Scalar and Vector Product MCQ Quiz - Objective Question with Answer for Scalar and Vector Product - Download Free PDF

Calculation:

Given,

The position vectors of points A, B, and C are $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ respectively, and $\overrightarrow{c}={\cos }^2\theta \overrightarrow{a}+{\sin }^2\theta \overrightarrow{b}$.

The expression to evaluate is: $(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{b}\times \overrightarrow{c})+(\overrightarrow{c}\times \overrightarrow{a})$.

First, substitute $\overrightarrow{c}$ into the equation:

$(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{b}\times ({\cos }^2\theta \overrightarrow{a}+{\sin }^2\theta \overrightarrow{b}))+({\cos }^2\theta \overrightarrow{a}+{\sin }^2\theta \overrightarrow{b})\times \overrightarrow{a}$.

Using the distributive property of the cross product:

$(\overrightarrow{a}\times \overrightarrow{b})+[(\overrightarrow{b}\times {\cos }^2\theta \overrightarrow{a})+(\overrightarrow{b}\times {\sin }^2\theta \overrightarrow{b})]+[({\cos }^2\theta \overrightarrow{a}\times \overrightarrow{a})+({\sin }^2\theta \overrightarrow{b}\times \overrightarrow{a})]$.

Since $\overrightarrow{b}\times \overrightarrow{b}=0$ and $\overrightarrow{a}\times \overrightarrow{a}=0$, we are left with:

$(\overrightarrow{a}\times \overrightarrow{b})+{\cos }^2\theta (\overrightarrow{b}\times \overrightarrow{a})+{\sin }^2\theta (-\overrightarrow{a}\times \overrightarrow{b})$.

Substitute $\overrightarrow{b}\times \overrightarrow{a}=-(\overrightarrow{a}\times \overrightarrow{b})$ into the expression:

$(\overrightarrow{a}\times \overrightarrow{b})+{\cos }^2\theta (-\overrightarrow{a}\times \overrightarrow{b})+{\sin }^2\theta (-\overrightarrow{a}\times \overrightarrow{b})$.

Factor out $\overrightarrow{a}\times \overrightarrow{b}$:

$\overrightarrow{a}\times \overrightarrow{b}[1-{\cos }^2\theta -{\sin }^2\theta ]$.

Since ${\cos }^2\theta +{\sin }^2\theta =1$, the expression becomes:

$\overrightarrow{a}\times \overrightarrow{b}[1-1]=0$.

∴ The final result is $\overrightarrow{0}$.

Hence, the correct answer is option 1. 

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.

Concept:

a.b = |a||b|cosθ

Calculation: 

Here,  $\overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{i}}+2\overrightarrow{\mathrm{j}}-3\overrightarrow{\mathrm{k}})$ and $\overrightarrow{b}=(3\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k})$ 

$$\begin{gathered} (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=(\overrightarrow{\mathrm{i}}+2\overrightarrow{\mathrm{j}}-3\overrightarrow{\mathrm{k}})+(\overrightarrow{3}\mathrm{i}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}}) \\ =\overrightarrow{4}\mathrm{i}+\overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}} \\ |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})|=\sqrt{4^2+1^2+(-1{)}^2} \\ =\sqrt{18} \\ =3\sqrt{2} \end{gathered}$$

$$\begin{gathered} (\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=(\overrightarrow{\mathrm{i}}+2\overrightarrow{\mathrm{j}}-3\overrightarrow{\mathrm{k}})-(\overrightarrow{3}\mathrm{i}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}}) \\ =-\overrightarrow{2}\mathrm{i}+\overrightarrow{3}\mathrm{j}-\overrightarrow{5}\mathrm{k} \\ |(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})|=\sqrt{(-2{)}^2+3^2+(-5{)}^2} \\ =\sqrt{38} \end{gathered}$$

Now, 

$$\begin{gathered} (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}).(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=(\overrightarrow{4}\mathrm{i}+\overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}})\cdot (-\overrightarrow{2}\mathrm{i}+\overrightarrow{3}\mathrm{j}-\overrightarrow{5}\mathrm{k})=-8+3+5 \\ |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})||(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})|\mathrm{c}\mathrm{o}\mathrm{s}\theta =0 \\ \Rightarrow \theta =\frac{\pi }{2} \end{gathered}$$

Concept:

$\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}$

Explanation:

$\overrightarrow{\mathrm{a}}\times (2\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}$

⇒ $(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}})2\overrightarrow{\mathrm{b}}-(2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$

Comparing both sides

$2(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}})=1;\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0$

So, $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are perpendicular.

i.e., angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is 90°

Option (1) is true.

Explanation:

Given:

$\overrightarrow{\mathrm{a}}=̂\mathrm{i}+̂\mathrm{j}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}=̂\mathrm{j}+̂\mathrm{k},$

Also, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ has same length

⇒ $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|$ = √2 

Let θ be the angle between the vectors.

⇒ Cosθ = $\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{0+1+0}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

(I) Let $\overrightarrow{c}=\hat{i}+\hat{k}$

$|\overrightarrow{c}|=\sqrt{2}$

Cosθ = $\frac{\overrightarrow{a}.\overrightarrow{c}}{|\overrightarrow{a}||\overrightarrow{c}|}=\frac{0+1+0}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

All the conditions are satisfied, so it can be vector $\overrightarrow{c}$

(II) Let $\overrightarrow{c}=\frac{-\hat{\mathrm{i}}+4\hat{\mathrm{j}}-\hat{\mathrm{k}}}{3}$

⇒ $|\overrightarrow{c}|=\frac{1}{3}\sqrt{1}8=\sqrt{2}$

Cosθ = $\frac{\overrightarrow{a}.\overrightarrow{c}}{|\overrightarrow{a}||\overrightarrow{c}|}$

= $\frac{\frac{-1}{3}\frac{4}{3}}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

All the conditions are satisfied, so it can be vector $\overrightarrow{c}$

∴ Option (c) is correct.

Concept:

Dot product of two vectors is defined as:

$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=\left|\mathrm{A}\right|\times \left|\mathrm{B}\right|\times \mathrm{c}\mathrm{o}\mathrm{s}\theta$

Cross/Vector product of two vectors is defined as:

$\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\left|\mathrm{A}\right|\times \left|\mathrm{B}\right|\times \mathrm{s}\mathrm{i}\mathrm{n}\theta \times \hat{\mathrm{n}}$

where θ is the angle between $\overrightarrow{\mathrm{A}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{B}}$

Calculation:

To Find: Value of $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{a}}$

Here angle between them is 0°

$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{a}}=\left|\mathrm{a}\right|\times \left|\mathrm{a}\right|\times \mathrm{s}\mathrm{i}\mathrm{n}0\times \hat{\mathrm{n}}=0$

Concept:

If $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}and\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ then $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \theta$

Calculation:

Given: $\overrightarrow{a}=2\hat{i}-6\hat{j}-3\hat{k}$ and $\overrightarrow{b}=4\hat{i}+3\hat{j}-\hat{k}$

$\left|\overrightarrow{a}\right|=7,\left|\overrightarrow{b}\right|=\sqrt{26}and\overrightarrow{a}\cdot \overrightarrow{b}=-7$

$\Rightarrow \cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|}=\frac{-7}{7\times \sqrt{26}}=-\frac{1}{\sqrt{26}}$

$\Rightarrow {\sin }^2\theta =1-{\cos }^2\theta =1-\frac{1}{26}=\frac{25}{26}$

The given two vectors are in parallel, therefore the vector product of these two vectors will be zero.

$\overrightarrow{a}\parallel \overrightarrow{b}\iff \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}$

$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& \lambda & 3\\ 3& 2& 9\end{array}\right|=0$

$=\left(9\lambda -6\right)\hat{i}-\left(9-9\right)\hat{j}+\left(2-3\lambda \right)\hat{k}=\overrightarrow{0}$

⇒ 9λ – 6 = 0 and 2 - 3λ = 0

Given: 

$\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{i}})+\hat{\mathrm{j}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{j}})+\hat{\mathrm{k}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{k}})$

Concept:

î × î  = ĵ × ĵ = k̂ × k̂  = 0 

î × ĵ = k̂ , ĵ × k̂ = î , k̂ × î = ĵ 

Calculation:

Let a = mî + nĵ +lk̂ 

According to the Question 

$\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{i}})+\hat{\mathrm{j}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{j}})+\hat{\mathrm{k}}\times (\overrightarrow{\mathrm{a}}\times \hat{\mathrm{k}})$

$\overrightarrow{u}$ = î  × (mî + nĵ +lk̂  × î) + ĵ ×  (mî + nĵ +lk̂  × ĵ) + k̂ × (mî + nĵ +lk̂  × k̂)

$\overrightarrow{u}$ =  î  × (-nk̂ + lĵ) + ĵ × (mk̂ -lî  ) + k̂ × (-mĵ + nî) 

$\overrightarrow{u}$ = nĵ  + lk̂ + mî +  lk̂ + mî + nĵ 

$\overrightarrow{u}$ = 2(mî + nĵ +lk̂ ) = 2$\overrightarrow{a}$

∴ The correct option is 3

Concept:

$\text{ Let }\overrightarrow{\mathrm{a}}={\mathrm{a}}_1\overrightarrow{\mathrm{i}}+{\mathrm{b}}_1\overrightarrow{\mathrm{j}}+{\mathrm{c}}_1\overrightarrow{\mathrm{k}},\overrightarrow{\mathrm{b}}={\mathrm{a}}_2\overrightarrow{\mathrm{i}}+{\mathrm{b}}_2\overrightarrow{\mathrm{j}}+{\mathrm{c}}_2\overrightarrow{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{c}}={\mathrm{a}}_3\overrightarrow{\mathrm{i}}+{\mathrm{b}}_3\overrightarrow{\mathrm{j}}+{\mathrm{c}}_3\overrightarrow{\mathrm{k}}\text{ be the three vectors }$

Condition for coplanarity:

$\overrightarrow{\mathbf{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=0$

$\Rightarrow \left|\begin{array}{lll}{\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\\ {\mathrm{a}}_3& {\mathrm{b}}_3& {\mathrm{c}}_3\end{array}\right|=0$

​Calculation:

Here, $\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}},2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}},\hat{\mathrm{i}}\lambda -\hat{\mathrm{j}}+\hat{\mathrm{k}}\lambda$ are coplanar

$\begin{array}{l}\Rightarrow \left|\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ \lambda & -1& \lambda \end{array}\right|=0\end{array}$

1(λ - 1) + 1(2λ + λ) + 1(-2 - λ) = 0

λ - 1 + 2λ + λ + -2 - λ = 0

3λ - 3 = 0

λ = 1

Hence, option (4) is correct.

Concept:

Let $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors

$\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\times \left|\overrightarrow{\mathrm{b}}\right|\times \cos \theta$

 $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\times \left|\overrightarrow{\mathrm{b}}\right|\times \sin \theta \times \hat{\mathrm{n}},\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\hat{\mathrm{n}}$ is the unit vector perpendicular to both $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$

Calculation:

Given: $\left|\overrightarrow{\mathrm{a}}\right|=3,\left|\overrightarrow{\mathrm{b}}\right|=4\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=6$

As we know, $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\times \left|\overrightarrow{\mathrm{b}}\right|\times \cos \theta$

⇒ 6 = 3 × 4 × cos θ 

⇒ cos θ = $\frac{6}{12}=\frac{1}{2}$

∴ θ = 60° 

As we know that, If $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors, then

$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\times \left|\overrightarrow{\mathrm{b}}\right|\times \sin \theta \times \hat{\mathrm{n}}$

$\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|=\left|\overrightarrow{\mathrm{a}}\right|\times \left|\overrightarrow{\mathrm{b}}\right|\times \left|\sin \theta \right|\times \left|\hat{\mathrm{n}}\right|=\left|\overrightarrow{\mathrm{a}}\right|\times \left|\overrightarrow{\mathrm{b}}\right|\times \sin \theta$      (∵ Magnitude of a unit vector is one)

$\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|$ = 3 × 4 × sin 60° 

$\therefore \left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|=3\times 4\times \frac{\sqrt{3}}{2}=6\sqrt{3}$

Concept:

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

If $\overrightarrow{\mathrm{u}}$ is a unit vector, then $\left|\overrightarrow{\mathrm{u}}\right|=1$.

Calculation:

It is given that $(\overrightarrow{\mathrm{x}}+2\overrightarrow{\mathrm{a}})\cdot (\overrightarrow{\mathrm{x}}-2\overrightarrow{\mathrm{a}})=12$.

⇒ ${\left|\overrightarrow{\mathrm{x}}\right|}^2-4{\left|\overrightarrow{\mathrm{a}}\right|}^2=12$

Since $\overrightarrow{\mathrm{a}}$ is a unit vector, we get:

⇒ ${\left|\overrightarrow{\mathrm{x}}\right|}^2-4=12$

⇒ ${\left|\overrightarrow{\mathrm{x}}\right|}^2=16$

⇒ $|\overrightarrow{\mathrm{x}}|$ = 4

Concept:

• Unit vector: a vector which has a magnitude of one.

Let $\overrightarrow{a}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

Magnitude of vector of a = $\left|\overrightarrow{a}\right|=\sqrt{x^2+y^2+z^2}$

Unit vector = $\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

• Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be the two vectors, then the vector $\overrightarrow{c}$ perpendicular to both  and

$\overrightarrow{a}=a_1\overrightarrow{i}+b_1\overrightarrow{j}+c_1\overrightarrow{k}$ and $\overrightarrow{b}=a_2\overrightarrow{i}+b_2\overrightarrow{j}+c_2\overrightarrow{k}$

$\therefore \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$

• If $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

Let vector $\overrightarrow{a}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $[\overrightarrow{b}=3\hat{\mathrm{i}}-4\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and vector $\overrightarrow{c}$ perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$

$\therefore \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 2& -1& 1\\ 3& -4& -1\end{array}\right|$

$=\overrightarrow{i}\left(1+4\right)-\overrightarrow{j}\left(-2-3\right)+\overrightarrow{k}\left(-8+3\right)$

$=5\overrightarrow{i}+5\overrightarrow{j}-5\overrightarrow{k}$

Unit vector = $\hat{c}=\frac{\overrightarrow{c}}{\left|c\right|}=\frac{5\overrightarrow{i}+5\overrightarrow{j}-5\overrightarrow{k}}{\sqrt{5^2+5^2+{\left(-5\right)}^2}}=\frac{5\overrightarrow{i}+5\overrightarrow{j}-5\overrightarrow{k}}{5\sqrt{3}}=\frac{\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}}{\sqrt{3}}$

Concept:  

$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{a}}=0$

$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}$

Calculation:

Given

$=\left(2\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}\right)\times \left(2\overrightarrow{\mathrm{a}}+3\overrightarrow{\mathrm{b}}\right)$

$=2\overrightarrow{\mathrm{a}}\times 2\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{a}}\times 3\overrightarrow{\mathrm{b}}-3\overrightarrow{\mathrm{b}}\times 2\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}\times 3\overrightarrow{\mathrm{b}}$

$=0+2\overrightarrow{\mathrm{a}}\times 3\overrightarrow{\mathrm{b}}-3\overrightarrow{\mathrm{b}}\times 2\overrightarrow{\mathrm{a}}-0$

$=6(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})+6(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$

$=12(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$

Additional Information

Properties of Scalar Product

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

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{a}}$ (Scalar product is commutative)

$\overrightarrow{\mathrm{a}}.\overrightarrow{0}=0$

$\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}$ (Distributive of scalar product over addition)

In terms of orthogonal coordinates for mutually perpendicular vectors, it is seen that $\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}.\overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{k}}.\overrightarrow{\mathrm{k}}=1$

Properties of Vector Product

Concept:

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

• Let the two vectors be $\overrightarrow{a}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{b}$ then dot Product of two vector: $\overrightarrow{a}.\overrightarrow{b}=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\mathbf{c}\mathbf{o}\mathbf{s}\theta$  Where, |$\overrightarrow{a}$| = Magnitude of vectors a and |$\overrightarrow{b}$| = Magnitude of vectors b and θ is angle between a and b 
• $\overrightarrow{i}.\overrightarrow{i}=\overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{i}.\overrightarrow{j}=\overrightarrow{j}.\overrightarrow{i}=\overrightarrow{i}.\overrightarrow{k}=\overrightarrow{k}.\overrightarrow{i}=\overrightarrow{j}.\overrightarrow{k}=\overrightarrow{k}.\overrightarrow{j}=0$