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

Concept:

a.b = |a||b|cosθ

Calculation: 

Here,  \(\rm \vec a = \left(\vec i + 2\vec j - 3\vec k\right)\) and \(\vec b = \left(3\vec i - \vec j + 2\vec k\right)\) 

\(\rm \left(\vec a + \vec b\right) = (\vec i + 2\vec j - 3\vec k )+(\vec 3i -\vec j +2\vec k)\\=\vec 4i + \vec j - \vec k\\ |\rm (\vec a + \vec b)|=\sqrt{4^2+1^2+(-1)^2}\\ =\sqrt{18}\\=3\sqrt2\)

\(\rm \left(\vec a - \vec b\right) = (\vec i + 2\vec j - 3\vec k )-(\vec 3i -\vec j +2\vec k)\\ =-\vec 2i + \vec3 j - \vec 5 k\\ |\rm (\vec a - \vec b)|=\sqrt{(-2)^2+3^2+(-5)^2}\\ =\sqrt{38}\\\)

Now, 

\(\rm \left(\vec a +\vec b\right) .\left(\vec a - \vec b\right) =(\vec 4i + \vec j - \vec k) \cdot (-\vec 2i + \vec3 j - \vec 5 k)= -8+3+5\\ |\left(\vec a + \vec b\right)||\left(\vec a - \vec b\right)|cosθ =0\\ \Rightarrow θ = \frac \pi 2\)

Concept:

\(\rm \vec a\times (\vec b\times \vec c)=(\vec a.\vec c)\vec b-(\vec a.\vec b)\vec c\)

Explanation:

\(\rm \vec a\times (2\vec b\times \vec c)=\vec b\)

⇒ \(\rm (\vec a.\vec c)2\vec b-(2\vec a.\vec b)\vec c=\vec b\)

Comparing both sides

\(\rm 2(\vec a.\vec c)=1; \vec a.\vec b=0\)

So, \(\rm \vec a\) and \(\rm \vec b\) are perpendicular.

i.e., angle between \(\rm \vec a\) and \(\rm \vec b\) is 90°

Option (1) is true.

Explanation:

Given:

\(\rm \vec a=̂ i+̂ j\ and \ \vec b=̂ j+̂ k,\)

Also, \(\vec a, \vec b, \vec c\) has same length

⇒ \(|\vec a| = |\vec b| = |\vec c|\) = √2 

Let θ be the angle between the vectors.

⇒ Cosθ = \(\frac{\vec a. \vec b}{|\vec a||\vec b|} = \frac{0+1+0}{\sqrt2} =\frac{1}{\sqrt2}\)

(I) Let \(\vec c =\hat i +\hat k\)

\(|\vec c| = \sqrt2\)

Cosθ = \(\frac{\vec a. \vec c}{|\vec a||\vec c|} = \frac{0+1+0}{\sqrt2} =\frac{1}{\sqrt2}\)

All the conditions are satisfied, so it can be vector \(\vec c\)

(II) Let \(\vec c =\rm \frac{-\hat i+4\hat j-\hat k}{3}\)

⇒ \(|\vec c| = \frac{1}{3}\sqrt18 = \sqrt2\)

Cosθ = \(\frac{\vec a. \vec c}{|\vec a||\vec c|} \)

= \(\frac{\frac{-1}{3}\frac{4}{3}}{\sqrt2 \sqrt2} = \frac{1}{2}\)

All the conditions are satisfied, so it can be vector \(\vec c\)

∴ Option (c) is correct.

Explanation:

Given:

\(⃗ a+2⃗ b\) and \(\rm 5⃗ a-4⃗ b\) are perpendicular vectors.

⇒ \((⃗ a+2⃗ b).\)\((\rm 5⃗ a-4⃗ b)\) = 0

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

⇒ \(5× 1 + 6⃗ a.⃗ b-8 =0\)

Now \(\vec a, \vec b\) are unit vectors

⇒ \(6 |\vec a||\vec b| Cosθ = 3\)

⇒ 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

Explanation:

Let P = \(\rm |\vec a\times \vec b|+√3|\vec a.\vec b|\)

= \(\rm |\vec a|.|\vec b||sinθ |+ √3|\vec a||\vec b||cosθ|\)

= \(\rm |\vec a||\vec b|[|sinθ |+ √3|cosθ|]\)

⇒ p will be max if (sin 3 + √ 3cos θ) is maximum.

Now, for its maxima,

\(\frac{d}{dθ } (sinθ +√3cosθ) =0\)

⇒ Cosθ -√3 sinθ =0

⇒ \(tanθ =\frac{1}{\sqrt3}\)

⇒ θ =30°

∴ Option (b) is correct

Concept:

Dot product of two vectors is defined as:

\({\rm{\vec A}}{\rm{.\vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times cos}}\;{\rm{\theta }}\)

Cross/Vector product of two vectors is defined as:

\({\rm{\vec A \times \vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times sin}}\;{\rm{\theta }} \times \rm \hat{n}\)

where θ is the angle between \({\rm{\vec A}}\;{\rm{and}}\;{\rm{\vec B}}\)

Calculation:

To Find: Value of \(\rm \vec{a} \times \vec{a}\)

Here angle between them is 0°

\({\rm{\vec a \times \vec a = }}\left| {\rm{a}} \right|{\rm{ \times }}\left| {\rm{a}} \right|{\rm{ \times sin}}\;{\rm{0 }} \times \rm \hat{n}=0\)

Concept:

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

Calculation:

Given: \(\vec a = 2\hat i - 6\hat j - 3\hat k\) and \(\vec b = 4\hat i + 3\hat j - \hat k\)

\(\left| {\vec a} \right| = 7,\;\left| {\vec b} \right| = \sqrt {26} \;and\;\vec a \cdot \;\vec b = - 7\)

\(\Rightarrow \;\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}} = \frac{{ - \;7}}{{7 \times \sqrt {26} }} = - \frac{1}{{\sqrt {26} }}\)

\( \Rightarrow \;{\sin ^2}\theta = 1 - {\cos ^2}\theta = 1 - \frac{1}{{26}} = \frac{{25}}{{26}}\)

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

\(\vec a\parallel \vec b \Leftrightarrow \vec a \times \vec b = \vec 0\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 1&\lambda &3\\ 3&2&9 \end{array}} \right| = 0\)

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

⇒ 9λ – 6 = 0 and 2 - 3λ = 0

Given: 

\(\rm\vec u = \hat i \times ( \vec a \times \hat i) + \hat j \times ( \vec a \times \hat j) + \hat k \times ( \vec a \times \hat k)\)

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 

\(\rm\vec u = \hat i \times ( \vec a \times \hat i) + \hat j \times ( \vec a \times \hat j) + \hat k \times ( \vec a \times \hat k)\)

\(\vec u \) = î  × (mî + nĵ +lk̂  × î) + ĵ ×  (mî + nĵ +lk̂  × ĵ) + k̂ × (mî + nĵ +lk̂  × k̂)

\(\vec u \) =  î  × (-nk̂ + lĵ) + ĵ × (mk̂ -lî  ) + k̂ × (-mĵ + nî) 

\(\vec u \) = nĵ  + lk̂ + mî +  lk̂ + mî + nĵ 

\(\vec u \) = 2(mî + nĵ +lk̂ ) = 2\(\vec a \)

∴ 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:

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

\(\Rightarrow \left|\begin{array}{lll} \rm a_{1} & \mathrm{b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{b}_{3} & \mathrm{c}_{3} \end{array}\right|=0 \)

​Calculation:

Here, \(\rm \hat i-\hat j+\hat k, 2\hat i+\hat j-\hat k, \hat iλ-\hat j+\hat k λ \) are coplanar

\(\begin{array}{l} \Rightarrow \left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ λ & -1 & λ \end{array}\right|=0\end{array}\)

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 \(\rm \vec a\;and\;\vec b\) are two vectors

\(\rm \vec a \cdot \;\vec b = \left| {\vec a} \right| × \left| {\vec b} \right| × \cos θ \)

 \(\rm \vec a × \;\vec b = \;\;\left| {\vec a} \right| × \left| {\vec b} \right| × \sin θ × \;\hat n,\;where\;\hat n\) is the unit vector perpendicular to both \(\rm \vec a\;and\;\vec b\)

Calculation:

Given: \(\rm \left| {\vec a} \right| = 3,\;\left| {\vec b} \right| = 4\;and \;\rm \vec a \cdot \;\vec b = 6\)

As we know, \(\rm \vec a \cdot \;\vec b = \left| {\vec a} \right| × \left| {\vec b} \right| × \cos θ \)

⇒ 6 = 3 × 4 × cos θ 

⇒ cos θ = \(\frac {6} {12} = \frac 1 2\)

∴ θ = 60° 

As we know that, If \(\rm \vec a\;and\;\vec b\) are two vectors, then

\(\rm \vec a × \;\vec b = \;\;\left| {\vec a} \right| × \left| {\vec b} \right| × \sin θ × \;\hat n\)

\(\rm \left| {\vec a × \;\vec b} \right| = \left| {\vec a} \right| × \left| {\vec b} \right| × \left| {\sin θ } \right| × \left| {\hat n} \right| = \;\left| {\vec a} \right| × \left| {\vec b} \right| × \sin θ \)      (∵ Magnitude of a unit vector is one)

\(\rm \left| {\vec a × \;\vec b} \right|\) = 3 × 4 × sin 60° 

\(\rm ∴ \rm \left| {\vec a × \;\vec b} \right| = 3 × 4 × \frac{\sqrt 3}{2} = 6\sqrt 3\)

Concept:

\(\rm \left(\vec a + \vec b\right) \cdot \left(\vec a - \vec b\right) = \left|\vec a\right|^2-\left|\vec b\right|^2\)

If \(\rm \vec u\) is a unit vector, then \(\rm \left|\vec u\right|=1\).

Calculation:

It is given that \((\rm \vec x + \rm 2\vec a) \cdot (\rm \vec x - \rm 2\vec a) = 12\).

⇒ \(\rm \left|\vec x\right|^2 - 4\left|\vec a\right|^2 = 12\)

Since \(\rm \vec a\) is a unit vector, we get:

⇒ \(\rm \left|\vec x\right|^2 - 4= 12\)

⇒ \(\rm \left|\vec x\right|^2 =16\)

⇒ \(|\rm \vec x |\) = 4

Concept:

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

Let \(\vec a = x\;\vec i + y\;\vec j + z\;\vec k\)

Magnitude of vector of a = \(\left| {\vec a} \right| = \sqrt {{x^2} + {y^2} + {z^2}} \)

Unit vector = \(\hat a = \frac{{\vec a}}{{\left| {\vec a} \right|}}\)

• Let \(\vec a\) and \(\vec b\) be the two vectors, then the vector \(\vec c\) perpendicular to both  and

\(\vec a = {a_1}\vec i + {b_1}\vec j + {c_1}\vec k\) and \(\vec b = {a_2}\vec i + {b_2}\vec j + {c_2}\vec k\)

\(\therefore \vec c = \vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k\;}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|\)

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) 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 \(\vec a = 2{\rm{\hat i}} - {\rm{\hat j}} + {\rm{\hat k}}\) and \([\vec b = 3{\rm{\hat i}} - 4{\rm{\hat j}} - {\rm{\hat k}}\) and vector \(\vec c\) perpendicular to both \(\vec a\) and \(\vec b\)

\(\therefore \vec c = \vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k\;}\\ 2&{ - 1}&1\\ 3&{ - 4}&{ - 1} \end{array}} \right|\)

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

\(= 5\vec i + 5\vec j - 5\vec k\)

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

Concept:  

\(\rm \overrightarrow{a} \times \overrightarrow{a} = 0\)

\(\rm \overrightarrow{a} \times \overrightarrow{b} = - \overrightarrow{b} \times \overrightarrow{a} \)

Calculation:

Given

\(\rm = \left( {2\vec a - 3\vec b} \right) \times \left( {2\vec a + 3\vec b} \right)\)

\(\rm = 2\overrightarrow{a} \times 2\overrightarrow{a} + 2\overrightarrow{a} \times 3\overrightarrow{b} - 3\overrightarrow{b} \times 2\overrightarrow{a} - 3\overrightarrow{b} \times 3\overrightarrow{b}\)

\(\rm = 0 + 2\overrightarrow{a} \times 3\overrightarrow{b} - 3\overrightarrow{b} \times 2\overrightarrow{a} - 0\)

\(\rm = 6 \: (\overrightarrow{a} \times \overrightarrow{b}) + 6\: (\overrightarrow{a} \times \overrightarrow{b} )\)

\(\rm = 12\: (\overrightarrow{a} \times \overrightarrow{b}) \)

Additional Information

Properties of Scalar Product

\(\rm \overrightarrow{a}.\overrightarrow{a} = \left |\overrightarrow{a} \right |^{2}\)

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

\(\rm \overrightarrow{a}.\overrightarrow{0} = 0\)

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

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

Properties of Vector Product

Concept:

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

• Let the two vectors be \(\vec a\;{\rm{and\;}}\vec b\) then dot Product of two vector: \(\vec a.\;\vec b = \;\left| {\bf{a}} \right|\left| {\bf{b}} \right|\;{\bf{cos}}\;{\bf{\theta }}\)  Where, |\(\vec a\)| = Magnitude of vectors a and |\(\vec b\)| = Magnitude of vectors b and θ is angle between a and b 
• \(\vec i.\vec i = \vec j.\vec j = \vec k.\vec k = 1{\rm{\;and\;}}\overrightarrow {\;i} .\vec j = \vec j.\vec i = \vec i.\vec k = \vec k.\vec i = \vec j.\vec k = \vec k.\vec j = 0\)

Calculation:

Let \(\vec a = \vec i,\vec b = \vec j,\vec c = \vec k\)

\(\therefore \left( {{\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}} \right).{\rm{\vec a}} = \left| {{\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}} \right|\left| {{\rm{\vec a}}} \right|\cos \theta \)

\(\Rightarrow \left( {{\rm{\vec i}} + \vec j + {\rm{\vec k}}} \right).{\rm{\vec i}} = \left| {{\rm{\vec i}} + \vec j + {\rm{\vec k}}} \right|\left| {{\rm{\vec i}}} \right|\cos \theta \)

⇒ 1 + 0 + 0 = √3 × 1 × cos θ

⇒ cos θ = 1/√3