Vector or Cross Product MCQ Quiz - Objective Question with Answer for Vector or Cross Product - Download Free PDF

Explanation:

Given:

⇒ (2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) = 0

⇒ \(\begin{bmatrix} \hat{i} &\hat{j}&\hat{k}\\\ 2&6&27\\\ 1&α&β\end{bmatrix}\ \) = 0

⇒ \(\hat{i} (6β -27α) - \hat{j}(2β-27)+\hat{k}(2α-6) =0\)

Comparing both sides, we get

6β  – 27α  = 0

⇒ 2β  = 9α 

2β - 27 = 0

⇒ β = 27/2

Also

⇒ 2α – 6 = 0 

α =3

Now, 

3α  + 2β = \(3\times 3 + 2 \times \frac{27}{2}\) =36

∴ Option (1) is correct

Concept:

Vector Perpendicular to Both Vectors:

• We are given two vectors: a + b and a - b, and we need to find a vector perpendicular to both of them.
• The method to find a vector perpendicular to both given vectors is by taking their cross product. The result will be a vector perpendicular to both.
• Once the cross product is found, we normalize the result (i.e., divide it by its magnitude) to find the unit vector perpendicular to both vectors.

Calculation:

Given vectors:

a = i + j + k

b = i + 2j + 3k

The vectors we need to take the cross product of are:

a + b = (i + j + k) + (i + 2j + 3k) = 2i + 3j + 4k

a - b = (i + j + k) - (i + 2j + 3k) = 0i - j - 2k

Now, we compute the cross product of (a + b) and (a - b):

(a + b) × (a - b) =

\( \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \\ \end{array}\right| \)

Expanding the determinant:

Result of cross product:

(a + b) × (a - b) = -2i + 4j - 2k

Now, let's find the magnitude of the resulting vector:

Magnitude = √((-2)² + 4² + (-2)²) = √(4 + 16 + 4) = √24 = 2√6

To find the unit vector, we divide the result by its magnitude:

Unit vector = (-2i + 4j - 2k) / 2√6

The unit vector is:

Unit vector = (-1/√6)i + (2/√6)j - (1/√6)k

Hence Option 4 is the correct answer. 

Formula Used:

Dot product of two vectors: \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}\), where \(\theta\) is the angle between the vectors.

Magnitude of a vector: \(|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\)

Sine of double angle: \(\sin{2\theta} = 2\sin{\theta}\cos{\theta}\)

Relation between sine and cosine: \(\sin^2{\theta} + \cos^2{\theta} = 1\)

Calculation:

Given:

Vector 1: \(\vec{a} = 4\hat{i} - \hat{j} + 2\hat{k}\)

Vector 2: \(\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}\)

⇒ \(\vec{a} \cdot \vec{b} = (4)(1) + (-1)(3) + (2)(-2) = 4 - 3 - 4 = -3\)

⇒ \(|\vec{a}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{16 + 1 + 4} = \sqrt{21}\)

⇒ \(|\vec{b}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}\)

⇒ \(\cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-3}{\sqrt{21} \sqrt{14}} = \frac{-3}{\sqrt{3 \cdot 7 \cdot 2 \cdot 7}} = \frac{-3}{7\sqrt{6}}\)

⇒ \(\sin^2{\theta} = 1 - \cos^2{\theta} = 1 - \frac{9}{49 \cdot 6} = 1 - \frac{3}{98} = \frac{95}{98}\)

⇒ \(\sin{\theta} = \sqrt{\frac{95}{98}} = \frac{\sqrt{95}}{7\sqrt{2}}\)

⇒ \(\sin{2\theta} = 2 \sin{\theta} \cos{\theta} = 2 \cdot \frac{\sqrt{95}}{7\sqrt{2}} \cdot \frac{-3}{7\sqrt{6}} = \frac{-6\sqrt{95}}{49\sqrt{12}} = \frac{-6\sqrt{95}}{49 \cdot 2\sqrt{3}} = \frac{-3\sqrt{95}}{49\sqrt{3}}\)

⇒ \(\sin{2\theta} = \frac{-3\sqrt{95} \cdot \sqrt{3}}{49 \cdot 3} = \frac{-\sqrt{285}}{49}\)

∴ The value of \(\sin{2\theta}\) is \(\frac{-\sqrt{285}}{49}\)

Hence option 3 is correct

Formula Used:

Unit vector along the internal bisector of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:

\(\hat{c} = \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}\)

Resultant vector \(\vec{R} = |\vec{R}| \hat{c}\) where \(\hat{c}\) is the unit vector along the bisector and \(|\vec{R}|\) is the magnitude of the resultant vector.

Calculation:

Given:

Vector 1: \(\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}\)

Vector 2: \(\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}\)

Magnitude of resultant vector: \(\sqrt{2}\)

⇒ \(|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\)

⇒ \(|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\)

⇒ Unit vector along bisector: \(\hat{c} = \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} + \frac{\hat{i} + 2\hat{j} + 2\hat{k}}{3}\)

⇒ \(\hat{c} = \frac{3\hat{i} + 3\hat{k}}{3} = \hat{i} + \hat{k}\)

Hence option 4 is correct

Answer : 3

Solution :

Let F be the vector coplanar to a̅ and b̅. Then,

r̅ = a̅ + mb̅

= \((\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathrm{m}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\)

= \((\hat{i}+2 \hat{j}+\hat{k})+m(\hat{i}-\hat{j}+\hat{k})\) ... (i)

Since the projection of r̅ along c̅ is \(\frac{1}{\sqrt{3}}, \frac{̅{x} \cdot ̅{c}}{|̅{c}|}= ± \frac{1}{\sqrt{3}}\)

\(\Rightarrow \frac{(1+m)+(2-m)-(1+m)}{\sqrt{3}}= ± \frac{1}{\sqrt{3}}\)

⇒ (1 + m) + (2 - m) - (1 + m) = ±1

∴ m = 3 or m = 1 

Substituting m = 3 in equation (i), we get

\(\bar{r}=\hat{i}(1+3)+\hat{j}(2-3)+\hat{k}(1+3)\)

\(\Rightarrow \bar{r}=4 \hat{i}-\hat{j}+4 \hat{k}\)

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°

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

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:

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:  

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

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

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:

Let \(\rm\vec{a}\) and \(\rm\vec{b}\) be the two vectors and the vector \(\rm\vec{c}\) perpendicular to both \(\rm\vec{a}\) and \(\rm\vec{b}\)

Hence  \(\rm\vec{c} = ​​\rm\vec{a} × \rm\vec{b}\)

Calculation:

Let vector \(\rm\vec{c}\) is perpendicular to both the vectors î - ĵ and î

Therefore, \(\rm\vec{c}\) = (î - ĵ) × î

= (î × î) -  (ĵ × î)

= 0 - (-k̂)

= k̂

Concept:

For three vectors \(\rm \vec A\), \(\rm \vec B\) and \(\rm \vec C\):

• Triple Cross Product: is defined as: \(\rm \vec A\times\left(\vec B\times\vec C\right)=\left(\vec A.\vec C\right)\vec B-\left(\vec A.\vec B\right)\vec C\)

Calculation:

Given that: \(\rm \left(\vec a\times\vec b\right)\times\vec c=\vec a\times\left(\vec b\times\vec c\right)\)

⇒ \(\rm -\vec c \times\left(\vec a\times\vec b\right)=\vec a\times\left(\vec b\times\vec c\right)\)

⇒ \(\rm -\left(\vec c.\vec b\right)\vec a+\left(\vec c.\vec a\right)\vec b=\left(\vec a.\vec c\right)\vec b-\left(\vec a.\vec b\right)\vec c\)

⇒ \(\rm \left(\vec a.\vec b\right)\vec c-\left(\vec c.\vec b\right)\vec a=0\)

⇒ \(\rm \vec b\times\left(\vec c\times\vec a\right)=0\), which is the required answer.

Concept:

Cross product of vectors:

•  \(|\vec a \times \vec b| = |\vec a| \cdot |\vec b| \cdot sin \theta \)
• 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 \(\vec a \times \;\vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|\)

Given: \(\;\vec a = 3\hat j + 4\hat k\;and\;\vec b = 6\hat i + 8\hat k\)

⇒ \(\vec a \times \;\vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{0}}&{{3}}&{{4}}\\ {{6}}&{{0}}&{{8}} \end{array}} \right|\)

⇒ \(\vec a \times \;\vec b = \hat i(24 -0) - \hat j(0 - 24) + \hat k(0 - 18) = 24\hat i + 24\hat j-18\hat k\)

⇒ \(|\vec a \times \vec b| = \rm \sqrt{24^{2}+24^{2}+18^{2}}=\sqrt{1476}=6\sqrt{41}\)

⇒ \(|\vec a| = 5 \ and \ |\vec b| = 10\)

As we know that, \(|\vec a \times \vec b| = |\vec a| \cdot |\vec b| \cdot sin \theta \)

⇒ \(\sin \theta = \frac{{\left| {\vec a \times \;\vec b} \right|}}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}} = \frac{{6\sqrt {41} }}{{50}} = \frac{{3\sqrt {41} }}{{25}}\)

Hence, the correct option is 3.

Concept:

Let \(\rm\vec{a}\) and \(\rm\vec{b}\) be the two vectors and the vector \(\rm\vec{c}\) perpendicular to both \(\rm\vec{a}\) and \(\rm\vec{b}\)

Hence  \(\rm\vec{c} = ​​\rm\vec{a} × \rm\vec{b}\)

Calculation:

Let vector \(\rm\vec{c}\) is perpendicular to both the vectors î + ĵ and ĵ + k̂

Let \(\rm \vec a \) = î + ĵ and \(\rm \vec b \) = ĵ + k̂

Therefore, \(\rm\vec{c} = ​​\rm\vec{a} × \rm\vec{b}\)

= (î + ĵ) × (ĵ + k̂)

= î × ĵ + î × k̂ + ĵ × ĵ + ĵ × k̂ 

= k̂ - ĵ + 0 + î 

= î - ĵ + k̂  

Concept:

Let \(\rm\vec {a}\) and \(\rm \vec {b}\) be the two vectors,

Dot product of two vectors  is given by:  \(\rm \vec{a}.\vec{b} = |\vec{a}||\vec{b}| cos θ\)

Cross product of two vectors  is given by: \(\rm \vec{a}× \vec{b} = |\vec{a}||\vec{b}| \sin θ\; \hat{n}\), Where \(\rm \hat n\) is a unit vector

Calculation:

Given: \(\rm |\vec{a}× \vec{b}| = |\vec{a} \cdot \vec{b}|\)

To Find: Angle between \(\rm \vec{a} \; \text {and} \; \vec{b}\)

\(\rm |\vec{a}× \vec{b}| = |\vec{a} \cdot \vec{b}|\)

\(⇒ \rm |\vec{a}||\vec{b}| \sin θ\;| \hat{n}| = |\vec{a}||\vec{b}| \cos θ\)

⇒ sin θ = cos θ                       (∵ |\(\rm \hat n\)| = 1)

⇒ tan θ = 1

∴ θ = 45° 

Concept:

Properties of vectors:

If \(\rm \vec a\) and \(\rm \vec b\) are two vectors parallel to each other then  \(\rm \vec a \times \vec b = 0\)

Cross product of parallel vectors are zero ⇔  \(\rm \vec a \times \vec a = 0,{\rm{\;}}\vec b \times \vec b = 0{\rm{\;and\;}}\vec c \times \vec c = 0\)

A cross or vector product is not commutative ⇔ \(\rm \vec a \times \vec b = - {\rm{\;}}\vec b \times \vec a\)

Calculation:

We have to find the value of \(\left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right)\)

\(\Rightarrow \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) = {\rm{\;\vec a\;}} \times {\rm{\;\vec a}} - {\rm{\;\vec a\;}} \times {\rm{\;\vec b}} + {\rm{\;\vec b\;}} \times {\rm{\;\vec a}} - {\rm{\;\vec b\;}} \times {\rm{\;\vec b}}\)

We know that \({\rm{\vec a\;}} \times {\rm{\;\vec b}} = - {\rm{\;\vec b\;}} \times {\rm{\;\vec a}}\)

\(\Rightarrow \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) = {\rm{\;}}0 - {\rm{\;\vec a\;}} \times {\rm{\;\vec b}} - {\rm{\;\vec a\;}} \times {\rm{\;\vec b}} - {\rm{\;}}0\)                \(\because \left( {{\rm{\vec a\;}} \times {\rm{\;\vec a}} = \;{\rm{\vec b\;}} \times {\rm{\;\vec b}} = 0} \right)\) 

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

∴ Option 2 is correct.