Scalar or Dot Product MCQ Quiz - Objective Question with Answer for Scalar or Dot Product - Download Free PDF

Calculation:

Given,

${\cos }^2(\alpha )+{\cos }^2(\beta )+{\cos }^2(\gamma )=1$

Using the identity ${\cos }^2(x)=1-{\sin }^2(x)$, we substitute:

$(1-{\sin }^2(\alpha ))+(1-{\sin }^2(\beta ))+(1-{\sin }^2(\gamma ))=1$

Simplifying the equation:

$3-({\sin }^2(\alpha )+{\sin }^2(\beta )+{\sin }^2(\gamma ))=1$

Rearrange to isolate the sine terms:

${\sin }^2(\alpha )+{\sin }^2(\beta )+{\sin }^2(\gamma )=2$

Now, calculate the dot product:

$\overrightarrow{a}\cdot \overrightarrow{b}={\sin }^2(\alpha )+{\sin }^2(\beta )+{\sin }^2(\gamma )=2$

∴ The value of $\overrightarrow{a}\cdot \overrightarrow{b}$is 2.

Hence, the correct answer is Option 4.

Calculation: 

$\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{j}})$

⇒ $\overrightarrow{\mathrm{b}}=\lambda (-2\hat{\mathrm{i}}-2\hat{\mathrm{j}}+2\hat{\mathrm{k}})$

⇒ $|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}|\therefore \sqrt{6}=\sqrt{12}|\lambda |\Rightarrow \lambda =\pm \frac{1}{\sqrt{2}}$

$(\lambda =\frac{1}{\sqrt{2}}\text{ rejected }\because \overrightarrow{\mathrm{b}}\text{ makes acute angle with y axis })$

⇒ $\overrightarrow{\mathrm{b}}=-\sqrt{2}(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

⇒ $\frac{(3\overrightarrow{a}+\sqrt{2b})\cdot \overrightarrow{c}}{|\overrightarrow{c}|}=3\sqrt{2}$

Hence, the correct answer is Option 1.

Explanation:

Given:

$⃗a+2⃗b$ and $5⃗\mathrm{a}-4⃗\mathrm{b}$ are perpendicular vectors.

⇒ $(⃗a+2⃗b).$$(5⃗\mathrm{a}-4⃗\mathrm{b})$ = 0

⇒ $5|⃗a{|}^2-4⃗a.⃗b+10⃗a.⃗b-8(⃗b{)}^2=0$

⇒ $5\times 1+6⃗a.⃗b-8=0$

Now $\overrightarrow{a},\overrightarrow{b}$ are unit vectors

⇒ $6|\overrightarrow{a}||\overrightarrow{b}|Cos\theta =3$

⇒ 1.1 cosθ =1/2

⇒Cosθ = 1/2 

Now

cos2 θ = Cos²θ -1

= 2× 1/4 -1

= $-\frac{1}{2}$

Now, 

cosθ + cos2θ = 1/2 -1/2 =0

∴The Correct answer is Option a

Concept:

$\begin{array}{rl}|\overrightarrow{a}-\overrightarrow{b}{|}^2& =|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}\\ & =|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}-4\overrightarrow{a}\cdot \overline{b}\\ & =|\overrightarrow{a}+\overrightarrow{b}{|}^2-4\overrightarrow{a}\cdot \overrightarrow{b}\end{array}$​

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}|.|\overrightarrow{\mathrm{b}}|\cos \theta$

Vector a is perpendicular to b if $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0$

Calculation:

$\left|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\right|=\left|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}\right|$

Squaring both sides,

$\begin{array}{rl}|\overrightarrow{a}-\overrightarrow{b}{|}^2& =|\overrightarrow{a}+\overrightarrow{b}{|}^2\end{array}$

Now we have,

$$\begin{gathered} \begin{array}{rl}|\overrightarrow{a}-\overrightarrow{b}{|}^2& =|\overrightarrow{a}+\overrightarrow{b}{|}^2-4\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}+\overrightarrow{b}{|}^2\end{array} \\ \Rightarrow -4\overrightarrow{a}\cdot \overrightarrow{b}=0 \\ \Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=0 \end{gathered}$$

∴  Vector a is perpendicular to b

Hence, option (3) is correct.

Calculation

Given: $\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ 

$(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot \mathbf{a}-\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{a}-\mathbf{b}\cdot \mathbf{b}$

⇒ $(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot \mathbf{a}-\mathbf{b}\cdot \mathbf{b}=|\mathbf{a}{|}^2-|\mathbf{b}{|}^2$

$|\mathbf{a}{|}^2=(1{)}^2+(1{)}^2+(1{)}^2=1+1+1=3$

$|\mathbf{b}{|}^2=(1{)}^2+(2{)}^2+(3{)}^2=1+4+9=14$

$|\mathbf{a}{|}^2-|\mathbf{b}{|}^2=3-14=-11$

⇒ $(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=-11$.

Hence option 2 is correct.

Concept:

Dot Product: it is also called the inner product or scalar product

Let the two vectors are $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$

Dot Product of two vectors is given by:  $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$ = |a||b| cos θ

Where |$\overrightarrow{\mathrm{a}}$| = Magnitudes of vectors a, |$\overrightarrow{\mathrm{b}}$| = Magnitudes of vectors b and θ is the angle between a and b

Formulas of Dot Product:

 $\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{k}}=1$

$\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{k}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{k}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{j}}=0$

Calculation:

Given that,

(â + b̂ + ĉ) = 0    ----(1)

We know that,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

⇒ (â + b̂ + ĉ)² = â ⋅ â + b̂ ⋅ b̂ + ĉ ⋅ ĉ + 2(â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) 

From equation (1), we get

⇒ (1 + 1 + 1) + 2 (â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) = 0

∴ (â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) = - 3/2        

Concept:

If  $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are mutually perpendicular unit vectors $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0=$$\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}=0$

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{a}}=|\overrightarrow{\mathrm{a}}{|}^2$,  $\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{b}}{|}^2$

Calculations:

Consider, $(3\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}})\cdot (5\overrightarrow{\mathrm{a}}-6\overrightarrow{\mathrm{b}})$

Given  $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are mutually perpendicular unit vectors.

So,  $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0=$$\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}=0$

And   $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are unit vectors

So, $|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=1$

$(3\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}})\cdot (5\overrightarrow{\mathrm{a}}-6\overrightarrow{\mathrm{b}})$

$=3\overrightarrow{\mathrm{a}}\cdot 5\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}}\cdot 5\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{a}}\cdot 6\overrightarrow{\mathrm{b}}-2\overrightarrow{\mathrm{b}}\cdot 6\overrightarrow{\mathrm{b}}$

= $15|\overrightarrow{\mathrm{a}}{|}^2-12|\overrightarrow{\mathrm{b}}{|}^2$

= 15 - 12 

= 3

Concept:

Dot Product of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is defined as $\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right|\cos \theta$, where $\left|\overrightarrow{\mathrm{A}}\right|$ is the magnitude of vector $\overrightarrow{\mathrm{A}}$.

$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{A}}={\left|\overrightarrow{\mathrm{A}}\right|}^2$.

Calculation:

We are given that "sum of two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is a vector $\overrightarrow{\mathrm{c}}$".

⇒ $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}$

Taking dot product of both sides with themselves, the magnitudes will still be equal:

⇒ $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}).(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=\left(\overrightarrow{\mathrm{c}}\right).\left(\overrightarrow{\mathrm{c}}\right)$

⇒ ${\left|\overrightarrow{\mathrm{a}}\right|}^2+{\left|\overrightarrow{\mathrm{b}}\right|}^2+2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}={\left|\overrightarrow{\mathrm{c}}\right|}^2$

Since $\left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right|=\left|\overrightarrow{\mathrm{c}}\right|=2$, we get:

⇒ $2^2+2^2+2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=2^2$

⇒ $4+4+2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=4$

⇒ $2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=-4$

Now, ${|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|}^2=(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}).(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})$

= ${\left|\overrightarrow{\mathrm{a}}\right|}^2+{\left|\overrightarrow{\mathrm{b}}\right|}^2-2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$

= 4 + 4 - (-4)

= 12

⇒ $|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{12}=2\sqrt{3}$.

Concept:

If $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors then $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\mathrm{c}\mathrm{o}\mathrm{s}\theta$

Note: If vectors $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are perpendicular to each other then $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0$

Calculation:

Given: $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}and\overrightarrow{b}=\hat{i}-\hat{j}-\hat{k}$

Let θ be the angle between the vector $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$

⇒ $|\overrightarrow{a}|=\sqrt{3}and|\overrightarrow{b}|=\sqrt{3}$

We know that, 

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\mathrm{c}\mathrm{o}\mathrm{s}\theta$

⇒ $(\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}-\hat{j}-\hat{k})=\sqrt{3}\times \sqrt{3}\times cos\theta$

⇒ 1 = 3 cos θ 

⇒ $\mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{1}{3}$ 

⇒ $\theta =\mathrm{c}\mathrm{o}{\mathrm{s}}^{-1}\frac{1}{3}$

Hence, option 3 is correct.

CONCEPT:

• $\overrightarrow{a}\cdot \overrightarrow{a}=|\overrightarrow{a}{|}^2$
• $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$
• If $\overrightarrow{a}$ is a unit vector then $|\overrightarrow{a}|=1$

CALCULATION:

Given: $(\overrightarrow{x}-\overrightarrow{a})\cdot (\overrightarrow{x}+\overrightarrow{a})=12$ and $\overrightarrow{a}$ is a unit vector

⇒ $(\overrightarrow{x}-\overrightarrow{a})\cdot (\overrightarrow{x}+\overrightarrow{a})=|\overrightarrow{x}{|}^2+\overrightarrow{x}\cdot \overrightarrow{a}-\overrightarrow{a}\cdot \overrightarrow{x}-|\overrightarrow{a}{|}^2=12$

As we know that, $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$

⇒ $(\overrightarrow{x}-\overrightarrow{a})\cdot (\overrightarrow{x}+\overrightarrow{a})=|\overrightarrow{x}{|}^2-|\overrightarrow{a}{|}^2=12$

As we know that, if $\overrightarrow{a}$ is a unit vector then $|\overrightarrow{a}|=1$

⇒ $|\overrightarrow{x}{|}^2=13\Rightarrow |\overrightarrow{x}|=\sqrt{13}$

Hence, correct option is 2.

Concept:

• The cross product of vector to itself = 0
• The cross product of collinear vectors = 0
• The dot product of collinear vectors = Product of their Magnitudes
• $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}$
• For dot product $(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})\cdot \overrightarrow{\mathrm{R}}=(\overrightarrow{\mathrm{P}}\cdot \overrightarrow{\mathrm{R}})+(\overrightarrow{\mathrm{Q}}\cdot \overrightarrow{\mathrm{R}})$
• For cross product $(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})\times \overrightarrow{\mathrm{R}}=(\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{R}})+(\overrightarrow{\mathrm{Q}}\times \overrightarrow{\mathrm{R}})$ 
• The unit vector in the direction of a $\overrightarrow{\mathrm{P}}=\hat{\mathrm{P}}=\frac{\overrightarrow{\mathrm{P}}}{\left|\overrightarrow{\mathrm{P}}\right|}$ 
• A vector $\overrightarrow{\mathrm{X}}$ in direction of $\overrightarrow{\mathrm{P}}$ = (Magnitude of $\overrightarrow{\mathrm{X}}$) × $\hat{\mathrm{P}}$

Calculation:

Given:

$\overrightarrow{\mathrm{a}}\times (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})=(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\times \overrightarrow{\mathrm{b}}$

⇒ $\overrightarrow{\mathrm{a}}\times (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})-(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\times \overrightarrow{\mathrm{b}}=0$

⇒ $\overrightarrow{\mathrm{a}}\times (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})+\overrightarrow{\mathrm{b}}\times (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})=0$

⇒ $\text{boldsymbol}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\times (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})=0$

It means the $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and $2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}$ are collinear vectors.

∴ $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|\times \frac{2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}}{|2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}|}$ 

⇒ $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\sqrt{29}\times \frac{2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}}{\sqrt{29}}$ 

⇒ $\text{boldsymbol}\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}$

$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\cdot (-7\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})=(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\cdot (-7\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})$

⇒ $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\cdot (-7\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})=(2\times (-7)+3\times 2+4\times 3)$ 

⇒ $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\cdot (-7\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})=(-14+6+12)$ 

⇒ $\text{boldsymbol}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\cdot (-7\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}})=4$

Concept:

Dot Product: it is also called the inner product or scalar product

Let the two vectors are $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$

Dot Product of two vectors is given by:  $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$ = |a||b| cos θ

Where |$\overrightarrow{\mathrm{a}}$| = Magnitudes of vectors a, |$\overrightarrow{\mathrm{b}}$| = Magnitudes of vectors b and θ is the angle between a and b

Formulas of Dot Product:

 $\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{k}}=1$

$\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{k}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{k}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{j}}=0$

Calculation:

Let $\overrightarrow{a}=x̂i+ŷj+ẑk$

$\Rightarrow |a|=\sqrt{x^2+y^2+z^2}$

$\overrightarrow{a}.̂i=(x̂i+ŷj+ẑk).̂i=x$

Similarly, $\overrightarrow{a}.̂j=y,\overrightarrow{a}.̂k=z$

Therefore $[(\overrightarrow{a}.̂i)̂i+(\overrightarrow{a}.̂j)̂j+(\overrightarrow{a}.̂k)̂k]$

= xî + yĵ + zk̂ 

Hence, required value of

 $[(\overrightarrow{a}.\hat{i})\hat{i}+(\overrightarrow{a}.\hat{j})\hat{j}+(\overrightarrow{a}.\hat{k})\hat{k}].\overrightarrow{a}$

= (xî + yĵ + zk̂ ).(xî + yĵ + zk̂ )

= x² + y² + z² = |a|²

CONCEPT:

• Projection of a vector $\overrightarrow{a}$ on other vector $\overrightarrow{b}$ is given by: $\overrightarrow{a}\cdot \hat{b}=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{b}|}$

CALCULATION:

Given: $\overrightarrow{a}=2\hat{i}+3\hat{j}+2\hat{k}$ and $\overrightarrow{b}=\overrightarrow{i}+2\overrightarrow{j}+\hat{k}$

Here, we have to find the projection of a vector $\overrightarrow{a}$ on other vector $\overrightarrow{b}$ is given by: $\overrightarrow{a}\cdot \hat{b}=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{b}|}$

⇒ $\overrightarrow{a}\cdot \overrightarrow{b}=2+6+2=10and|\overrightarrow{b}|=\sqrt{6}$

⇒ $\overrightarrow{a}\cdot \hat{b}=\frac{10}{\sqrt{6}}=\frac{5\sqrt{6}}{3}$

Hence, option 3 is correct.

CONCEPT:

The scalar product of two vectors $\overrightarrow{a}and\overrightarrow{b}$is given by $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \theta$

If the vectors $\overrightarrow{a}and\overrightarrow{b}$ are perpendicular then $\overrightarrow{a}\cdot \overrightarrow{b}=0$

CALCULATION:

Given: $\overrightarrow{a}and\overrightarrow{b}$ are two vectors such that $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|=4$

⇒ $|\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{a}-\overrightarrow{b}{|}^2$

⇒ $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b})=(\overrightarrow{a}-\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})$

⇒ $|\overrightarrow{a}{|}^2+\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{a}+|\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2-\overrightarrow{a}\cdot \overrightarrow{b}-\overrightarrow{b}\cdot \overrightarrow{a}+|\overrightarrow{b}{|}^2$

∵ $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$

⇒ $|\overrightarrow{a}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2$

⇒ $4\overrightarrow{a}\cdot \overrightarrow{b}=0$

⇒ $\overrightarrow{a}\cdot \overrightarrow{b}=0$

So, $\overrightarrow{a}$ must be perpendicular to $\overrightarrow{b}.$

Hence, correct option is 3.