Vector Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Vector Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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:

• \(\vec a\) and \(\vec b\) are two vectors parallel to each other ⇔  \(\vec a \times \vec b = 0\)
• Cross product of parallel vectors are zero ⇔  \(\vec a \times \vec a = 0,{\rm{\;}}\vec b \times \vec b = 0{\rm{\;and\;}}\vec c \times \vec c = 0\)
• A cross or vector product is not commutative ⇔ \(\vec a \times \vec b = - {\rm{\;}}\vec b \times \vec a\)

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 3 is correct.

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

Concept:

Length of the vector \(\rm a\hat{i} + b\hat{j} + c\hat{k} \) from origin is \(\rm \sqrt{a^{2} + b^{2} + c^{2}}\)

Calculation:

Length of the vector k(2î -  ĵ -  2k̂) from origin is 

\(\rm \sqrt{(2k)^{2} + (-k)^{2} + (-2k)^{2}}\)

= \(\rm \sqrt{4k^{2} + k^{2} + 4k^{2}}\) 

= \(\rm \sqrt{9k^{2} }\) 

= 3k

Length is 6 units given

3k = 6

k = 6/3

k = 2

Hence option 2 is correct.

Concept:

If θ is the angle between x and y then x.y = xycosθ 

Calculation:

GIven, a + b + c = 0

⇒ a + b = -c, squaring both sides we get

⇒  (a + b)² = (-c)²

⇒ |a + b|2 = |c|2

⇒ |a|² + |b|² +2|a||b|cosθ = |c|², where θ = angle betwee a and b

Putting the values we get,

3² + 5² + 2× 3× 5 cosθ = 7²

⇒ 9 + 25 + 30cosθ = 49

⇒ 30cosθ = 49 - 34 = 15

⇒ cosθ = \(1\over 2\)

∴ θ = 60° 

Concept:

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

Let the two vectors are \(\rm \vec a\) and \(\rm \vec b\)

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

Where |\(\rm \vec a\)| = Magnitudes of vectors a, |\(\rm \vec b\)| = Magnitudes of vectors b and θ is the angle between a and b

Formulas of Dot Product:

 \(\rm \vec i.\vec i = \vec j.\vec j = \vec k.\vec k = 1\)

\(\rm \vec i.\vec j = \vec j.\vec i = \vec i.\vec k = \vec k.\vec i =\vec j.\vec k= \vec k.\vec j = 0\)

Calculation:

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:

For unit vectors:

\(|\vec{a}|=\vec{a}\cdot\vec{a}=1\)

\(|\vec{b}|=\vec{b}\cdot\vec{b}=1\)

When two vectors \(\vec{A}\)and \(\vec{B}\) are perpendicular to each other then the dot product is zero i.e.

\(\vec{A}\cdot\vec{B}=0\)

Calculation:

Given:

\(\vec{A}=â + 2b̂\), \(\vec{B}=5â - 4b̂\)

\(\vec{A}\cdot\vec{B}=0\)  [∵ perpendicular to each other]

\((â + 2b̂)\cdot(5â - 4b̂)=0\)

\(5\vec{a}\cdot\vec{a}-4\vec{a}\cdot\vec{b}+10\vec{a}\cdot\vec{b}-8\vec{b}\cdot\vec{b}=0\)

\(-3+6\vec{a}\cdot\vec{b}=0\)

\(\vec{a}\cdot\vec{b}=\frac{3}{6}=\frac{1}{2}\)

\(|\vec{a}|\cdot|\vec{b}|\cdot\cos θ=\frac{1}{2}\)

∴ θ = π/3

Concept:

If vectors \(\rm \vec a\;and\;\vec b\) are perpendicular then \(\rm \vec a \cdot \;\vec b = 0\)

Calculation:

Given: \(\rm 2\hat i - 5\hat j - \hat k\) and \(\rm -\hat i + 4 \hat j + λ\hat k\) are perpendicular

Let \(\rm \vec a = 2\hat i - 5\hat j - \hat k\) and \(\rm \vec b = -\hat i + 4 \hat j + λ\hat k\)

We know that, If vectors \(\rm \vec a\;and\;\vec b\) are perpendicular then \(\rm \vec a \cdot \;\vec b = 0\)

\(\rm \vec a \cdot \vec b = ( 2\hat i - 5\hat j - \hat k) \cdot (-\hat i + 4 \hat j + λ\hat k) = 0\)

⇒ -2 - 20 - λ = 0 

⇒ -22 - λ = 0 

∴ λ = -22

Concept:

If \({\rm{\vec a\;and\;\vec b}}\) are two vectors parallel to each other then \({\rm\vec{a} = λ \vec{b}}\) or \(\rm \vec{a} × \vec{b} =0\)

Calculation:

Given:

 \(\rm x\vec{i} - 2\vec{j} + 3\vec{k}\) and \(\rm 2\vec{i} - 4\vec{j} + y\vec{k}\) are parallel vectors,

Therefore, \(\rm x\vec{i} - 2\vec{j} + 3\vec{k} = λ (\rm 2\vec{i} - 4\vec{j} + y\vec{k})\)

Equating the coefficient of \(\rm \vec{i},\vec{i} \;and\; \vec{k}\)

⇒ x = 2λ                    .... (1)

⇒ -2 = -4λ 

∴ λ = 1/2

Put the value of λ in equation (1), we get

x = 2 × (1/2)

So, x = 1

Concept:

Scaler triple product of the vectors:

The scaler triple product of the vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) is given by:

Coplaner vectors:

Three vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) are said to be coplaner if the scaler triple product \({\bf{\bar A}} \cdot \left[ {{\bf{\bar B}} \times {\bf{\bar C}}} \right] = 0.\)

Solution:

Let the given vectors be \({\rm{\bar A}} = {\rm{a}}\hat i + \;\hat j + {\rm{\;}}\hat k,\;\bar B = \;\hat i + \;b\hat j + \hat k\) and \({\rm{\bar C}} = {\rm{\;}}\hat i + \;\hat j + {\rm{c}}\hat k\).

It is given that the vectors are coplaner therefore the scaler triple product \({\rm{\bar A}} \cdot \left[ {{\rm{\bar B}} \times {\rm{\bar C}}} \right] = 0.\)

Therefore,

\(\left| {\begin{array}{*{20}{c}} {\rm{a}}&1&1\\ 1&{\rm{b}}&1\\ 1&1&{\rm{c}} \end{array}} \right| = 0\)

Perform the coloumn operation C₁ – C₂ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&1&1\\ {1 - {\rm{b}}}&{\rm{b}}&1\\ 0&1&{\rm{c}} \end{array}} \right| = 0\)

Now perform another coloum operation C₂ – C₃ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&0&1\\ {1 - {\rm{b}}}&{{\rm{b}} - 1}&1\\ 0&{1 - {\rm{c}}}&{\rm{c}} \end{array}} \right| = 0\)

Take (1 - a)(1 - b)(1 - c) common. Note that it is given that a,b,c ≠ 0 therefore this action is jstified.

\(\left( {1 - {\rm{a}}} \right)\left( {1 - {\rm{b}}} \right)\left( {1 - {\rm{c}}} \right)\left| {\begin{array}{*{20}{c}} { - 1}&0&{\frac{1}{{1 - {\rm{a}}}}}\\ 1&{ - 1}&{\frac{1}{{1 - {\rm{b}}}}}\\ 0&1&{\frac{{\rm{c}}}{{1 - {\rm{c}}}}} \end{array}} \right| = 0\)

Since a, b, c ≠ 0 therefore (1 - a)(1 - b)(1 - c) ≠ 0.Therefore the determinant has to be zero.

\( - 1\left( {\frac{{\rm-{c}}}{{1 - {\rm{c}}}} - {\rm{}}\frac{1}{{1 - {\rm{b}}}}} \right) + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{{\rm{c}}}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

Simplify the above expression as follows:

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} - {\rm{\;}}1 + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 1\)