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

Concept:

1. Circumcentre: The point of intersection of perpendicular bisectors of all the sides of a triangle is known as the circumcentre.

2. Orthocentre: The point of intersection of all the 3 altitudes of a triangle is known as the orthocentre.

3. Centroid: The point of intersection of all the 3 medians of a triangle is known as the centroid.

4. The orthocentre, centroid and circumcentre of any triangle are collinear. The centroid divides the distance from orthocentre to the circumcentre in the ratio 2 : 1.

Calculation:

Here, O is the circumcentre, G is the centroid and O’ is the orthocentre of a triangle ABC.

We know that, the centroid divides the distance from orthocentre to the circumcentre in the ratio 2 : 1.

i.e G divides the line segment joining O and O’ in the ratio of 2 : 1.

$\begin{array}{l}\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\left(\overrightarrow{A}-\overrightarrow{O}\right)+\left(\overrightarrow{B}-\overrightarrow{O}\right)+\left(\overrightarrow{C}-\overrightarrow{O}\right)\\ \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\left(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}\right)-3\times \overrightarrow{O}\end{array}$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=3\times \left[\left(\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}\right)-\overrightarrow{O}\right]$

$\begin{array}{l}\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=3\times \left[\overrightarrow{G}-\overrightarrow{O}\right]\\ 3\times \overrightarrow{G}=2\times \overrightarrow{O}+\overrightarrow{O^{\prime }}\end{array}$

Concept:

Magnitude of a vector:

Magnitude of a vector $\overline{a}=ai+bj+ck$ is given by $|\overline{a}|=\sqrt{a^2+b^2+c^2}$.

Area of triangle:

If the vectors $\overline{a}\text{ and }\overline{b}$ form adjacent sides of the triangle then the area of the triangle is given by: $A={\displaystyle \frac{1}{2}}|\overline{a}\times \overline{b}|$

Cross product of the vectors:

For two vectors $\overline{a}=ai+bj+ck$ and $\overline{b}=di+ej+fk$ the cross product is given by: $\overline{a}\times \overline{b}=\left|\begin{array}{ccc}i& j& k\\ a& b& c\\ d& e& f\end{array}\right|$

Calculation:

Let the vectors be $\overline{a}=-2i+3j+2k$ and $\overline{b}=-4i+5j+2k$.

First we will calculate the cross product as follow:

$\begin{array}{rl}\overline{a}\times \overline{b}& =\left|\begin{array}{ccc}i& j& k\\ -2& 3& 2\\ -4& 5& 2\end{array}\right|\\ & =i(6-10)-j(-4+8)+k(-10+12)\\ & =-4i-4j+2k\end{array}$

Therefore, the magnitude of the cross product is:

$\begin{array}{rl}|\overline{a}\times \overline{b}|& =\sqrt{16+16+4}\\ & =6\end{array}$

Using the formula for the area of the triangle, the area is given by:

$\begin{array}{rl}A& ={\displaystyle \frac{1}{2}}|\overline{a}\times \overline{b}|\\ & ={\displaystyle \frac{1}{2}}\times 6\\ & =3\end{array}$

Thus, the area of the triangle is 3 square units.

Concept:

Triangle Law of Vector Addition: Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

⇒$\overrightarrow{\mathrm{R}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Parallelogram law of vector addition: 

If two vectors are acting simultaneously at a point, then it can be represented both in magnitude and direction by the adjacent sides drawn from a point. Therefore, the resultant vector is completely represented both in direction and magnitude by the diagonal of the parallelogram passing through the point

$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Calculation:

Here, vector a and b are adjacent sides of parallelogram

a = 2î - 4ĵ + 5k̂ and b = î - 2ĵ - 3k̂.

Let vector c and d are digonals of parallelogram 

Using parallelogram law, 

$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

= (2î - 4ĵ + 5k̂) + (î - 2ĵ - 3k̂)

=.$3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathrm{k}}$

Now using triangle law,

$$\begin{gathered} \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \\ \Rightarrow \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \end{gathered}$$

= (2î - 4ĵ + 5k̂) - (î - 2ĵ - 3k̂)

= $\hat{\mathrm{i}}-2\hat{\mathrm{j}}+8\hat{\mathrm{k}}$

$$\begin{gathered} \overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{d}}=(3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathrm{k}}).(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+8\hat{\mathrm{k}}) \\ =3+12+16 \\ =31 \end{gathered}$$

Hence, option (3) is correct.

Concept:

Referring to the diagram,

From the law of addition of vectors,

$\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}$ and $\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}$

Calculation:

$2\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{CD}$

$\overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}[\overrightarrow{BD}=-\overrightarrow{CD}]$

$\overrightarrow{AD}=\frac{(3+5)i+(0-2)j+(4+4)k}{2}=4i-j+4k$

$\overrightarrow{|AD|}=\sqrt{16+16+1}=\sqrt{33.}$

Concept:

If two points A and B have position vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ respectively, then the vector $\overrightarrow{\mathrm{A}\mathrm{B}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}$.

For two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ at an angle θ to each other:

• Dot Product is defined as: $\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}|\cos \theta$.
• Resultant Vector is equal $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$.
• Work: The work (W) done by a force ($\overrightarrow{\mathrm{F}}$) in moving (displacing) an object along a vector $\overrightarrow{\mathrm{D}}$ is given by: W = $\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{D}}=|\overrightarrow{\mathrm{F}}||\overrightarrow{\mathrm{D}}|\cos \theta$.

Calculation:

Let's say that the forces acting on the particle are $\overrightarrow{\mathrm{P}}$ = 2î - 5ĵ + 6k̂ and $\overrightarrow{\mathrm{Q}}$ = -î + 2ĵ - k̂.

∴ The resulting force acting on the particle will be $\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}$.

⇒ $\overrightarrow{\mathrm{F}}$ = (2î - 5ĵ + 6k̂) + (-î + 2ĵ - k̂)

⇒ $\overrightarrow{\mathrm{F}}$ = î - 3ĵ + 5k̂.

Since the particle is moved from the point 4î - 3ĵ - 2k̂ to the point 6î + ĵ - 3k̂, the displacement vector $\overrightarrow{\mathrm{D}}$ will be:

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

= (6î + ĵ - 3k̂) - (4î - 3ĵ - 2k̂)

⇒ ​$\overrightarrow{\mathrm{D}}$ = 2î + 4ĵ - k̂.

And finally, the work done W will be:

W = $\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{D}}$ = (î - 3ĵ + 5k̂).(2î + 4ĵ - k̂)

⇒ W = (1)(2) + (-3)(4) + (5)(-1)

⇒ W = 2 - 12 - 5 =

∴ -15 units.

Concept:

Calculations:

Given, triangle whose vertices are at A = (3, -1, 2) = $3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}}$

B = (1, -1, -3) =  $\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}-3\overrightarrow{\mathrm{k}}$

and C = (4, -3, 1) =  $4\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}$

Let A, B , and C be the vertices of the $\mathrm{△}ABC$ , then the area of that triangle = ${\displaystyle \frac{1}{2}}\times |\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}|$

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

⇒$\overrightarrow{\mathrm{A}\mathrm{B}}=-2\overrightarrow{\mathrm{i}}+0\overrightarrow{\mathrm{k}}-5\overrightarrow{\mathrm{k}}$

$\overrightarrow{\mathrm{A}\mathrm{C}}=(4\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})-(3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}})$

⇒ $\overrightarrow{\mathrm{A}\mathrm{C}}=\overrightarrow{\mathrm{i}}-2\overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}}$

Now, $\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}=$$\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ -2& 0& -5\\ 1& -2& -1\end{array}\right|$

⇒$\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}=\overrightarrow{\mathrm{i}}(-10)-\overrightarrow{\mathrm{j}}(2+5)+\overrightarrow{\mathrm{k}}(4-0)$

⇒$\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}=-10\overrightarrow{\mathrm{i}}-7\overrightarrow{\mathrm{j}}+4\overrightarrow{\mathrm{k}}$

⇒ $|\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}|=\sqrt{(-10{)}^2+(-7{)}^2+(4{)}^2}$

⇒$|\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}|=\sqrt{165}$

The area of that triangle = ${\displaystyle \frac{1}{2}}\times |\overrightarrow{AB}\times \overrightarrow{AC}|$

⇒ The area of that triangle = ${\displaystyle \frac{1}{2}}\times \sqrt{165}$
Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is ${\displaystyle \frac{\sqrt{165}}{2}}$

Concept:

I. If a particle is displace from point A to B under the influence of force $\overrightarrow{F}$. Then the work done in displacing the particle from point A to B is given by: $W=\overrightarrow{F}\cdot \overrightarrow{d}$

II. 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 $\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

Calculation:

Given: Particle is displace from point A = (2, - 6, 1) to the point B = (5, 9, 7) under the influence of force $\overrightarrow{F}=2\hat{i}+\hat{j}+\hat{k}$

So, the displacement of particle is given by:

$\overrightarrow{d}=\overrightarrow{AB}=\left(5\hat{i}+9\hat{j}+7\hat{k}\right)-\left(2\hat{i}-6\hat{j}+\hat{k}\right)=3\hat{i}+15\hat{j}+6\hat{k}$

As we know that, If a particle is displace from point A to B under the influence of force $\overrightarrow{F}$. Then the work done in displacing the particle from point A to B is given by: $W=\overrightarrow{F}\cdot \overrightarrow{d}$

$W=\overrightarrow{F}\cdot \overrightarrow{d}=\left(2\hat{i}+\hat{j}+\hat{k}\right)\cdot \left(3\hat{i}+15\hat{j}+6\hat{k}\right)=6+15+6=27units$

Calculation

$\overrightarrow{AC}=5\hat{i}-1\hat{j}+7\hat{k}$

$\overrightarrow{BD}=\hat{i}+2\hat{j}+3\hat{k}$

Area = |AC × BD|

⇒ $\left|\begin{array}{ccc}\hat{i}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 5& -1& 7\\ 1& 2& 3\end{array}\right|$

⇒ $\frac{1}{2}|-17\hat{\mathrm{i}}-8\hat{\mathrm{j}}+11\hat{\mathrm{k}}|=\frac{1}{2}\sqrt{474}$

Hence option(2} is correct

Concept:

The work done $\mathrm{W}=(\overrightarrow{{\mathrm{F}}_{\mathrm{r}}})\cdot (\overrightarrow{\mathrm{d}\mathrm{x}})$ 

where $\overrightarrow{{\mathrm{F}}_{\mathrm{r}}}$ is the resultant force, and  $\overrightarrow{\mathrm{d}\mathrm{x}}$ is the displacement 

The unit vector in the direction of a $\overrightarrow{\mathrm{P}}=\hat{\mathrm{P}}=\frac{\overrightarrow{\mathrm{P}}}{\left|\overrightarrow{\mathrm{P}}\right|}$ 

Calculation:

The initial position of the particle (A) = 2i - j - 3k

The initial position of the particle (B) = 5i - j + k

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

First force F₁ = magnitude of the 1^st force × unit vector in the given direction

⇒ $\overrightarrow{{\mathrm{F}}_1}=5\times \frac{6\mathrm{i}+2\mathrm{j}+3\mathrm{k}}{|6\mathrm{i}+2\mathrm{j}+3\mathrm{k}|}$ 

⇒ $\text{boldsymbol}\overrightarrow{{\mathrm{F}}_1}=\frac{30\mathrm{i}+10\mathrm{j}+15\mathrm{k}}{7}$ 

Second force F₂ = magnitude of the 2^nd force × unit vector in the given direction

⇒ $\overrightarrow{{\mathrm{F}}_2}=3\times \frac{3\mathrm{i}-2\mathrm{j}+6\mathrm{k}}{|3\mathrm{i}-2\mathrm{j}+6\mathrm{k}|}$ 

⇒ $\text{boldsymbol}\overrightarrow{{\mathrm{F}}_2}=\frac{9\mathrm{i}-6\mathrm{j}+18\mathrm{k}}{7}$ 

Third force F₃ = magnitude of the 3^rd force × unit vector in the given direction

⇒ $\overrightarrow{{\mathrm{F}}_3}=1\times \frac{2\mathrm{i}-3\mathrm{j}-6\mathrm{k}}{|2\mathrm{i}-3\mathrm{j}-6\mathrm{k}|}$ 

⇒ $\text{boldsymbol}\overrightarrow{{\mathrm{F}}_3}=\frac{2\mathrm{i}-3\mathrm{j}-6\mathrm{k}}{7}$ 

Resultant Force $\overrightarrow{{\mathrm{F}}_{\mathrm{r}}}=\overrightarrow{{\mathrm{F}}_1}+\overrightarrow{{\mathrm{F}}_2}+\overrightarrow{{\mathrm{F}}_3}$ 

⇒ $\overrightarrow{{\mathrm{F}}_{\mathrm{r}}}=\frac{30\mathrm{i}+10\mathrm{j}+15\mathrm{k}}{7}+\frac{9\mathrm{i}-6\mathrm{j}+18\mathrm{k}}{7}+\frac{2\mathrm{i}-3\mathrm{j}-6\mathrm{k}}{7}$

⇒ $\text{boldsymbol}\overrightarrow{{\mathrm{F}}_{\mathrm{r}}}=\frac{41\mathrm{i}+\mathrm{j}+27\mathrm{k}}{7}$ 

The work done by the forces:

$\mathrm{W}=(\overrightarrow{{\mathrm{F}}_{\mathrm{r}}})\cdot (\overrightarrow{\mathrm{d}\mathrm{x}})$

⇒ $\mathrm{W}=\left(\frac{41\mathrm{i}+\mathrm{j}+27\mathrm{k}}{7}\right)\cdot (3\mathrm{i}+4\mathrm{k})$ 

⇒ $\mathrm{W}=\left(\frac{123+0+108}{7}\right)=\frac{231}{7}$ 

⇒ $\text{boldsymbol}\mathrm{W}=33$ units

CONCEPT:

If vectors $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ are coplanar then $\left[abc\right]=0$ where $\left[abc\right]=\overrightarrow{a}\cdot \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

CALCULATION:

Given: Vectors $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}and\overrightarrow{c}=3\hat{i}+\lambda \hat{j}+5\hat{k}$ are coplanar

As we know that, if vectors $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ are coplanar then $\left[abc\right]=0$ where $\left[abc\right]=\overrightarrow{a}\cdot \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

Here, a1 = 2, b1 = - 1, c1 = 1

Similarly, a2 = 1, b2 = 2, c2 = 3

Similarly, a3 = 3, b3 = λ, c3 = 5

$\Rightarrow [abc]=\left|\begin{array}{ccc}2& 1& 3\\ -1& 2& \lambda \\ 1& 3& 5\end{array}\right|=10-5\lambda$

∵ The given vectors are coplanar

⇒ 10 - 5λ = 0

⇒ λ = 2

Hence, option A is the correct answer.