Vector Algebra Test

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Vector Algebra Test - 43

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Question 1
1 / -0

If $$2\overrightarrow { a }\cdot \overrightarrow { b } =\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| $$ then the angle between $$\vec { a }$$ and $$ \vec { b }$$ is

A
$${30}^{o}$$

B
$${0}^{o}$$

C
$${90}^{o}$$

D
$${60}^{o}$$

Solution

$$2\overrightarrow { a }\cdot \overrightarrow { b } = \left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| $$
But we know that $$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos\theta$$
$$\therefore \cos\theta=\dfrac{1}{2}$$
$$\Rightarrow \theta=60^o$$

Here $$\theta$$ is the angle between both the vectors
Question 2
1 / -0

For non-zero vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ if $$|\overrightarrow{a}+\overrightarrow{b}| < |\overrightarrow{a}-\overrightarrow{b}|$$, then $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ are

A
Collinear

B
Perpendicular to each other

C
Inclined at an acute angle

D
Inclined at an obtuse angle

Solution

$$|\vec{a}+\vec{b}|<|\vec{a}-\vec{b}|$$
$$a^{2}+b^{2}+2ab\cos\theta<a^{2}+b^{2}-2ab\cos\theta$$ where $$\theta$$ is the included angle between the two vectors.
Hence
$$4ab\cos\theta<0$$
Since $$a,b>0$$ (magnitude of a vector is always a positive quantity).
Hence
$$\cos\theta<0$$
Or
$$\theta\in \left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)$$. Hence $$\theta$$ is an obtuse angle.
Question 3
1 / -0

Let $$\vec {a} = i + 2j + k, \vec {b} = i - j + k$$ and $$\vec {c} = i + j - k$$, a vector in the plane $$\vec {a}$$ and $$\vec {b}$$ whose projection on $$\vec {c}$$ is $$\dfrac {1}{\sqrt {3}}$$ is _____

A
$$3i + j - 3k$$

B
$$4i + j - 4k$$

C
$$i + j - 2k$$

D
$$2i + j + 2k$$

Solution

A vector in the plane of $$\vec{a}$$ and $$\vec{b}$$ would be $$m\vec{a} + n\vec{b}$$
i.e. $$(m + n)i + (2m - n)j + (m + n)k$$
Its projection on $$\vec{c}$$ will be $$\cfrac{(m\vec{a} + n\vec{b}).\vec{c}}{|c|}$$
$$\therefore \cfrac{1}{\sqrt{3}} = \cfrac{[(m + n)i + (2m - n)j + (m + n)k].(i + j - k)}{\sqrt{3}}$$
$$\Rightarrow 1 = m + n + 2m - n - m - n$$
$$\Rightarrow 1 = 2m - n$$
This relation is satisfied by $$m = 1, n = 1$$
Thus, the required vector can be $$2i + j + 2k$$
Question 4
1 / -0

Point $$P$$ is the endpoint of vector $$OP$$ and point $$Q$$ is the endpoint of vector $$OQ$$ as shown in the above figure. When the vectors $$\overline {OP}$$ and $$\overline {OQ}$$ are added, calculate the length of the resultant vector.

A
$$1.41$$

B
$$2.24$$

C
$$2.65$$

D
$$3.00$$

E
$$8.60$$

Solution

Given in figure, $$P (-3,7)$$ and $$Q(4,-7)$$ and point $$O (0,0)$$
Then vector $$PO=$$ $$\overline{PO}=(0,0)-(3,-7)=(3,-7)$$
And vector $$OQ =$$ $$\overline{OQ}=(0,0)-(4,-5)=(-4,5)$$
Then length of vector $$PQ =$$ $$\overline{PQ}=\sqrt{(-4+3)^{2}+(5-7)^{2}}=\sqrt{1+4}\sqrt{5}=2.24$$
Question 5
1 / -0

The value of $$x$$ if $$x(\hat { i } +\hat { j } +\hat { k } )$$ is a unit vector is

A
$$\pm \cfrac { 1 }{ \sqrt { 3 } } $$

B
$$\pm \sqrt { 3 } $$

C
$$\pm 3$$

D
$$\pm \cfrac{1}{3}$$

Solution

If $$x(\hat i+\hat j+\hat k)$$ is an unit vector then
$$|x(\hat i+\hat j+\hat k)|=1$$
$$\Rightarrow \bigg|x\sqrt{1^2+1^2+1^2} \bigg|=1$$
$$\Rightarrow \sqrt 3|x|=1$$
$$\Rightarrow |x|=\dfrac{1}{\sqrt 3}$$
$$\therefore x=\pm \dfrac{1}{\sqrt 3}$$
Question 6
1 / -0

If $$\displaystyle \vec{a}\:and\:\vec{b}$$ are the vectors $$\displaystyle \overrightarrow{AB}\:and\:\overrightarrow{BC}$$ determined by the adjacent sides of a regular hexagon. What are the vectors determined by the side CD and DE taken in order?

A
$$\vec b-\vec a,-\vec a$$

B
$$\vec a-\vec b,-\vec a$$

C
$$\vec b-\vec a,-\vec a +\vec b$$

D
None of these

Solution

If $$\overrightarrow { a } $$ and $$\overrightarrow { b }$$ are the vectors determined by the adjacent sides of a regular hexagon.
Let $$ABCDEF$$ be a regular hexagon such that
$$\overrightarrow { AB } =\overrightarrow { a } ,\quad \overrightarrow { BC } =\overrightarrow { b } $$
Now $$\overrightarrow { AC } =\overrightarrow { AB } +\overrightarrow { BC } =\overrightarrow { a } +\overrightarrow { b } \\ \overrightarrow { AD } =2\overrightarrow { BC } \quad [\because \quad AD=2BC\quad and\quad AD\parallel BC]\\ \therefore \overrightarrow { AD } =2\overrightarrow { b } $$
In $$\\ \Delta ADC,\overrightarrow { AC } +\overrightarrow { CD } =\overrightarrow { AD } \\ \overrightarrow { CD } =\overrightarrow { AD } -\overrightarrow { AC } =2\overrightarrow { b } -\left( \overrightarrow { a } +\overrightarrow { b } \right) \\ \quad \quad \quad =\overrightarrow { b } -\overrightarrow { a } \\ \overrightarrow { DE } =-\overrightarrow { AB } =-\overrightarrow { a } \\ \quad \quad $$
Hence correct option is $$A$$
$$\overrightarrow { b } -\overrightarrow { a } $$ ,$$-\overrightarrow { a } \\ \quad \quad $$
Question 7
1 / -0

If $$p$$ , $$q$$ and $$r$$ are three non-coplanar vectors such that $$p + q + r = \alpha s$$ and $$q + r + s = \beta p $$. then $$ p +q +r +s $$ is equal to

A
$$(\alpha+1)s$$

B
$$\alpha p$$

C
$$\beta q$$

D
$$(\alpha + \beta) r$$

Solution

Given
$$p+q+r=\alpha s$$----(1)
$$q+r+s=\beta p$$----(2)

$$p+q+r+s$$ $$=\alpha s+s$$( from 1)
$$=(\alpha+1)s$$
Question 8
1 / -0

$$ABCDEF$$ is a regular hexagon whose centre is $$O$$. The $$\overline { AB } +\overline { AC } +\overline { AD } +\overline { AE } +\overline { AF } $$ is

A
$$2\overline { AO } $$

B
$$3\overline { AO } $$

C
$$5\overline { AO } $$

D
$$6\overline { AO } $$

Solution

Since it is a regular hexagon,
$$\overline { AB } =\overline { ED } ,\overline { AF } =\overline { CD } $$

Now, $$\overline { AB } +\overline { AC } +\overline { AD } +\overline { AE } +\overline { AF } =\overline { AE } +\overline { ED } +\overline { AC } +\overline { CD } +\overline { AD } =3\overline { AD } $$

Now, since $$O$$ is the centre of the hexagon, $$O$$ will be the midpoint of $$AD$$
$$\therefore 3\overline{AD}=6\overline { AO } $$
Question 9
1 / -0

Two $$\vec a$$ and $$\vec b$$ are at an angle of $$60^o$$ with each other Their resultant makes an angle of $$45^o$$ with $$\vec a$$ If magnitude $$|\vec b|= 2$$ units then $$|\vec a|$$ is:

A
$$\sqrt 3$$

B
$$\sqrt 3- 1$$

C
$$\sqrt 3 +1$$

D
$$\dfrac {\sqrt 3}{2}$$

Solution

The vectors make $$60°$$ with each other, and the resultant vector makes $$45°$$ with $$a$$
The other angle $$=15°$$
Using the side rule;
$$a/\sin { 15 } =2/\sin { 45 } \\ a=2\sin { 15 } /\sin { 45 } \\ a=0.732$$
Hence correct answer is $$(B)$$ $$\sqrt { 3 } -1$$
Question 10
1 / -0

The magnitude of resultant of three vectors of magnitude $$1,2$$ and $$3$$ whose direction are those of the sides of an equilateral triangle taken in order is:

A
$$0$$

B
$$2\sqrt 2 $$ units

C
$$4\sqrt 3 $$ units

D
$$\sqrt 3 $$ units

Solution

The $$1$$ magnitude vector in each cancels out to give only $$1$$ and $$2$$ magnitude vectors which are $$120$$ degrees apart, which gives a resultant of $$\sqrt{3}$$ units

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 43

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Vector Algebra Test - 43

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SHARING IS CARING

Question 1

$${30}^{o}$$

$${0}^{o}$$

$${90}^{o}$$

$${60}^{o}$$

Solution

Question 2

Collinear

Perpendicular to each other

Inclined at an acute angle

Inclined at an obtuse angle

Solution

Question 3

$$3i + j - 3k$$

$$4i + j - 4k$$

$$i + j - 2k$$

$$2i + j + 2k$$

Solution

Question 4

$$1.41$$

$$2.24$$

$$2.65$$

$$3.00$$

$$8.60$$

Solution

Question 5

$$\pm \cfrac { 1 }{ \sqrt { 3 } } $$

$$\pm \sqrt { 3 } $$

$$\pm 3$$

$$\pm \cfrac{1}{3}$$

Solution

Question 6

$$\vec b-\vec a,-\vec a$$

$$\vec a-\vec b,-\vec a$$

$$\vec b-\vec a,-\vec a +\vec b$$

None of these

Solution

Question 7

$$(\alpha+1)s$$

$$\alpha p$$

$$\beta q$$

$$(\alpha + \beta) r$$

Solution

Question 8

$$2\overline { AO } $$

$$3\overline { AO } $$

$$5\overline { AO } $$

$$6\overline { AO } $$

Solution

Question 9

$$\sqrt 3$$

$$\sqrt 3- 1$$

$$\sqrt 3 +1$$

$$\dfrac {\sqrt 3}{2}$$

Solution

Question 10

$$0$$

$$2\sqrt 2 $$ units

$$4\sqrt 3 $$ units

$$\sqrt 3 $$ units

Solution

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