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

Calculation:

Given, \(\vec a=\vec b\times \vec c\)

⇒ \(\vec a\) is perpendicular to \(\vec c\)

⇒ \(\vec{\mathrm{a}} \cdot \vec{\mathrm{c}}\) = 0

∴ \(|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|^2=|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\)

= \(|\overrightarrow{\mathrm{c}}|^2+4-0\)

Now, \(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}\)

⇒ \(|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|\)

⇒ \(2=3|\overrightarrow{\mathbf{c}}| \sin \alpha\)

⇒ \(|\overrightarrow{\mathrm{c}}|=\frac{2}{3} \operatorname{cosec} \alpha \quad \alpha\in\left[0,\frac{\pi}{3}\right]\)

⇒ \(|\overrightarrow{\mathrm{c}}|_{\min }=\frac{2}{3} \times \frac{2}{\sqrt{3}} \quad \operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \infty\right)\)

⇒ \(27|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|_{\min }^2=27\left(\frac{16}{27}+4\right)=124\)

∴ The minimum value of \(\rm 27|\vec c-\vec a|^2\) is equal to 124.

The correct answer is Option 3.

Concept:

The magnitude of a vector is the length of the vector. The magnitude of the vector p is denoted as \(∣ \overrightarrow{p}∣ = \sqrt{a^2+b^2+c^2}\)

The unit vector in the direction of \(\overrightarrow{p}\) is \(\hat p=\frac{1}{∣ p∣ }\overrightarrow{p}\)

Calculation:

Statement I: Sum of the two vectors is i + j + k

Let \(\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}\). Calculating its value, we have

\(\overrightarrow{c}=(2i-j+2k)+(-i+j-k)\)

 = i + k

Statement I is incorrect.

Statement II: If \(\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}\), then \(∣ \overrightarrow{c}∣ \)

\(\overrightarrow{c}=i+k\)

\(∣ \overrightarrow{c}∣ = \sqrt{1^2+0^2+1^2}=\sqrt{1+0+1} =\sqrt2 \)

Statement II is correct.

Statement II: Unit vector in the direction of  is \(\overrightarrow{a}+\overrightarrow{b}\) is \(\frac{i}{\sqrt 2 }+\frac{k}{\sqrt 2 }\)

We know that, the unit vector in the direction of \(\overrightarrow{c}\) is \(\hat c=\frac{1}{∣ c∣ }\overrightarrow{c}\)

So, \(\hat c=\frac{1}{\sqrt 2 }{(i+k)}\)

\(\hat c=\frac{i}{\sqrt 2 }+\frac{k}{\sqrt 2 }\)

Statement III is correct.

∴ Statements II and III are correct.

Concept:

The magnitude of a vector is the length of the vector. The magnitude of the vector p is denoted as \(∣ \overrightarrow{p}∣ = \sqrt{a^2+b^2+c^2}\)

The unit vector in the direction of \(\overrightarrow{p}\) is \(\hat p=\frac{1}{∣ p∣ }\overrightarrow{p}\)

Calculation:

Statement I: Sum of the two vectors is i + j + k

Let \(\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}\). Calculating its value, we have

\(\overrightarrow{c}=(2i-j+2k)+(-i+j-k)\)

 = i + k

Statement I is incorrect.

Statement II: If \(\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}\), then \(∣ \overrightarrow{c}∣ \)

\(\overrightarrow{c}=i+k\)

\(∣ \overrightarrow{c}∣ = \sqrt{1^2+0^2+1^2}=\sqrt{1+0+1} =\sqrt2 \)

Statement II is correct.

Statement II: Unit vector in the direction of  is \(\overrightarrow{a}+\overrightarrow{b}\) is \(\frac{i}{\sqrt 2 }+\frac{k}{\sqrt 2 }\)

We know that, the unit vector in the direction of \(\overrightarrow{c}\) is \(\hat c=\frac{1}{∣ c∣ }\overrightarrow{c}\)

So, \(\hat c=\frac{1}{\sqrt 2 }{(i+k)}\)

\(\hat c=\frac{i}{\sqrt 2 }+\frac{k}{\sqrt 2 }\)

Statement III is correct.

∴ Statements II and III are correct.

Concept:

Vector Law of Addition:

Explanation:

Applying the triangle law of addition

\(⇒ \overrightarrow{PA}=\overrightarrow{PC}+\overrightarrow{CA}\)      -----(1)

\(⇒ \overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{CB}\)      --------(2)

Adding (1) and (2), We get,

⇒ \(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{CA}+\overrightarrow{PC}+\overrightarrow{CB}\)

⇒ \(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{PC}+(\overrightarrow{CA}+\overrightarrow{CB})\)

Since, \(\overrightarrow{CA}=-\overrightarrow{CB}\)

⇒ \(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{PC}\)   

∴ \(\overrightarrow{PA}-\overrightarrow{PC}=\overrightarrow{PC}-\overrightarrow{PB}\)

Calculation:

\(\vec{u} = 3\hat{i} - \hat{j}, \quad \vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}\)

\(\Rightarrow \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} = \cos \theta\)

\(\Rightarrow \frac{5}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}}\)

\(\Rightarrow \lambda^2 = 9 \quad \Rightarrow \lambda = 3 \quad (\because \lambda > 0)\)

\(\vec{v} = \vec{v}_1 + \vec{v}_2\)

\(\Rightarrow |\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 + 2 \vec{v}_1 \cdot \vec{v}_2\)

\(\Rightarrow 14 = |\vec{v}_1|^2 + |\vec{v}_2|^2 + 0 \quad (\because \vec{v}_1 \perp \vec{v}_2)\)

\(\Rightarrow |\vec{v}_1|^2 + |\vec{v}_2|^2 = 14\)

Hence, the correct answer is Option 2.

Calculation:

Given, \(\vec a=\vec b\times \vec c\)

⇒ \(\vec a\) is perpendicular to \(\vec c\)

⇒ \(\vec{\mathrm{a}} \cdot \vec{\mathrm{c}}\) = 0

∴ \(|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|^2=|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\)

= \(|\overrightarrow{\mathrm{c}}|^2+4-0\)

Now, \(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}\)

⇒ \(|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|\)

⇒ \(2=3|\overrightarrow{\mathbf{c}}| \sin \alpha\)

⇒ \(|\overrightarrow{\mathrm{c}}|=\frac{2}{3} \operatorname{cosec} \alpha \quad \alpha\in\left[0,\frac{\pi}{3}\right]\)

⇒ \(|\overrightarrow{\mathrm{c}}|_{\min }=\frac{2}{3} \times \frac{2}{\sqrt{3}} \quad \operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \infty\right)\)

⇒ \(27|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|_{\min }^2=27\left(\frac{16}{27}+4\right)=124\)

∴ The minimum value of \(\rm 27|\vec c-\vec a|^2\) is equal to 124.

The correct answer is Option 3.