Vector Algebra Test

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Vector Algebra Test - 19

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Question 1
1 / -0

If $$\vec{a} = (2, 1, -1), \vec{b} = (1,-1,0), \vec{c} = (5, -1, 1) $$ , then what is the unit vector parallel to $$ \vec{a} + \vec{b} - \vec{c} $$ in the opposite direction ?

A
$$\dfrac{\hat{i}+\hat{j}-2 \hat{k}}{3}$$

B
$$\dfrac{\hat{i}-2 \hat{j} + 2 \hat{k}}{3}$$

C
$$\dfrac{2 \hat{i} - \hat{j} + 2 \hat{k}}{3}$$

D
None of the above.

Solution

GIven : $$\vec{a} = (2, 1, -1), \vec{b} = (1,-1,0), \vec{c} = (5, -1, 1) $$
$$\vec { a } +\vec { b } -\vec { c } =\hat { (2i+ } \hat { j- } \hat { k)+ } (\hat { i- } \hat { j) } -(5\hat { i- } \hat { j } +\hat { k) } =-2\hat { i } +\hat { j } -2\hat { k } $$

Magnitude of the vector $$=\sqrt { 4+1+4 } =3$$

So unit vector in the opposite direction $$=-\left( \dfrac { -2\hat { i } +\hat { j } -2\hat { k } }{ 3 } \right) =\left( \dfrac { 2\hat { i } -\hat { j } +2\hat { k } }{ 3 } \right) $$
Hence, C is correct.
Question 2
1 / -0

The position vectors of the points $$A$$ and $$B$$ are respectively $$3\hat { i } -5\hat { j } +2\hat { k } $$ and $$\hat { i } +\hat { j } -\hat { k } $$. What is the length of $$AB$$?

A
$$11$$

B
$$9$$

C
$$7$$

D
$$6$$

Solution

Now, for vectors
$$A=a_{ 1 }\hat { i } +b_{ 1 }\hat { j } +c_{ 1 }\hat { k }$$
$$ B=a_{ 2 }\hat { i } +b_{ 2 }\hat { j } +c_{ 2 }\hat { k }$$

length of $$AB=\sqrt { (a_{ 2 }-a_{ 1 })^{ 2 }+(b_{ 2 }-b_{ 1 })^{ 2 }+(c_{ 2 }-c_{ 1 })^2 } $$

So, for given vectors
Length of AB $$= \sqrt { (3-1)^{ 2 }+(-5-1)^{ 2 }+(2+1)^{ 2 } } $$

$$ =\sqrt { 4+36+9 }$$

$$ =7$$
Question 3
1 / -0

The unit vector perpendicular to the vector $$\hat{i}-\hat{j}$$ and $$\hat{i}+\hat{j}$$ forming a right handed system, is

A
$$\hat{k}$$

B
$$-\hat{k}$$

C
$$\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}$$

D
$$\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$$

Solution

Let $$\vec{a}=\hat{i}-\hat{j}$$

and $$\vec{b}=\hat{i}+\hat{j}$$

Let the unknown vector be $$\vec{c}$$

Given $$\vec{c}$$ $$\perp \vec{a}$$ and $$\vec{b}$$ and $$|c|=1$$

$$\Rightarrow \vec{a}\times\vec{b}=(\hat{i}-\hat{j})\times(\hat{i}+\hat{j})=2\hat{k}$$

$$\therefore \vec{c}=\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$$

$$\Rightarrow \hat{c}=\dfrac{2\hat{k}}{2}$$

$$\Rightarrow \hat{c}=\hat{k}$$

Correct option is A
Question 4
1 / -0

Point $$(4, 0)$$ lies on ________.

A
$$\vec {XO}$$

B
$$\vec {YO}$$

C
$$\vec {OX}$$

D
$$\vec {OY}$$

Solution

$$\vec {XO}$$ is positive x-axis, so $$(4, 0)$$ lies on it.
Question 5
1 / -0

If the points $$A$$ and $$B$$ are $$\left( 1,2,-1 \right)$$ and $$ \left( 2,1,-1 \right)$$ respectively, then $$ \vec { AB } $$ is

A
$$\hat { i } +\hat { j } $$

B
$$\hat { i } -\hat { j } $$

C
$$2\hat { i } +\hat { j } -\hat { k } $$

D
$$\hat { i } +\hat { j } +\hat { k } $$

Solution

$$\vec{ AB } = $$ $$\left < 2-1, 1-2, -1+1 \right >$$

$$= \left < 1,-1, 0 \right >$$

$$\therefore\quad\vec{AB} = $$ $$\hat{ i } - \hat{ j} $$
Question 6
1 / -0

If $$\vec {a} . \hat {i} = \vec {a} . (\hat {i} + \hat {j}) = \vec {a} (\hat {i} + \hat {j} + \hat {k})$$, thus $$\vec {a}=$$

A
$$\hat {i}$$

B
$$\hat {i} + j$$

C
$$\hat {k} - \hat {j}$$

D
$$\hat {i} + \hat {j} + \hat {k}$$

Solution

Let $$\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$$
And $$\vec{a}.\hat{i}=\vec{a}.(\hat{i}+\hat{j}+\hat{k})=\vec{a}.(i+j)$$
$$\Rightarrow x.1=x+y+z=x+y$$
$$\Rightarrow y+z=0$$ and $$y=0 \Rightarrow z=0$$
Hence,
$$\vec{a}=k\hat{i}$$ where $$k$$ can be any real no
from options , $$\vec{a}=i$$
Question 7
1 / -0

The Polygon Law of Vector Addition is simply an extension of ____________.

A
Parallelogram Law of Vector Addition

B
Triangular Law of Vector Addition

C
Both A and B

D
None of the above

Solution

The Polygon Law of Vector Addition is simply an extension of Triangular Law of Vector Addition. Here we take into consideration more than two sides unlike in triangular law of vector addition.
Question 8
1 / -0

If for unit vectors $$\bar{a}$$ and $$\bar{b}$$, $$\bar{a}+2\bar{b}$$ and $$5\bar{a}-4\bar{b}$$ are each-other, then $$(\bar{a} \wedge \bar{b})=$$ _____________.

A
$$\cos^{-1}\displaystyle\frac{1}{3}$$

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

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

D
$$\cos^{-1}\displaystyle\frac{2}{7}$$

Solution

$$\vec{a}$$ and $$\vec{b}$$ are unit vector , and  $$5\vec{a}-4\vec{b}$$, $$\vec{a}+2\vec{b}$$ are perpendicular so
$$\Rightarrow$$ $$ (5\vec{a} -4\vec{b}) .(\vec{a} +\vec{2b}) = 0 $$

$$\Rightarrow$$ $$5a^2 +10ab -4ab -8b^2$$ = 0

$$\Rightarrow$$   $$5a^2 +6ab -8b^2$$ = 0....equation 1

now we know $$\vec{a}$$ and $$\vec{b} $$ are unit vector so
$$(\vec{a}.\vec{b}) = \dfrac{1}{2}$$

so angle between $$\vec{a}$$ , $$\vec{b}$$ is $$\dfrac{\pi}{3}$$
Question 9
1 / -0

The number of vectors of unit length perpendicular to vectors $$\bar{a} = ( 1, 1, 0)$$ & $$\bar{b}= (0, 1, 1)$$ is :

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\infty$$

Solution

There exists two unit length vectors perpendicular to both of the vectors $$\vec{a}$$ and $$\vec{b}$$ and it is given by $$\dfrac{\vec{a}\times \vec{b}}{|\vec{a}\times\vec{b}|}$$ and $$\dfrac{\vec{b}\times \vec{a}}{|\vec{a}\times\vec{b}|}$$.
So, the number of vector is $$2$$.
Question 10
1 / -0

For $$A(1, -2, 4), B(5, -1, 7), C(3, 6, -2), D(4, 5, -1)$$, the projection of $$\overline {AB}$$ on $$\overline {CD}$$ is ________.

A
$$(2\sqrt {3}, -2\sqrt {3}, 2\sqrt {3})$$

B
$$\dfrac {3}{13} (4, 1, 3)$$

C
$$(1, -1, 1)$$

D
$$(2, -2, 2)$$

Solution

$$\overline{AB}=\overline{B}-\overline{A}$$
$$A(1, -2, 4), B(5, -1, 7)$$ ....given
$$\overline{A}=\overline i-2\overline j +4\overline k $$ and
$$\overline{B}=5\overline i-\overline j +7\overline k $$
$$\therefore \overline{AB}=(5-1)\overline i+(-1+2)\overline k+(7-4)\overline k$$
$$\therefore \overline {AB}=4\overline i+\overline j+3\overline k$$
$$\overline {AB} = (4, 1, 3)$$
$$C(3, 6, -2), D(4, 5, -1)$$
$$\overline{C}=3\overline i+6\overline j -2\overline k $$ and
$$\overline{D}=4\overline i+5\overline j -\overline k $$
$$\therefore \overline{CD}=(4-3)\overline i+(5-6)\overline k+(-1+2)\overline k$$
$$\therefore \overline {CD}=\overline i+\overline j-\overline k$$
$$\overline{CD} = (1, -1, 1)$$
Now projection of $$\overline {AB}$$ on $$\overline CD =$$
$$\left (\dfrac {\overline {AB} \cdot \overline {CD}}{|CD|}\right ) \cdot \hat {CD}$$
$$= \left (\dfrac {4 - 1 + 3}{\sqrt {4}}\right ) \cdot \dfrac {(1, -1, 1)}{\sqrt {3}}$$
$$= (2, -2, 2)$$.

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Vector Algebra Test - 19

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Vector Algebra Test - 19

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Question 1

$$\dfrac{\hat{i}+\hat{j}-2 \hat{k}}{3}$$

$$\dfrac{\hat{i}-2 \hat{j} + 2 \hat{k}}{3}$$

$$\dfrac{2 \hat{i} - \hat{j} + 2 \hat{k}}{3}$$

None of the above.

Solution

Question 2

$$11$$

$$9$$

$$7$$

$$6$$

Solution

Question 3

$$\hat{k}$$

$$-\hat{k}$$

$$\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}$$

$$\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$$

Solution

Question 4

$$\vec {XO}$$

$$\vec {YO}$$

$$\vec {OX}$$

$$\vec {OY}$$

Solution

Question 5

$$\hat { i } +\hat { j } $$

$$\hat { i } -\hat { j } $$

$$2\hat { i } +\hat { j } -\hat { k } $$

$$\hat { i } +\hat { j } +\hat { k } $$

Solution

Question 6

$$\hat {i}$$

$$\hat {i} + j$$

$$\hat {k} - \hat {j}$$

$$\hat {i} + \hat {j} + \hat {k}$$

Solution

Question 7

Parallelogram Law of Vector Addition

Triangular Law of Vector Addition

Both A and B

None of the above

Solution

Question 8

$$\cos^{-1}\displaystyle\frac{1}{3}$$

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

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

$$\cos^{-1}\displaystyle\frac{2}{7}$$

Solution

Question 9

$$1$$

$$2$$

$$3$$

$$\infty$$

Solution

Question 10

$$(2\sqrt {3}, -2\sqrt {3}, 2\sqrt {3})$$

$$\dfrac {3}{13} (4, 1, 3)$$

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

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

Solution

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