Types of Vectors MCQ Quiz in मल्याळम - Objective Question with Answer for Types of Vectors - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

We Have,

If, \(\left| {\overrightarrow a } \right|\) = \(\left| {\overrightarrow b } \right|\) = \(\left| {\overrightarrow c } \right|\)

⇒ a = b = c .... (1)

And,

\(\overrightarrow a \, + \,\overrightarrow b \, = \,\,\overrightarrow c \) .... (2)

Using the concept of, Parallelogram Law of Vector Addition: It is used to add two vector quantities.

\(\overrightarrow A \, + \,\overrightarrow B \, = \,\,\overrightarrow R\)

R2 = A2 + B2 +2AB cos θ 

From equation (2),

c2 = a2 + b2 +2ab cos θ

Since,

a = b = c

Hence,

a² = a² + a² +2a² cos θ

or, a2 = a2 (1 + 1 +2cos θ)

or, 1 = 2 + 2 cos θ

or, 2 cos θ = -1

or, cos θ = (-0.5)

Hence, θ = cos^-1 (0.5) = 120° = \(\frac{{2\pi }}{3}\)

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

Concept:

Two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are perpendicular if and only if, their dot product is equal to zero.

\(\overrightarrow{A}.\overrightarrow{B}=0\)

Solution:

Given: \(\overrightarrow{a}= 2\hat{i}+2\hat{j}+3\hat{k}\) , \(\overrightarrow{b}= -\hat{i}+2\hat{j}+\hat{k}\) and \(\overrightarrow{c}= 3\hat{i}+\hat{j}\)

Then, \(\overrightarrow{a}+λ \overrightarrow{b} \) \(= (2\hat{i}+2\hat{j}+3\hat{k})+λ (-\hat{i}+2\hat{j}+\hat{k})\)

\(= (2-λ )\hat{i}+(2+2λ )\hat{j}+(3+λ )\hat{k}\)

Given that, \(\overrightarrow{a}+λ \overrightarrow{b}\) is perpendicular to \(\overrightarrow{c}\)

Then, \((\overrightarrow{a}+λ \overrightarrow{b}).\overrightarrow{c}=0\)

⇒ \([(2-λ )\hat{i}+(2+2λ )\hat{j}+(3+λ )\hat{k}].(3\hat{i}+\hat{j}) =0 \)

⇒ 3(2 - λ ) + (2 + 2λ ) = 0 

⇒ 6 - 3λ + 2 + 2λ = 0

⇒ 8 - λ = 0

⇒ λ = 8

∴ The correct option is (2)

Concept:

Vector \(\vec{r}\) of magnitude |p| in the direction of \(\vec{q}\) is given by

\(\vec{r}\ =\ |p|.\hat{q}\ =\ |p|.\frac{\hat{q}}{|q|}\)

​If  \(\vec{a} = {a_1}\hat{i} \ + \ {a_2}\hat{j} \ + \ {a_3}\hat{k}\) then magnitude of ​\(\vec{a}\) is written as​​

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

Calculation:

Let the required vector is \(\vec{c}\)

\(\vec{a}\ =\ \hat{i}\ +\ 2\hat{j}\ +\ 2\hat{k}\)

\(\vec{b}\ =\ 3\hat{i}\ +\ 5\hat{j}\ +\ \sqrt{2}\hat{k}\)

A vector in the direction a and magnitude |b| is given by

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

\(\vec{c}\ =\ \sqrt{3^2\ +\ 5^2\ +\ (\sqrt{2})^2}.\frac{\hat{\hat{i}\ +\ 2\hat{j}\ +\ 2\hat{k}}}{\sqrt{1^2\ +\ 2^2\ +\ 2^2}}\)

\(\vec{c}\ =\ 6.\frac{\hat{\hat{i}\ +\ 2\hat{j}\ +\ 2\hat{k}}}{3}\)

\(\vec{c}\ =\ 2({\hat{\hat{i}\ +\ 2\hat{j}\ +\ 2\hat{k}}})\)

Hence, option 4 is correct.

Concept:

If \(\rm\overrightarrow{c}\) perpendicular to both the vectors \(\rm\overrightarrow{a}\) and  \(\rm\overrightarrow{b}\) then \(\rm \vec c = \vec a \times \vec b\)

Unit vector, \(\rm \hat{c}= \frac{\overrightarrow{c}}{\left | \overrightarrow{c} \right |}\)  

Calculation :

Here the given vectors are, \(\overrightarrow{a}=\rm \hat{i}-2\hat{j}+3\hat{k}\)  and \(\overrightarrow{b}=\rm 2\hat{i}+3\hat{j}-\hat{k}\) , 

Let, vector \(\rm\overrightarrow{c}\) are perpendicular to both the vectors, 

So,  \(\rm \overrightarrow{c}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & 3\\ 2& 3 & -1 \end{vmatrix}\) 

⇒ \(\rm \overrightarrow{c}= \hat{i}(2-9)-\hat{j}(-1-6)+\hat{k}(3+4)\) 

⇒ \(\rm \overrightarrow{c}= -7\hat{i}+7\hat{j}+7\hat{k}\) 

∴ \(\rm \left | \overrightarrow{c} \right | = \sqrt{(-7)^{2}+7^{2}+7^{2}}\) = \(7\sqrt{3}\) 

Hence the unit vector perpendicular to both the vectors are, 

\(\rm \hat{c}= \frac{\overrightarrow{c}}{\left | \overrightarrow{c} \right |}\) = \(\rm\frac{-7\hat{i}+7\hat{j}+7\hat{k}}{7\sqrt{3}}\) 

⇒ \(\rm \hat{c}= \frac{1}{\sqrt{3}}\left ( -\hat{i}+\hat{j}+\hat{k} \right )\) 

The correct option is 1. 

Calculation:

\(\rm\overrightarrow{a}\) is a non-zero vector of modulus a and λ is a non-zero scalar and \(\rm\lambda\overrightarrow{a}\) is a unit vector i.e. \(\rm|\lambda \overrightarrow{a}|\) = 1

⇒ |λ|\(\rm|\overrightarrow{a}|\) = 1   (∵ |mn| = |m|⋅|n|)

⇒ |λ|⋅a = 1   (∵ \(\rm|\overrightarrow{a}|\) = a(given))

⇒ a = \(\frac{1}{|\lambda|}\)

The correct answer is option "3"

Concept:

If three vectors a, b and c are coplanar vectors, then

their scalar triple product is zero.

That is, a.(b × c) = 0

Calculation:

Given, 

The vectors 2î - 3ĵ + 4k̂, î + 2ĵ - k̂ and mî - ĵ + 2k̂ are coplanar

⇒  (2î - 3ĵ + 4k̂).[(î + 2ĵ - k̂) × (mî - ĵ + 2k̂)] = 0

⇒  (2î - 3ĵ + 4k̂).[(4  - 1)î - (2 + m)ĵ + (-1 - 2m) k̂] = 0

⇒  (2î - 3ĵ + 4k̂).(3î - (2 + m)ĵ - (1 + 2m) k̂) = 0

⇒  (2)(3) + (-3)(-(2 + m)) + (4)(-(1 + 2m)) = 0

⇒ 6 + 6 + 3m - 4 - 8m = 0

⇒ 8 - 5m = 0

⇒ m = 8/5 

∴ The correct answer is option (4).

CONCEPT:

Two vectors \(\vec a \ and \ \vec b\)are said to be equal if and only if they have same direction and same magnitude.

CALCULATION:

Given: \(\vec a = {a_1}\hat i + 3\hat j + {a_3}\hat k\;and\;\vec b = 2\hat i + {b_2}\hat j + \;\hat k\)  are two vectors such that \(\vec a = \vec b\)

∵ \(\vec a = \vec b\)

As we know that, two vectors \(\vec a \ and \ \vec b\)are said to be equal if and only if they have same direction and same magnitude.

So, in order to have same magnitude the corresponding coefficients of \(\hat i, \hat j \ and\ \hat k\) of both the vectors should be same.

⇒ a₁ = 2, b₂ = 3 and a₃ = 1

Hence, option C is the correct answer.

​Concept:

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

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

Collinear vectors: Two vectors are collinear if they lie on the same line or parallel lines.

If \(\rm \vec{a}\) and \(\rm \vec{b}\) are collinear vectors then \(\rm \vec{b} = λ \vec{a}\)​

Calculation:

Given:

vectors \(\rm \vec{a}, \vec{b}\) are collinear,

Therefore, \(\rm \vec{b} = λ \vec{a}\)

⇒ \(\rm \vec{b} = λ (2\hat{i} +6\hat{j} - 3\hat{k})\)

Given: magnitude of b = |b| = 14

⇒ \(|λ (2\hat{i} +6\hat{j} - 3\hat{k})| = 14\)

⇒ \(λ| (2\hat{i} +6\hat{j} - 3\hat{k})| = 14\)

⇒ \(λ \times \sqrt {2^2+6^2+(-3)^2} = 14\)

⇒ 7λ = 14

∴ λ = 2

Now, \(\rm \vec{b} = λ \vec{a} = λ (2\hat{i} +6\hat{j} - 3\hat{k}) = 2 (2\hat{i} +6\hat{j} - 3\hat{k})\)

Hence \(\rm \vec{b}= (4\hat{i} +12\hat{j} - 6\hat{k})\)

Calculation:

Given, \(\vec a,\;\vec b\) and \(\vec c\)   all unit vectors        ---(A)

And

\({\left| {\vec a + \vec b} \right|^2} = {\left| {\vec b + \vec c} \right|^2} = {\left| {\vec c + \vec a} \right|^2} = 8\)    ---(1)

Take,

\({\left| {\vec a + \vec b} \right|^2} = 8\)

\({\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + 2a \cdot b = 8\)

1 + 1 + 2 ab = 8 (ab \(\vec a\) is unit vector)

2ab = 6

ab = 3

take,

\({\left| {\vec b + \vec c} \right|^2} = 8\)

\({\left| {\vec b} \right|^2} + {\left| {\vec c} \right|^2} + 2bc = 8\)

b c = 3

Similarly, c a = 3

Take,

\(\vec x = \left| {2\vec a + \vec b + \vec c} \right|\)

On squaring both sides we get,

\({{\vec x}^2} = {\left| {2\vec a + \vec b + \vec c} \right|^2}\)

\({{\vec x}^2} = 4{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + {\left| {\vec c} \right|^2} + 2\left| {2a \cdot b + 2a \cdot c + b \cdot c} \right|\)

\({{\vec x}^2} = 4 + 1 +1+ 2\left( {6 + 6 + 3} \right)\)

\(\vec x = \sqrt {36} \) 

i.e |2â + b̂ + ĉ| = 6