Vector Algebra MCQ Quiz - Objective Question with Answer for Vector Algebra - Download Free PDF

Concept:

Vector Operations and Identities:

• Dot Product: A scalar defined as 𝒂 · 𝒃 = |𝒂||𝒃|cosθ. It measures the projection of one vector on another.
• Cross Product: A vector defined as 𝒂 × 𝒃 = |𝒂||𝒃|sinθ 𝒏̂, perpendicular to the plane of 𝒂 and 𝒃.
• Vector Triple Product Identity: 𝒂 × (𝒃 × 𝒄) = (𝒂 · 𝒄)𝒃 − (𝒂 · 𝒃)𝒄
• Unit Vector: A vector of magnitude 1. If |𝒂| = 1, then 𝒂 is a unit vector.
• Regular Tetrahedron: A solid with 6 equal edges and 4 equilateral triangle faces. Angle between adjacent edges is 60°.

Calculation:

Given,

Statement A: 𝒂, 𝒃, 𝒄 form sides BC, CA, AB of ΔABC

⇒ 𝒂 + 𝒃 + 𝒄 = 0

⇒ Square both sides: (𝒂 + 𝒃 + 𝒄)² = 0

⇒ 𝒂² + 𝒃² + 𝒄² + 2(𝒂·𝒃 + 𝒃·𝒄 + 𝒄·𝒂) = 0

⇒ Let |𝒂| = |𝒃| = |𝒄| = 1

⇒ 3 + 2(𝒂·𝒃 + 𝒃·𝒄 + 𝒄·𝒂) = 0

⇒ 𝒂·𝒃 + 𝒃·𝒄 + 𝒄·𝒂 = −3/2

Statement B: 𝒂, 𝒃, 𝒄 are adjacent sides of regular tetrahedron

⇒ Angle between adjacent edges = 60°

⇒ 𝒂·𝒃 = 𝒃·𝒄 = 𝒄·𝒂 = cos 60° = 1/2

Statement C: 𝒂 × 𝒃 = 𝒄 ; 𝒃 × 𝒄 = 𝒂

⇒ Take LHS: 𝒂 × 𝒃 = 𝒄

⇒ Then 𝒃 × 𝒄 = 𝒂, and 𝒄 × 𝒂 = 𝒃 must also hold

⇒ Hence, 𝒂 × 𝒃 = 𝒃 × 𝒄 = 𝒄 × 𝒂

Statement D: 𝒂, 𝒃, 𝒄 are unit vectors & 𝒂 + 𝒃 + 𝒄 = 0

⇒ As before, square both sides:

⇒ 𝒂² + 𝒃² + 𝒄² + 2(𝒂·𝒃 + 𝒃·𝒄 + 𝒄·𝒂) = 0

⇒ 3 + 2(𝒂·𝒃 + 𝒃·𝒄 + 𝒄·𝒂) = 0

⇒ 𝒂·𝒃 + 𝒃·𝒄 + 𝒄·𝒂 = −3/2

∴ Correct matching is: I-(T), II-(P), III-(Q), IV-(R)

Concept:

If two points A and B have position vectors \(\rm \vec A\) and \(\rm \vec B\) respectively, then the vector \(\rm \vec {AB}=\vec B-\vec A\).

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

• Dot Product is defined as: \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
• Resultant Vector is equal \(\rm \vec A + \vec B\).
• Work: The work (W) done by a force (\(\rm \vec F\)) in moving (displacing) an object along a vector \(\rm \vec D\) is given by: W = \(\rm \vec F.\vec D=|\vec F||\vec D|\cos \theta\).

Calculation:

Let's say that the forces acting on the particle are \(\rm \vec P\) = 2î - 5ĵ + 6k̂ and \(\rm \vec Q\) = -î + 2ĵ - k̂.

∴ The resulting force acting on the particle will be \(\rm \vec F=\vec P+\vec Q\).

⇒ \(\rm \vec F\) = (2î - 5ĵ + 6k̂) + (-î + 2ĵ - k̂)

⇒ \(\rm \vec F\) = î - 3ĵ + 5k̂.

Since the particle is moved from the point 4î - 3ĵ - 2k̂ to the point 6î + ĵ - 3k̂, the displacement vector \(\rm \vec D\) will be:

\(\rm \vec D=\vec{AB}=\vec B-\vec A\)

= (6î + ĵ - 3k̂) - (4î - 3ĵ - 2k̂)

⇒ ​\(\rm \vec D\) = 2î + 4ĵ - k̂.

And finally, the work done W will be:

W = \(\rm \vec F.\vec D\) = (î - 3ĵ + 5k̂).(2î + 4ĵ - k̂)

⇒ W = (1)(2) + (-3)(4) + (5)(-1)

⇒ W = 2 - 12 - 5 =

∴ -15 units.

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:

 \(\rm \vec{i} - a\vec{j} + 5\vec{k}\) and \(\rm 3\vec{i} - 6\vec{j} + b\vec{k}\) are parallel vectors,

Therefore, \(\rm \vec{i} - a\vec{j} + 5\vec{k}= λ (\rm 3\vec{i} - 6\vec{j} + b\vec{k})\)

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

⇒ 1 = 3λ, ∴ λ = 1/3            

⇒ -a = -6λ 

⇒ 5 = bλ                 .... (1)

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

5 = b × (1/3)

So, b = 15

Concept:

Magnitude of vector \(\rm \vec{z} = a\hat i + b\hat j + c\hat k\) then magnitude of vector is given by \( \left | \vec z \right |=\sqrt{(a^2+b^2 + c^2)}\)

Calculation:

Given: Let \(\rm \vec{z} = 5\vec a\)where \(\rm \vec{a} = 2\hat i + 3\hat j + 7\hat k\)

⇒ \(\vec z = \rm 5\vec{a} = 10\hat i + 15\hat j + 35\hat k\)

As we know that, if \(\rm \vec{z} = a\hat i + b\hat j + c\hat k\) then \( \left | \vec z \right |=\sqrt{(a^2+b^2 + c^2)}\)

⇒ |\(\rm \vec{z}| = \sqrt {10^2 + 15^2 +35^2} = \sqrt{1550}\)

⇒  \(|\vec z | = 5\sqrt{62}\)

Hence, option 1 is correct.

Concept:

Equal Vectors
Two or more vectors are said to be equal when their magnitude is equal and also their direction is the same.

Calculation:

Given:  \(\rm \vec a = 3\hat i - 2\hat j +a_3\hat k\)and \(\rm \vec b =\rm \vec 3\hat i - 2\hat j +\hat k\) 

\(\rm \vec a = \vec b\)

 \(\rm 3\hat i - 2\hat j +a_3\hat k =\rm \vec 3\hat i - 2\hat j +\hat k\)

∴ a₃ = 1

Concept:

Conditions of collinear vector:

• Three points with position vectors \(\vec a,\;\vec b\;and\;\vec c\) are collinear if and only if the vectors \(\left( {\vec a - \vec b} \right)\) and \(\left( {\vec a\; - \vec c} \right)\) are parallel. ⇔ \(\left( {\vec a - \vec b} \right) = λ \left( {\vec a\; - \vec c} \right)\)
• If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Solution:

We know that, If the points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Given  \(\widehat i + 2\widehat j + 3\widehat k\), \(λ \widehat i + 4\widehat j + 7\widehat k\), \(- 3\widehat i - 2\widehat j - 5\widehat k\) are collinear

∴ \(\left| {\begin{array}{*{20}{c}} { 1}&{ 2}&3\\ λ&4&7\\ -3&-2&-5 \end{array}} \right| = 0\)

⇒ 1 (-20 + 14) – (2) (-5λ + 21) + 3 (-2λ + 12) = 0

⇒ -6 + 10λ – 42 - 6λ + 36  = 0

⇒ 4λ = 12

∴ λ = 3

Concept:

Let \(\rm {\rm{\vec a}} = {\rm{x\;\vec i}} + {\rm{y\;\vec j}} + {\rm{z\;\vec k}}\) then magnitude of the vector of a = \(\left| {{\rm{\vec a}}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {\rm{\;}}{{\rm{y}}^2} + {{\rm{z}}^2}} \)

Calculation:

Let \(\rm \vec{a}\) =  p(2î - ĵ + 2k̂)

Given, \(\left| {{\rm{\vec a}}} \right| = 3\)

⇒ \(\rm \sqrt{4p^2 + p^2+4p^2} = 3\)

⇒ \(\rm \sqrt{9p^2} = 3\)

⇒ 3p = 3

∴ p = 1

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 \(\rm \vec A = x \hat i-y\hat j +z\hat k\), then \(\rm |\vec A| = \sqrt {x^2 +y^2+z^2}\)

Calculation:

Given A = \(\rm 5 \hat i-2\hat j +4\hat k\) and B = \(\rm \hat i+3\hat j -7\hat k\)

\(\rm \vec{AB} = \vec B - \vec A\)

\(\rm \vec{AB}\) = \(\rm \hat i+3\hat j -7\hat k - (5 \hat i-2\hat j +4\hat k)\)

\(\rm \vec{AB}\) = \(\rm -4\hat i+5\hat j -11\hat k\)

Now \(\rm |\vec {AB}| = \sqrt{(-4)^2 +5^2+(-11)^2}\)

\(\rm |\vec {AB}| = \sqrt{16 +25+121}\)

\(\rm |\vec {AB}| = \sqrt{162}\) = 9√2

Concept:

Three or more points are collinear, if slope of any two pairs of points is same.

Here, \(\rm 5\hat i-2\hat j, 8\hat i-3\hat j, a\hat i-12\hat j \)

Let, A = (5, -2), B = (8, -3), C = (a, -12)

Now, slope of AB = Slope of BC = Slope of AC ....(∵ points are collinear)

 \(\rm \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{a-8}\\ ⇒ \frac{-1}{3}=\frac{-9}{a-8}\)

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.

Concept:

For two vectors \(\vec m \ and \ \vec n \) to be collinear,​ \(\vec m\; = \;λ \vec n\) where λ is a scalar.

Calculation:

Given that, the vectors \(4\hat i + \hat j - 3\hat k\) & \(p\hat i + q\hat j - 2\hat k\) are collinear.

Since two vectors \(\vec m \ and \ \vec n \) are collinear then \(\vec m\; = \;λ \vec n\) where λ is a scalar.

⇒ \(4\hat i + \hat j - 3\hat k\;\ = λ × (\;p\hat i + q\hat j - 2\hat k)\)

⇒ \(4\hat i + 1\hat j - 3\hat k\;\ = λ p \hat i + λq \hat j - 2λ \hat k\)

⇒ λp = 4,  λq = 1 and -2λ = -3

⇒  λ = 3/2 

So, by substituting λ = 3/2 in  λp = 4 and λq = 1, we get

⇒ (3/2)p = 4 and (3/2)q = 1

⇒ p = 8/3 and q  = 2/3

∴  \(\frac{8}{3}, \frac{2}{3}\)is the correct answer.

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 \(\vec b = 4\hat i + 3\hat j - \hat k\)

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

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

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

Concept:

Let the angle between \(\vec a\) and \(\vec b\)is \(\rm \theta\)

\(\rm \vec a.\vec b = 2ab cos\;\theta\)

Calculations:

consider, the angle between \(\vec a\) and \(\vec b\)is \(\rm \theta\)

Given, \(\vec a + \vec b + \vec c = \vec 0 \)

⇒\(\vec a + \vec b = - \vec c \)

⇒\(\rm |\vec a + \vec b| = |- \vec c |\)

Squaring on both side, we get

⇒\(\rm |\vec a + \vec b|^2 = |- \vec c |^2\)

⇒\(\rm |\vec a|^2 +2\;\vec a.\vec b+ |\vec b|^2 = |- \vec c |^2\)

⇒\(\rm |\vec a|^2 +|\vec b|^2+2\;ab\cos\;\theta = |- \vec c |^2\)

⇒\(\rm (3)|^2 +(5)^2+2\;(3)(5)\cos\;\theta = (7)^2\)

⇒\(\rm 30\cos\;\theta = 15\)

⇒\(\rm \cos\;\theta = \dfrac 12\)

⇒ \(\rm \theta\) = π / 3

Hence, If \(\vec a + \vec b + \vec c = \vec 0,\;|\vec a| = 3,\;|\vec b| = 5\) and \(|\vec c| = 7\), then the angle between \(\vec a\) and \(\vec b\)is π / 3

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:

 \(\rm \vec{i} - a\vec{j} + 5\vec{k}\) and \(\rm 3\vec{i} - 6\vec{j} + b\vec{k}\) are parallel vectors,

Therefore, \(\rm \vec{i} - a\vec{j} + 5\vec{k}= λ (\rm 3\vec{i} - 6\vec{j} + b\vec{k})\)

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

⇒ 1 = 3λ, ∴ λ = 1/3            

⇒ -a = -6λ 

⇒ 5 = bλ                 .... (1)

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

5 = b × (1/3)

So, b = 15

Calculation:

 \(\rm \vec a =\hat i +\hat j +\hat k,\; \vec b =\hat i -\hat j + \hat k\) and c = î - ĵ - k̂

Given:  vector \(\rm \vec v\) in the plane of \(\rm \vec a\) and \(\rm \vec b\) 

Therefore, \(\rm \vec v = \vec a + λ \vec b\)

⇒ \(\rm \vec v =(\hat i +\hat j +\hat k ) \; + λ (\hat i -\hat j + \hat k)\)

= (1 + λ)î + (1 - λ)ĵ  + (1 + λ)k̂                .... (1)

Projection of \(\rm \vec v\) on \(\rm \frac {\vec c} {|\vec c|}=\frac 1 {\sqrt 3}\)

⇒ \(\rm \vec v=\rm \frac {\vec c} {|\vec c|}=\frac 1 {\sqrt 3}\) 

⇒ \(\frac {(1 + λ) - (1 - λ) - (1 + λ)}{\sqrt3} = \frac {1}{\sqrt 3}\)

⇒ -(1 - λ) = 1

∴ λ = 2

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

\(\rm \vec v\) = 3î - ĵ + 3k̂