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

Calculation:

Given,

The vectors $\overrightarrow{a},\overrightarrow{b},(\overrightarrow{a}\times \overrightarrow{b})$ are unit vectors.

Since $\overrightarrow{a}$ and  $\overrightarrow{b}$ are unit vectors, we know:

$|\overrightarrow{a}|=1\text{and}|\overrightarrow{b}|=1$.

The magnitude of the cross product $(\overrightarrow{a}\times \overrightarrow{b})$ is given by:

$|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin \theta =1\times 1\times \sin \theta =\sin \theta$.

Since $(|\overrightarrow{a}\times \overrightarrow{b}|=1$, we have:

$\sin \theta =1$, so $\theta ={90}^{\circ }$, meaning $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular.

The dot product $(\overrightarrow{a}\cdot \overrightarrow{b})$ is:

$\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta =1\times 1\times \cos {90}^{\circ }=0$.

∴ The value of$(\overrightarrow{a}\cdot \overrightarrow{b})$ is 0.

Hence, the correct answer is Option 1.

Calculation:

Given,

The position vectors of points A, B, and C are $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ respectively, and $\overrightarrow{c}={\cos }^2\theta \overrightarrow{a}+{\sin }^2\theta \overrightarrow{b}$.

The expression to evaluate is: $(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{b}\times \overrightarrow{c})+(\overrightarrow{c}\times \overrightarrow{a})$.

First, substitute $\overrightarrow{c}$ into the equation:

$(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{b}\times ({\cos }^2\theta \overrightarrow{a}+{\sin }^2\theta \overrightarrow{b}))+({\cos }^2\theta \overrightarrow{a}+{\sin }^2\theta \overrightarrow{b})\times \overrightarrow{a}$.

Using the distributive property of the cross product:

$(\overrightarrow{a}\times \overrightarrow{b})+[(\overrightarrow{b}\times {\cos }^2\theta \overrightarrow{a})+(\overrightarrow{b}\times {\sin }^2\theta \overrightarrow{b})]+[({\cos }^2\theta \overrightarrow{a}\times \overrightarrow{a})+({\sin }^2\theta \overrightarrow{b}\times \overrightarrow{a})]$.

Since $\overrightarrow{b}\times \overrightarrow{b}=0$ and $\overrightarrow{a}\times \overrightarrow{a}=0$, we are left with:

$(\overrightarrow{a}\times \overrightarrow{b})+{\cos }^2\theta (\overrightarrow{b}\times \overrightarrow{a})+{\sin }^2\theta (-\overrightarrow{a}\times \overrightarrow{b})$.

Substitute $\overrightarrow{b}\times \overrightarrow{a}=-(\overrightarrow{a}\times \overrightarrow{b})$ into the expression:

$(\overrightarrow{a}\times \overrightarrow{b})+{\cos }^2\theta (-\overrightarrow{a}\times \overrightarrow{b})+{\sin }^2\theta (-\overrightarrow{a}\times \overrightarrow{b})$.

Factor out $\overrightarrow{a}\times \overrightarrow{b}$:

$\overrightarrow{a}\times \overrightarrow{b}[1-{\cos }^2\theta -{\sin }^2\theta ]$.

Since ${\cos }^2\theta +{\sin }^2\theta =1$, the expression becomes:

$\overrightarrow{a}\times \overrightarrow{b}[1-1]=0$.

∴ The final result is $\overrightarrow{0}$.

Hence, the correct answer is option 1. 

Calculation:

Given,

$3\overrightarrow{a}-4\overrightarrow{b}+\overrightarrow{c}=0$

$\overrightarrow{c}=4\overrightarrow{b}-3\overrightarrow{a}$

The vector$\overrightarrow{AB}$ is:

$\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}$

The vector $\overrightarrow{BC}$ is:

$\overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}$

Substituting $\overrightarrow{c}=4\overrightarrow{b}-3\overrightarrow{a}$:

$\overrightarrow{BC}=(4\overrightarrow{b}-3\overrightarrow{a})-\overrightarrow{b}$

$\overrightarrow{BC}=3\overrightarrow{b}-3\overrightarrow{a}$

Step 4: Now, $\overrightarrow{BC}=3(\overrightarrow{b}-\overrightarrow{a})$, which gives:

$AB:BC=1:3$

∴ The correct ratio is AB : BC = 1 : 3 , 

Hence, the correct answer is Option 2. 

Calculation:

Given,

The vector $\overrightarrow{d}=(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}$

Statement I: $\overrightarrow{d}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$.

We use the vector triple product identity: $(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}=(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b}\cdot \overrightarrow{c})\overrightarrow{a}$.

This shows that $\overrightarrow{d}$ is a linear combination of $\overrightarrow{a}$ and $\overrightarrow{b}$, hence $\overrightarrow{d}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$.

Therefore, Statement I is correct.

Statement II: $\overrightarrow{d}$ is perpendicular to $\overrightarrow{c}$.

To check this, compute the dot product $\overrightarrow{d}\cdot \overrightarrow{c}$. Using the vector triple product identity, we find:

$\overrightarrow{d}\cdot \overrightarrow{c}=(\overrightarrow{a}\cdot \overrightarrow{c})(\overrightarrow{b}\cdot \overrightarrow{c})-(\overrightarrow{b}\cdot \overrightarrow{c})(\overrightarrow{a}\cdot \overrightarrow{c})=0$,

which means $\overrightarrow{d}$ is perpendicular to $\overrightarrow{c}$.

Therefore, Statement II is correct.

∴ Both Statement I and Statement II are correct.

Hence, the correct answer is option  3. 

Calculation:

Given,

${\cos }^2(\alpha )+{\cos }^2(\beta )+{\cos }^2(\gamma )=1$

Using the identity ${\cos }^2(x)=1-{\sin }^2(x)$, we substitute:

$(1-{\sin }^2(\alpha ))+(1-{\sin }^2(\beta ))+(1-{\sin }^2(\gamma ))=1$

Simplifying the equation:

$3-({\sin }^2(\alpha )+{\sin }^2(\beta )+{\sin }^2(\gamma ))=1$

Rearrange to isolate the sine terms:

${\sin }^2(\alpha )+{\sin }^2(\beta )+{\sin }^2(\gamma )=2$

Now, calculate the dot product:

$\overrightarrow{a}\cdot \overrightarrow{b}={\sin }^2(\alpha )+{\sin }^2(\beta )+{\sin }^2(\gamma )=2$

∴ The value of $\overrightarrow{a}\cdot \overrightarrow{b}$is 2.

Hence, the correct answer is Option 4.

Concept:

Conditions of collinear vector:

• Three points with position vectors $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ are collinear if and only if the vectors $\left(\overrightarrow{a}-\overrightarrow{b}\right)$ and $\left(\overrightarrow{a}-\overrightarrow{c}\right)$ are parallel. ⇔ $\left(\overrightarrow{a}-\overrightarrow{b}\right)=\lambda \left(\overrightarrow{a}-\overrightarrow{c}\right)$
• If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

Solution:

We know that, If the points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) be collinear then $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

Given  $\hat{i}+2\hat{j}+3\hat{k}$, $\lambda \hat{i}+4\hat{j}+7\hat{k}$, $-3\hat{i}-2\hat{j}-5\hat{k}$ are collinear

∴ $\left|\begin{array}{ccc}1& 2& 3\\ \lambda & 4& 7\\ -3& -2& -5\end{array}\right|=0$

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

⇒ -6 + 10λ – 42 - 6λ + 36  = 0

⇒ 4λ = 12

∴ λ = 3

Concept:

Let $\overrightarrow{\mathrm{a}}=\mathrm{x}\overrightarrow{\mathrm{i}}+\mathrm{y}\overrightarrow{\mathrm{j}}+\mathrm{z}\overrightarrow{\mathrm{k}}$ then magnitude of the vector of a = $\left|\overrightarrow{\mathrm{a}}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}$

Calculation:

Let $\overrightarrow{\mathrm{a}}$ =  p(2î - ĵ + 2k̂)

Given, $\left|\overrightarrow{\mathrm{a}}\right|=3$

⇒ $\sqrt{4{\mathrm{p}}^2+{\mathrm{p}}^2+4{\mathrm{p}}^2}=3$

⇒ $\sqrt{9{\mathrm{p}}^2}=3$

⇒ 3p = 3

∴ p = 1

Concept:

Dot product of two vectors is defined as:

$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=\left|\mathrm{A}\right|\times \left|\mathrm{B}\right|\times \mathrm{c}\mathrm{o}\mathrm{s}\theta$

Cross/Vector product of two vectors is defined as:

$\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\left|\mathrm{A}\right|\times \left|\mathrm{B}\right|\times \mathrm{s}\mathrm{i}\mathrm{n}\theta \times \hat{\mathrm{n}}$

where θ is the angle between $\overrightarrow{\mathrm{A}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{B}}$

Calculation:

To Find: Value of $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{a}}$

Here angle between them is 0°

$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{a}}=\left|\mathrm{a}\right|\times \left|\mathrm{a}\right|\times \mathrm{s}\mathrm{i}\mathrm{n}0\times \hat{\mathrm{n}}=0$

Concept:

If $\overrightarrow{\mathrm{A}}=\mathrm{x}\hat{\mathrm{i}}-\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}$, then $|\overrightarrow{\mathrm{A}}|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}$

Calculation:

Given A = $5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}}$ and B = $\hat{\mathrm{i}}+3\hat{\mathrm{j}}-7\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{A}\mathrm{B}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}$

$\overrightarrow{\mathrm{A}\mathrm{B}}$ = $\hat{\mathrm{i}}+3\hat{\mathrm{j}}-7\hat{\mathrm{k}}-(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{A}\mathrm{B}}$ = $-4\hat{\mathrm{i}}+5\hat{\mathrm{j}}-11\hat{\mathrm{k}}$

Now $|\overrightarrow{\mathrm{A}\mathrm{B}}|=\sqrt{(-4{)}^2+5^2+(-11{)}^2}$

$|\overrightarrow{\mathrm{A}\mathrm{B}}|=\sqrt{16+25+121}$

$|\overrightarrow{\mathrm{A}\mathrm{B}}|=\sqrt{162}$ = 9√2

Concept:

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

Here, $5\hat{\mathrm{i}}-2\hat{\mathrm{j}},8\hat{\mathrm{i}}-3\hat{\mathrm{j}},\mathrm{a}\hat{\mathrm{i}}-12\hat{\mathrm{j}}$

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

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

 $$\begin{gathered} \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{\mathrm{a}-8} \\ \Rightarrow \frac{-1}{3}=\frac{-9}{\mathrm{a}-8} \end{gathered}$$

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.

Concept:

For two vectors $\overrightarrow{m}and\overrightarrow{n}$ to be collinear,​ $\overrightarrow{m}=\lambda \overrightarrow{n}$ 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 $\overrightarrow{m}and\overrightarrow{n}$ are collinear then $\overrightarrow{m}=\lambda \overrightarrow{n}$ where λ is a scalar.

⇒ $4\hat{i}+\hat{j}-3\hat{k}=\lambda \times (p\hat{i}+q\hat{j}-2\hat{k})$

⇒ $4\hat{i}+1\hat{j}-3\hat{k}=\lambda p\hat{i}+\lambda q\hat{j}-2\lambda \hat{k}$

⇒ 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 $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}and\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ then $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \theta$

Calculation:

Given: $\overrightarrow{a}=2\hat{i}-6\hat{j}-3\hat{k}$ and $\overrightarrow{b}=4\hat{i}+3\hat{j}-\hat{k}$

$\left|\overrightarrow{a}\right|=7,\left|\overrightarrow{b}\right|=\sqrt{26}and\overrightarrow{a}\cdot \overrightarrow{b}=-7$

$\Rightarrow \cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|}=\frac{-7}{7\times \sqrt{26}}=-\frac{1}{\sqrt{26}}$

$\Rightarrow {\sin }^2\theta =1-{\cos }^2\theta =1-\frac{1}{26}=\frac{25}{26}$

Concept:

Let the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$is $\theta$

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=2\mathrm{a}\mathrm{b}\mathrm{c}\mathrm{o}\mathrm{s}\theta$

Calculations:

consider, the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$is $\theta$

Given, $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$

⇒$\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$

⇒$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|-\overrightarrow{\mathrm{c}}|$

Squaring on both side, we get

⇒$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}{|}^2=|-\overrightarrow{\mathrm{c}}{|}^2$

⇒$|\overrightarrow{\mathrm{a}}{|}^2+2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{b}}{|}^2=|-\overrightarrow{\mathrm{c}}{|}^2$

⇒$|\overrightarrow{\mathrm{a}}{|}^2+|\overrightarrow{\mathrm{b}}{|}^2+2\mathrm{a}\mathrm{b}\cos \theta =|-\overrightarrow{\mathrm{c}}{|}^2$

⇒$(3){|}^2+(5{)}^2+2(3)(5)\cos \theta =(7{)}^2$

⇒$30\cos \theta =15$

⇒$\cos \theta ={\displaystyle \frac{1}{2}}$

⇒ $\theta$ = π / 3

Hence, If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},|\overrightarrow{a}|=3,|\overrightarrow{b}|=5$ and $|\overrightarrow{c}|=7$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$is π / 3

Concept:

If $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors parallel to each other then $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=0$

Calculation:

Given:

 $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}$ and $3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}}$ are parallel vectors,

Therefore, $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}=\lambda (3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}})$

Equating the coefficient of $\overrightarrow{\mathrm{i}},\overrightarrow{\mathrm{j}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{k}}$

⇒ 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:

 $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}},\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and c = î - ĵ - k̂

Given:  vector $\overrightarrow{\mathrm{v}}$ in the plane of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ 

Therefore, $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$

⇒ $\overrightarrow{\mathrm{v}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda (\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

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

Projection of $\overrightarrow{\mathrm{v}}$ on $\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}=\frac{1}{\sqrt{3}}$

⇒ $\overrightarrow{\mathrm{v}}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}=\frac{1}{\sqrt{3}}$ 

⇒ $\frac{(1+\lambda )-(1-\lambda )-(1+\lambda )}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

⇒ -(1 - λ) = 1

∴ λ = 2

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

$\overrightarrow{\mathrm{v}}$ = 3î - ĵ + 3k̂