Types of Vectors MCQ Quiz - Objective Question with Answer for Types of Vectors - Download Free PDF

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:

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:

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:  \(\rm \vec a = 5\hat i - 3\hat j +a_3\hat k\) and  \(\rm \vec b =\rm \vec 5\hat i - 3\hat j -2\hat k\) 

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

 \(\rm 5\hat i - 3\hat j +a_3\hat k =\rm \vec 5\hat i - 3\hat j -2\hat k\)

∴ a3 = - 2

Explanation:

 \(\vec a\) and \(\vec b\) are unit vectors

So, |\(\vec a\)| = |\(\vec b\)| = 1

Now, 

|\(\rm \vec a+ \vec b\)| = 1 if

\(|\vec a|^2+|\vec b|^2+2|\vec a||\vec b|\cosθ=1\)

i.e., \(1+1+2.1.1\cosθ=1\)

i.e., \(\cosθ=-\frac12\)

i.e., θ = \(\frac{2\pi}{3}\)

Hence \(\rm \vec a+ \vec b\) is unit vector, if angle between \(\vec a\) and \(\vec b\) is \(\frac{2\pi}{3}\)

Option (4) is true.

Explanation:

\(\rm \vec A, \vec B\) and \(\rm \vec C\) are unit vectors.

So, \(\rm |\vec A|=| \vec B|=|\vec C|\) = 1

Also, \(\rm \vec A \) is perpendicular to both \(\rm \vec B\) and \(\rm \vec C\).

So, \(\rm \vec A \) is parallel to \(\rm \vec B\times \vec C \)

i.e., \(\rm \vec A =λ(\rm \vec B\times \vec C )\) where λ is any scaler.

Given, angle between \(\rm \vec B\) and \(\rm \vec C\) is 30°.

So, \(\rm \vec B\times \vec C=|\vec B||\vec C|\sin\theta\)

⇒ \(\rm \frac1{λ}\vec A=1.1.\frac12\)

⇒ \(\rm |\vec A|=\frac{|λ|}2\) (Taking mod both sides)

⇒ \(\rm1=\frac{|λ|}2\)

⇒ |λ| = 2

⇒ λ = ± 2

Hence we get

\(\rm \vec A =\pm(\rm \vec B\times \vec C )\)

Option (3) is true.

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:

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 \({\rm{\vec a\;and\;\vec b}}\) 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:

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:

Unit vector in the direction of vector \(\rm \vec{z}\)is given by \(\hat z = \rm \frac{\vec{z}}{|z|}\).

Calculation:

Given: \(\rm \vec{a}= 3i -4j\)

⇒ Unit vector in the direction of vector \(\rm \vec{a}\) is given by \(\hat a = \rm \frac{\vec{a}}{|a|}\).

⇒  \(\rm \hat{a} = \rm \frac{3i-4j}{\sqrt{3^2+(-4)^2}} = \rm \frac{3\hat i-4\hat j}{5}\)

A vector in the direction of vector \(\rm \vec{a}\) that has magnitude 10 is given by \(10 \ \hat a\)

⇒ \(10 \ \hat a = 6\hat i - 8 \hat j\)

Hence, option 2 is correct.

Concept used:

a, b, and c vectors are coplanar when \([\vec{a}, \vec{b},\vec{c}] = 0\)

Calculation:

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

⇒ 2(10 + 3a) + 1(5 + 9) + 1(a - 6) = 0

⇒ 20 + 6a + 14 + a - 6 = 0

⇒ 7a = - 28

∴ a = - 4

∴ The vector 2î - ĵ + k̂, î + 2ĵ - 3k̂ and 3î + aĵ + 5k̂ are coplanar when a = - 4

CONCEPT:

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

CALCULATION:

Given: \(3\hat i + \hat j - 2\hat k\) and \(\hat i + \lambda \hat j - 3\hat k\) are perpendicular

\(\Rightarrow \vec a \cdot \;\vec b = \left( {3\hat i + \hat j - 2\hat k} \right) \cdot \left( {\hat i + \lambda \hat j -3\hat k} \right) = \; 3 + \lambda + 6 = 9 + \lambda\)

∵ \(\hat i + \lambda \hat j - 3\hat k\) and \(\hat i + \lambda \hat j - 3\hat k\) are perpendicular

⇒ λ + 9 = 0

⇒ λ = - 9

Hence, option A is the correct answer.

Concept:

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

\(\begin{array}{l} \vec{a}=x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k}\\ \vec{b}=x_{2} \hat{i}+y_{2} \hat{j}+z_{2} \hat{k} \\ ​​\vec{c}=\vec{a} × \vec{b}=\left|\begin{array}{ccc} \hat{i} & j & \hat{k} \\ x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \end{array}\right| \end{array}\)

Calculation:

Here, \({\rm{\vec a}} = - {\rm{\hat i}} + {\rm{\hat j}} + {\rm{\hat k}}\) and \({\rm{\vec b}} = {\rm{\hat i}} - {\rm{\hat j}} + {\rm{\hat k\;}}\)

Vector perpendicular to both vectors will be \(\rm \vec{a} \times \vec{b}\)

\(\begin{aligned} \vec{a} \times \vec{b} &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -1 & 1 \end{array}\right| \\ &=\hat{i}(1+1)-\hat{j} (-1-1)+\hat{k}(1-1) \\ &=2 \hat{i}+2 \hat{j} \end{aligned}\)

\(\begin{aligned} |\vec{a} \times \vec{b}| =\sqrt{(2)^{2}+(2)^{2}} =\sqrt{8} =2 \sqrt{2}\\ \end{aligned} \)

\(\text {Unit vector} =\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\\ =\frac{2 \hat{i}+2 \hat{j}}{2 \sqrt{2}} \\ =\frac{i+\hat{j}}{\sqrt{2}}\)

Hence, option (1) is correct.

Explanation:

Given, λî + 2λĵ + 2λk̂ ​is a unit vector.

∴ \(\left|λ \hat{i} +2λ \hat{j}+2λ \hat{k}\right|=1\)

⇒ \(\sqrt{λ ^{2}+(2λ )^{2}+(2λ )^{2}}=1\)

⇒ \(\sqrt{λ ^{2}+4λ ^{2}+4λ ^{2}}=1\)

⇒ \(\sqrt{9λ ^{2}}=1\)

⇒ 3λ = 1

⇒ \(\lambda=\frac{1}{3} \)