$

from (1), $$\vec{a}\times \vec{b}=0$$

is $$(-2\hat{j})[(a-1)\hat{i}+(b+1)\hat{j}+c\hat{k}]=0$$
Refer image 2.
$$\Rightarrow +2(a-1)\hat{k}-2(b+1)0+(-2)c\hat{i}$$
$$\Rightarrow -2c\hat{i}+2(a-1)\hat{k}=0$$ (from (i))

this is possible when $$a-1=0$$ or $$\boxed{c=0}$$
$$\Rightarrow \boxed{a=1}$$

Since b is missing, so b can be any real numbers
$$\boxed{b \in R}$$
$$\therefore \boxed{a=1},\boxed{c=0}$$ and $$\boxed{b\in E}$$
Question 5
1 / -0

The three points whose position vectors are $$\overline{i}+2\overline{j}+3\overline{k,}$$ $$3\overline{i}+4\overline{j}+7\overline{k,}$$ and $$-3\overline{i}-2\overline{j}-5\overline{k}$$

A
Form the vertices of an equilateral triangle

B
Form the vertices of an right angled triangle

C
Are collinear

D
Form the vertices of an isosceles triangle

Solution

$$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$
$$\vec{b}=3\hat{i}+4\hat{j}+7\hat{k}$$
$$\vec{c}=3\hat{i}-2\hat{j}-5\hat{k}$$
$$\vec{ca}=4\hat{i}+4\hat{j}+8\hat{k}$$
$$\vec{cb}=6\hat{i}+6\hat{j}+12\hat{k}$$
$$\vec{cb}=\frac{3}{2}\left ( \vec{ca} \right )$$
So, $$\vec{a}, \vec{c}, \vec{b}$$ are collinear.
Question 6
1 / -0

If $$\vec{a},\vec{b},\vec{c}$$ are the position vectors of points lie on a line, then $$\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=$$

A
$$0$$

B
$$ \vec{b}$$

C
$$1$$

D
$$\vec{a}$$
Question 7
1 / -0

The product of the d.r's of a line perpendicular to the plane passing through the points $$(4,0,0),(0,2,0)$$ and $$( 1,0,1)$$ is

A
$$6$$

B
$$2$$

C
$$0$$

D
$$1$$

Solution

Let three points be $$A(4,0,0) B(0,2,0) C(1,0,1)$$.
The dr's of $$AB$$ are $$(4,-2,0)$$
The dr;s of $$BC$$ are $$(1,-2,1)$$.
Since, both these lines lie on the plane, the normal to the plane will also be a normal to these lines.
Hence, the dr's of the normal can be obtained by the cross product of these $$2$$ lines.
$$n = AB \times BC $$
$$(-1,2,-3)$$
Hence, the product is $$6$$.
Question 8
1 / -0

If $$R(a+2,a+3,a+4)$$ divides the line segment joining $$P(2, 3, 4)$$ and $$Q(4, 5, 6) $$ in the ratio $$-3:2$$, then the value of the parameter which represents $$a$$ is

A
$$3$$

B
$$2$$

C
$$6$$

D
$$-1$$

Solution

If a point P divides the line segment joining $$A=(x_1,y_1,z_1)$$ and $$B=(x_2,y_2,z_2)$$ in the ratio $$m:n$$.
Then the point P is given by $$(\cfrac{mx_2+nx_1}{m+n}, \cfrac{my_2+ny_1}{m+n},\cfrac{mz_2+nz_1}{m+n})$$.
In the given problem, we have
$$a+2=\cfrac{-3\times4+2\times2}{-1}$$.
On solving, we get $$a=6.$$
Question 9
1 / -0

lf the equation of the plane perpendicular to the $$\mathrm{z}$$ -axis and passing through the point $$(2, -3,4)$$ is $$ax+by+cz=d$$ then $$\displaystyle \dfrac{a+b+c}{d}=$$

A
$$4$$

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

C
$$3$$

D
$$\displaystyle \dfrac{1}{4}$$

Solution

Since, the plane is perpendicular to the z axis, $$a = b=0$$

The plane passes through (2,-3,4).

So $$d = 4c$$

$$ \dfrac{a+b+c}{d} = \dfrac{c}{4c} = \dfrac{1}{4} $$
Question 10
1 / -0

The equation to the plane bisecting the line segment joining $$(-3, 3, 2), (9, 5, 4)$$ and perpendicular to the line segment is

A
$$x-y+4_{Z}-13=0$$

B
$$2x-2y+7z-23=0$$

C
$$x-7y+2_{Z}-1=0$$

D
$$6x+y+z-25=0$$

Solution

The point of bisection is $$(3,4,3)$$.
The d.r's of the line joining the $$2$$ points are $$(9+3,5-3,4-2)=(12,2,2)$$
These are also the d.r's of the normal to the plane.
Hence, the equation of the plane is $$12x + 2y +2z = d$$
Since, it passes through $$(3,4,3)$$.
$$\therefore 12(3)+2(4)+2(3)=d$$
$$\therefore d=50$$
Hence, equation of the plane is
$$12x + 2y + 2z = 50$$
$$6x + y + z = 25$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 21

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Three Dimensional Geometry Test - 21

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SHARING IS CARING

Question 1

$$(2,-3)$$

$$(-2,3)$$

$$(-2,-3)$$

$$(2,3)$$

Solution

Question 2

Only integrals values of $$\lambda$$

No value of $$\lambda$$

All real values of $$\lambda$$

Only rational values of $$\lambda$$

Question 3

$$-3$$

$$7$$

$$-14$$

$$-7$$

Solution

Question 4

$$a=b=c=1$$

$$a=b=c=0$$

$$a=1,\ b,c$$ in $$R$$

$$a=1, c=0,\ b\in R$$

Solution

Question 5

Form the vertices of an equilateral triangle

Form the vertices of an right angled triangle

Are collinear

Form the vertices of an isosceles triangle

Solution

Question 6

$$0$$

$$ \vec{b}$$

$$1$$

$$\vec{a}$$

Question 7

$$6$$

$$2$$

$$0$$

$$1$$

Solution

Question 8

$$3$$

$$2$$

$$6$$

$$-1$$

Solution

Question 9

$$4$$

$$\displaystyle \dfrac{3}{4}$$

$$3$$

$$\displaystyle \dfrac{1}{4}$$

Solution

Question 10

$$x-y+4_{Z}-13=0$$

$$2x-2y+7z-23=0$$

$$x-7y+2_{Z}-1=0$$

$$6x+y+z-25=0$$

Solution

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