Magnitude and Directions of a Vector MCQ Quiz in తెలుగు - Objective Question with Answer for Magnitude and Directions of a Vector - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Let \(\vec m = 3\vec x + 4\vec y - a\vec z\;and\;\vec n = 2b\vec x - 3\vec y\; + \;5\vec z\)

For two vectors to be collinear,

\(\vec m\; = \;\lambda \vec n\)

So, \(3\vec x\; + \;4\vec y - a\vec z = \lambda \left( {2b\vec x - 3\vec y\; + \;5\vec z} \right)\)

On comparing,

3 = 2λb, 4 = -3λ and -a = 5λ

λ = (-4)/3

a = 20/3

And b = (-9)/8

a – b = 20/3 - ((-9)/8)

Calculation:

Here, \(\left| {{\rm{\vec c}}} \right| = 7\)

\({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\\⇒ {\rm{\vec a}} + {\rm{\vec b}}=-{\rm{\vec c}}\\ \Rightarrow |{\rm{\vec a}}+{\rm{\vec b}}|=|{\rm{-\vec c}}|=​​​​|{\rm{\vec c}}|\)

∴\(\left| {{\rm{\vec a}} + {\rm{\vec b}}} \right| = 7\)

Hence, option (1) is correct. 

Concept:

\(\begin{aligned} |\vec{b}+\vec{c}|^{2} &=|\vec{b}|^{2}+|\vec{c}|^{2}+2 |\vec{b} |\cdot |\vec{c}|\cos θ \end{aligned}\)

Calculation:

Here, \(\left| {{\rm{\vec a}}} \right| = 3,{\rm{\;}}\left| {{\rm{\vec b}}} \right| = 5\) and \(\left| {{\rm{\vec c}}} \right| = 7\)

\({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\\⇒ {\rm{\vec b}} + {\rm{\vec c}}=-{\rm{\vec a}}\)

\(\begin{aligned} |\vec{b}+\vec{c}|^{2} &=|\vec{b}|^{2}+|\vec{c}|^{2}+2 |\vec{b} |\cdot |\vec{c}|\cos θ \end{aligned}\)

⇒\(\begin{aligned} |-\vec{a}|^{2} &=|\vec{b}|^{2}+|\vec{c}|^{2}+2 |\vec{b} |\cdot |\vec{c}|\cos θ \end{aligned}\)

⇒ 3² = 5² + 7² + 2(5)(7) cos θ 

⇒ 9 - 25 - 49 = 70 cos θ 

⇒ cos θ = (-65/70) = (-13/14)

Hence, option (4) is correct. 

Concept:

If \(\vec{\text{a}}\) is unit vector, then |\(\vec{\text{a}}\)| = 1

For any vector \(\vec{\text{a}}\), \(\vec{\text{a}}\).\(\vec{\text{a}}\) = |\(\vec{\text{a}}\)|²

For any vector \(\vec{\text{a}}\), \(\vec{\text{a}}\) × \(\vec{\text{a}}\) = 0

For any two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), \(\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{a}}\)

For any two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), \(\vec{\text{a}}×\vec{\text{b}}=-(\vec{\text{b}}×\vec{\text{a}})\)

If  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) are three coplanar vectors, then their scalar triple product is 0

\(\vec{\text{a}}.(\vec{\text{b}} \times \vec{\text{c}})= \vec{\text{b}}.(\vec{\text{c}} \times \vec{\text{a}})= \vec{\text{c}}.(\vec{\text{a}} \times \vec{\text{b}}) = 0\)

If any two vectors in scalar triple product are equal, then scalar triple product is 0.

Calculation:

Given,  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) be unit vectors lying on the same plane. 

⇒  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) are coplanar vectors. 

that is \(\vec{\text{a}}.(\vec{\text{b}} \times \vec{\text{c}})= \vec{\text{b}}.(\vec{\text{c}} \times \vec{\text{a}})= \vec{\text{c}}.(\vec{\text{a}} \times \vec{\text{b}}) = 0\)

and  |\(\vec{\text{a}}\)| = 1,  |\(\vec{\text{b}}\)| = 1 and  |\(\vec{\text{c}}\)| = 1

Now, \(\{(3\vec{\text{a}} + 2\vec{\text{b}}) × (5\vec{\text{a}} − 4\vec{\text{c}})\}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)

Using distributive property,

\(\{(3\vec{\text{a}}× 5\vec{\text{a}} + 2\vec{\text{b}} × 5\vec{\text{a}} − 3\vec{\text{a}}×4\vec{\text{c}}- 2\vec{\text{b}}×4\vec{\text{c}})\}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)

⇒ \(\{(15\vec{\text{a}}× \vec{\text{a}} + 10\vec{\text{b}} × \vec{\text{a}} − 12\vec{\text{a}}×\vec{\text{c}}- 8\vec{\text{b}}×\vec{\text{c}})\}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)

⇒ \(\{(15(0) + 10(\vec{\text{b}} × \vec{\text{a}}) − 12(\vec{\text{a}}×\vec{\text{c}})- 8(\vec{\text{b}}×\vec{\text{c}})\}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)     (∵  \(\vec{\text{a}}\) × \(\vec{\text{a}}\) = 0)

⇒ \(\{ 10(\vec{\text{b}} × \vec{\text{a}}) − 12(\vec{\text{a}}×\vec{\text{c}})- 8(\vec{\text{b}}×\vec{\text{c}})\}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)  

Again using distributive property,

⇒ \(\{ 10(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{b}} − 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}- 8(\vec{\text{b}}×\vec{\text{c}})⋅\vec{\text{b}}\} + \{ 20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}} − 24(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{c}}- 16(\vec{\text{b}}×\vec{\text{c}})⋅\vec{\text{c}}\}\) 

As, if any two vectors in scalar triple product are equal, then scalar triple product is 0.

⇒ \(\{ 10(0) − 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}- 8(0)\} + \{ 20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}} − 24(0)- 16(0)\}\) 

⇒ \(20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}}− 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}\) 

As,  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) are coplanar vectors. 

⇒ 20(0) - 12(0) = 0

∴ The correct option is (4).

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 \(\vec z = \hat i + 5\hat j + \lambda \hat k\) and \( \left | \vec z \right |= 6\)

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

⇒ \(\rm |\vec{z}| =\sqrt{(1^2+5^2 + λ^2)}\)

⇒ \(\rm 6 =\sqrt{1+25 + λ^2}\)

By squaring both the sides, we get

⇒ 36 = 26 + λ²

⇒ λ2 = 10 

Hence, option 1 is correct.

Concept:

If vector v = ai + bj + ck, then magnitude of vector v is \( \sqrt {a^2+b^2+c^2}\).

Two vectors a and b are said to be parallel vectors if one is a scalar multiple of the other. i.e., a = λb, where 'λ' is a scalar.

Formulae

(a × b) = -(b × a)

Calculation:

We have,

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

⇒ a × (3i + 2j + 4k) - (3i + 2j + 4k) × b = 0

⇒ a × (3i + 2j + 4k) + b × (3i + 2j + 4k) = 0             [(a × b) = -(b × a)]

⇒ (a + b) × (3i + 2j + 4k) = 0

So, (a + b) and (3i + 2j + 4k) are parallel vectors.

⇒ (a + b) = λ(3i + 2j + 4k)

⇒ \(\mid a+ b\mid\) = \(\mid λ \sqrt {3^2+2^2+4^2}\mid\)

⇒ \(\mid λ \sqrt {9+4+16}\mid=\sqrt {29}\)

⇒ \(\mid λ \sqrt {29}\mid=\sqrt {29}\)

⇒ λ = ± 1

a + b = ± (3i + 2j + 4k)

Now,

(a + b).(2i - 7j + 3k) =  ± (3i + 2j + 4k).(2i - 7j + 3k)

 = ± (6 - 14 + 12)

 = ± 4

∴ The possible value of (a + b).(2i - 7j + 3k) is 4.

Given:

\(\vec{a}=\frac{1}{{√ 3}}̂ i + \frac{1}{{√ 3}}̂ j + \frac{1}{{√ 3}}̂{k}\)

Formula:

The magnitude of vector aî + bĵ + ck̂ is given by √(a² + b² +c²)

Calculation:

Magnitude of \(\vec{a}=\frac{1}{{√ 3}}̂ i + \frac{1}{{√ 3}}̂ j + \frac{1}{{√ 3}}̂{k}\) 

= √[(1/√3)2 + (1/√3)2 + (1/√3)2]

= 1

Concept:

The unit vector in the direction of a vector \(\vec{a}\) is given by \(\frac{\vec{a}}{|\vec a|}\).

Calculation:

Let \(\vec a =\) î − 2ĵ + 2k̂

The unit vector in the direction of a vector \(\vec{a}\) is given by \(\frac{\vec{a}}{|\vec a|}\).

= \(\frac{\rm\hat{i}-2\rm\hat{j}+2\rm\hat{k}}{\sqrt{1+2^2+2^2}}\)

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

∴ Vector in the direction of the vector î − 2ĵ + 2k̂ that has magnitude 9 is

9 ×  \(\frac{\rm\hat{i}-2\rm\hat{j}+2\rm\hat{k}}{3}\)

= 3(î − 2ĵ + 2k̂)

The vector in the direction of the vector î − 2ĵ + 2k̂ that has magnitude 9 is  3(î − 2ĵ + 2k̂).

The correct answer is option 3.

Concept:

The direction cosines of the vector aî + bĵ + ck̂ are given by α = \(\rm \frac{a}{\sqrt{a^2+b^2+c^2}}\), β = \(\rm \frac{b}{\sqrt{a^2+b^2+c^2}}\) and γ = \(\rm \frac{c}{\sqrt{a^2+b^2+c^2}}\).

Calculation:

For the given vector î + 2ĵ - k̂, a = 1, b = 2 and  = -1.

The direction cosines of the vector are:

α = \(\rm \frac{1}{\sqrt{1^2+2^2+(-1)^2}}\), β = \(\rm \frac{2}{\sqrt{1^2+2^2+(-1)^2}}\) and γ = \(\rm \frac{-1}{\sqrt{1^2+2^2+(-1)^2}}\)

⇒ α = \(\rm \frac{1}{\sqrt{6}}\), β = \(\rm \frac{2}{\sqrt{6}}\) and γ = \(\rm \frac{-1}{\sqrt{6}}\)

CONCEPT:

If \(\vec a = \;{a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then magnitude of \(\vec a\) is given by: \(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2} \;\)

CALCULATION:

Given: \(\vec a = \;2\hat i + 2\hat j\ - 4\hat k \ and\;\vec b = 4\hat i + 2\hat j + 3 \hat k\)

Here, we have to find the value of \(\left| {2\vec a - \;\vec b} \right|\)

⇒ \(2\vec a - \vec b = 2(\;2\hat i + 2\hat j\ - 4\hat k \ ) - ( 4\hat i + 2\hat j + 3 \hat k)\)

⇒ \(2\vec a - \vec b = 2\hat j - 11\hat k\)