Tag: scalar product

Questions Related to scalar product

Consider the following statements A and B given below and identify the correct answer:
A) lf $\vec{\mathrm{A}}$ is a vector, then the magnitude of the vector is given by $\sqrt{\vec{A}\times \vec{A}}$
B) lf $\vec{a}=m\vec{b}$ where 'm' is a scalar, the value of 'm' is equal to $\frac{\vec{a} \cdot  \vec{b}}{b^{2}}$

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. both A & B are correct

  2. A is correct but B is wrong

  3. A is wrong but B is correct

  4. both A and B are wrong

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Correct Option: C
Explanation:

Magnitude of a vector A can be given as $ \sqrt{\overrightarrow{A}.\overrightarrow{A}} $. 
Hence, statement A is wrong .
Also, if $ \overrightarrow{a} = m \overrightarrow{b}$, taking dot product with b vector on both sides,
$ \overrightarrow{a}.\overrightarrow{b} = m \  b^2 \Rightarrow m = \dfrac {\overrightarrow{a}.\overrightarrow{b}}{b^2} $
Hence , statement B is correct.

lf vectors $\vec{\mathrm{A}}$ and $\vec{\mathrm{B}}$ are given by $\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$ then which of the following is/are correct?
$a)\vec{\mathrm{A}}$ and $\vec{\mathrm{B}}$ are mutually perpendicular
$\mathrm{b})$ Product of $\vec{\mathrm{A}}\times\vec{\mathrm{B}}$ is same as $\vec{\mathrm{B}}\times\vec{\mathrm{A}}$
$\mathrm{c})$ The magnitude of $\vec{\mathrm{A}}$ and $\vec{\mathrm{B}}$ are equal
$\mathrm{d})$ The magnitude of $\vec{\mathrm{A}}.\vec{\mathrm{B}}$ is zero

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. a, d are correct

  2. b, c are correct

  3. c, d are correct

  4. b, a are correct

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Correct Option: A
Explanation:

$\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$  and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$  is given. 

Now scalar product of this two vector is $\vec{\mathrm{A}} . \vec{\mathrm{B}} = 5\times 6-6\times 2-3\times 6=0$     
So they are mutually perpendicular.

lf $\vec{a}=2\hat{i}+6n\hat{j}+m\hat{k}$ and $\vec{b}=\hat{i}+18\hat{j}+3\hat{k}$ are parallel to each other then the values of $m,n$ are:

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. 6,6

  2. 6,1

  3. -1,6

  4. -1,-6

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Correct Option: A
Explanation:

Since, the vector a is parallel to b, the corresponding coefficients of all the 3 components must bear the same ratio
i.e $ \dfrac{2}{1} = \dfrac{6n}{18} = \dfrac{m}{3} $
Or, $6n = 36, n = 6$
And $m = 6$

$\vec{A}$ and $\vec{B}$ are two vectors in a plane at an angle of $60^{0}$ with each other. $\vec{C}$ is another vector perpendicular to the plane containing vectors $\vec{A}$ and $\vec{B}$. Which of the following relations is possible?

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $\vec{A}+\vec{B}=\vec{C}$

  2. $\vec{A}+\vec{C}=\vec{B}$

  3. $\vec{A}\times\vec{B}=\vec{C}$

  4. $\vec{A}\times\vec{C}=\vec{B}$

Show answer
Correct Option: C
Explanation:

Vector C is perpendicular to both vectors A and B. Hence, it can be equal to their cross product.
Options A and B make vector C in the plane of vectors A  and B which is not possible.
In option D, this is not possible as vector B is not perpendicular to  vector A

If $\vec{A} = 2\hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$, sketch vectors graphically and find the component of $\vec{A}$ along $\vec{B}$ and perpendicular to $\vec{B}$.

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{4}{2}(\hat{i}+\hat{j})$

  2. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{3}{2}(\hat{i}+\hat{j})$

  3. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{1}{2}(\hat{i}+\hat{j})$

  4. Component of $A$ along $B$; $\dfrac{1}{3}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{3}{2}(\hat{i}+\hat{j})$

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Correct Option: B

Given $\vec{A} = 2\hat{i} + p\hat{j} + q\hat{k}$ and $\vec{B}=5\hat{i}+7\hat{j} + 3\hat{k}$. If $\vec{A}|| \vec{B}$, then the values of $p$ and $q$ are, respectively,

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $\dfrac{14}{5}$ and $\dfrac{6}{5}$

  2. $\dfrac{14}{3}$ and $\dfrac{6}{5}$

  3. $\dfrac{6}{5}$ and $\dfrac{1}{3}$

  4. $\dfrac{3}{4}$ and $\dfrac{1}{4}$

Show answer
Correct Option: A
Explanation:
Given $\vec { A } =2\uparrow +p\hat { j } +q\hat { k } \quad \quad \vec { B } =5\uparrow +7\hat { j } +3\hat { k } $
$A\parallel B\Rightarrow A\times B=0$

Take $det{AB}$ and simplifying,

$\therefore$  $3p-7q=0$ and $6-5q=0$  and  $14-5p=0$
                               $\Rightarrow q=\dfrac { 6 }{ 5 } $                   $p=\dfrac { 14 }{ 5 } $

If the two vectors $\vec{A} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = \hat{i} + 2 \hat{j} - n \hat{k}$ are perpendicular, then the value of $n$ is:-

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

Given,

$\begin{array}{l} \overrightarrow { A } =2\widehat { i } +3\widehat { j } +4\widehat { k }  \ \overrightarrow { B } =\widehat { i } +2\widehat { j } -n\widehat { k }  \end{array}$
They are perpendicular,
$\begin{array}{l} \overrightarrow { A } =2\widehat { i } +3\widehat { j } +4\widehat { k }  \ \overrightarrow { B } =\widehat { i } +2\widehat { j } -n\widehat { k }  \ \therefore \overrightarrow { A } .\overrightarrow { B } =0 \ \Rightarrow \left( { 2\widehat { i } +3\widehat { j } +4\widehat { k }  } \right) .\left( { \widehat { i } +2\widehat { j } -n\widehat { k }  } \right) =0 \ \Rightarrow 2\widehat { i } .\widehat { i } +3\widehat { j } .2\widehat { i } -4\widehat { k } .n\widehat { k } =0\, \, \, \, \, \, \left[ { \because \widehat { i } .\widehat { j } =j.\widehat { k } =\widehat { i } .\widehat { k } =0 } \right]  \ \Rightarrow 2+6-4n=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because \widehat { i } .\widehat { i } =\widehat { j } .\widehat { j } =\widehat { k } .\widehat { k } =1 } \right]  \ \Rightarrow 4n=8 \ \therefore n=2 \end{array}$
Hence, Option $B$ is correct.

Given $\overline { a } + \overline { b } + \vec { c } + \overline { d } = 0$ , which of the following statements is/are not a correct statement?

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $\vec { a } , \vec { b } , \vec { c }$ and $\vec { d }$ must be a null vector.

  2. The magnitude of $( \vec { a } + \vec { c } )$ equals the magnitude of $a( \vec { b } + \vec { d } )$

  3. The magnitude of $\vec { a }$ can never be greater than the sum of the magnitudes of $\vec { b } , \vec { c }$ and $\vec { d }$

  4. $\vec{b}$+$\vec{c}$ must He in the plane of $\vec{a}$ and $\vec{d}$ if $\vec{a}$ and $\vec{d}$ are not collinear and in the line of $\vec{a}$ and $\vec{d}$, if they are collinear.

Show answer
Correct Option: A
Explanation:

In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

Simple example:- a=î,b=2î,c=–3î,d=0

(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

(c) Correct
a + b + c + d = 0
a = – (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d |
| a |  ≤  | a | + | b | + | c |  …. (#)

Equation (#) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.

(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.

If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

This is the explanation of correct solution.