Properties of Vectors MCQ Quiz in తెలుగు - Objective Question with Answer for Properties of Vectors - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Let \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

Calculation:

Let \(\vec{u}~and~\vec{v}\) be two non-zero vectors.

Given: The magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference.

\(\Rightarrow \left| \vec{u}+~\vec{v} \right|=~\left| \vec{u}-~\vec{v} \right|\)

\(\Rightarrow {{\left| \vec{u}+~\vec{v} \right|}^{2}}=~{{\left| \vec{u}-~\vec{v} \right|}^{2}}\)

As we know that, if \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

\(\Rightarrow \left( \vec{u}+~\vec{v} \right)\cdot \left( \vec{u}+~\vec{v} \right)=\left( \vec{u}-~\vec{v} \right)\cdot \left( \vec{u}-~\vec{v} \right)\)

\(\Rightarrow {{\left| {\vec{u}} \right|}^{2}}+{{\left| {\vec{v}} \right|}^{2}}+2~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta =~{{\left| {\vec{u}} \right|}^{2}}+{{\left| {\vec{v}} \right|}^{2}}-2~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta \)

\(\Rightarrow 4\times ~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta =0\)

∵ \(\vec{u}~and~\vec{v}\) be two non-zero vectors ⇒ \(\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\ne 0\)

⇒ cos θ = 0 ⇒ θ = 90°

Concept:

Conditions of collinear vector:

• Three points with position vectors \(\vec a,\;\vec b\;and\;\vec c\) are collinear if and only if the vectors \(\left( {\vec a - \vec b} \right)\) and \(\left( {\vec a\; - \vec c} \right)\) are parallel. ⇔ \(\left( {\vec a - \vec b} \right) = \lambda \left( {\vec a\; - \vec c} \right)\)
• If the points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Calculation:

Let the given points be A (−1, −1, 2), B (2, K, 5), C (3, 11, 6).

Then \(\overrightarrow {{\rm{AB}}} = \left( {2 + 1} \right){\rm{\vec i}} + \left( {{\rm{k}} + 1} \right){\rm{\vec j}} + \left( {5 - 2} \right){\rm{\vec k}} = 3{\rm{\vec i}} + {\rm{\;}}\left( {{\rm{k}} + 1} \right){\rm{\vec j\;}} + 3{\rm{\vec k}}\)

And \(\overrightarrow {{\rm{AC}}} = \left( {3 + 1} \right){\rm{\vec i}} + \left( {3 + 1} \right){\rm{\vec j}} + \left( {2 - 2} \right){\rm{\vec k}} = 4{\rm{\vec i}} + {\rm{\;}}4{\rm{\vec j\;}} + 4{\rm{\vec k}}\)

Now, if A, B and C are collinear, then \(\overrightarrow {{\rm{AB}}} {\rm{\;}} = {\rm{\;\lambda \;}}\overrightarrow {{\rm{AC}}} \)

\(\Rightarrow 3{\rm{\vec i}} + {\rm{\;}}\left( {{\rm{k}} + 1} \right){\rm{\vec j\;}} + 3{\rm{\vec k}} = \;{\rm{\lambda \;}}\left( {4{\rm{\vec i}} + {\rm{\;}}4{\rm{\vec j\;}} + 4{\rm{\vec k}}} \right)\)

Comparing the coefficient of vector i, we get

⇒ 3 = 4 λ

∴ λ = 3/4

Now, comparing the coefficient of vector j, we get

⇒ (k + 1) = 4 λ

⇒ k + 1 = 4 × (3/4)

⇒ k + 1 = 3

∴ k = 2

Alternate solution:

We know that, If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Given (-1, -1, 2), (2, k, 5) and (3, 3, 6) are collinear

∴ \(\left| {\begin{array}{*{20}{c}} { - 1}&{ - 1}&2\\ 2&k&5\\ 3&3&6 \end{array}} \right| = 0\)

⇒ -1 (6k – 15) – (-1) (12 – 15) + 2 (6 – 3k) = 0

⇒ -6k + 15 – 3 + 12 – 6k = 0

⇒ 12k = 24

∴ k = 2

Dot product of two vectors \( = \vec a \cdot \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\) cos α

If \(\vec a\) and \(\vec b\) are non zero vectors than if Dot product is to be 0 then.

Cos α = 0

α = 90°

\( \Rightarrow \vec a \bot \vec b\)

Concept:

Area of parallelogram determined by the the vectors \(\vec a\) and \(\vec b\) is |\(\vec a\) × \(\vec b\)|.

Explanation:

Given

\(\vec a\) = î + 2ĵ +3k̂ and \(\vec b\) = 3î - 2ĵ + k̂

So, 

\(\vec a\) × \(\vec b\) = \(\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\3&-2&1\end{vmatrix}\)

    = î(2 + 6) + ĵ(9 - 1) + k̂(-2 - 6) = 8î + 8ĵ - 8k̂

Hence area of the parallelogram 

= |\(\vec a\) × \(\vec b\)| = \(\sqrt{8^2+8^2+(-8)^2}\) = \(\sqrt{64+64+64}\)  = 8√3

Option (1) is true.

Concept:

The position vector of the point that divides the line joining position vectors

\(\rm\vec{p}\) and \(\rm\vec{q}\) in the ratio m:n is given by \(\frac{m\vec{q}+n\vec{p}}{m+n}\).

Calculation:

Given position vectors are  \(2\rm \vec{a}-3\vec{b}\)  and  \( \rm \vec{a}+\vec{b}\) .

∴ The position vector of the point which divides the line joining the above points in the ratio 3: 1 is,

= \(\frac{(2\vec{a}-3\vec{b}) +3(\vec{a}+\vec{b})}{1+3}\)

=  \(\frac{5\vec{a}}{4}\)

The position vector of the point which divides the join of points \(\rm2\vec{a}−3\vec{b}\) and \(\rm\vec{a}+\vec{b}\) in the ratio 3 ∶ 1 is \(\frac{5\vec{a}}{4}\).

The correct answer is option 4.

Concept:

For three vectors \(\rm \vec A\), \(\rm \vec B\) and \(\rm \vec C\) to be co-planar, the volume of the parallelepiped formed by them must be 0.​ i.e. \(\rm [\vec A\ \vec B\ \vec C]\) = 0.

Triple Scalar Product (Box Product): is defined as: \(\rm [\vec A\ \vec B\ \vec C]=\vec A.(\vec B\times\vec C)=\begin{vmatrix} \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3 \\\rm c_1 & \rm c_2 & \rm c_3 \end{vmatrix}\).

Calculation:

Let the three vectors be \(\rm \vec A=2\hat i - \hat j + \hat k\), \(\rm \vec B=\hat i +2 \hat j -3 \hat k\) and \(\rm \vec C=3\hat i +\lambda \hat j + 5\hat k\). For the three vectors to be co-planar, their Box Product must be 0.

⇒ \(\rm [\vec A\ \vec B\ \vec C]=0\)

⇒ \(\rm \begin{vmatrix} 2& -1 &\ \ \ 1 \\ \rm 1 &\ \ \ 2 & -3 \\ \rm 3 &\ \ \ \lambda & \ \ \ 5 \end{vmatrix}=0\)

⇒ 2[(2)(5) - (-3)(λ)] - (-1)[(5)(1) - (3)(-3)] + 1[(1)(λ) - (2)(3)] = 0

⇒ 2(10 + 3λ) + 1(5 + 9) + 1(λ - 6) = 0

⇒ 20 + 6λ + 14 + λ - 6 = 0

⇒ 7λ = -28

⇒ λ = -4.

Additional Information

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

• Dot Product is defined as \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
• Cross Product is defined as \(\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta\) where \(\rm \vec n\) is the unit vector perpendicular to the plane containing \(\rm \vec A\) and \(\rm \vec B\).

Volume of a parallelepiped, with vectors \(\rm \vec A\), \(\rm \vec B\) and \(\rm \vec C\) as its sides, is given by the box product of the three vectors.

• Volume = \(\rm [\vec A\ \vec B\ \vec C]\).

For three vectors \(\rm \vec A\), \(\rm \vec B\) and \(\rm \vec C\):

• Triple Cross Product: is defined as: \(\rm \vec A\times(\vec B\times\vec C)=(\vec A.\vec C)\vec B-(\vec A.\vec B)\vec C\).

Concept:

Diagonals of a parallelogram bisect each other
Calculation:

Given:

Let A(1,1,1), B(1,3,5), C(7,9,11) and D(x,y,z) be the vertices of a parallelogram

As we know,

Diagonals of a parallelogram bisect each other

∴ midpoint of AC = midpoint of BD

\(\left( {\frac{{1 + 7}}{2},\frac{{1 + 9}}{2},\frac{{1 + 11}}{2}} \right) = \left( {\frac{{1 + x}}{2},\frac{{3 + y}}{2},\frac{{5 + z}}{2}} \right)\)

Comparing both sides, we get

1 + 7 = 1 + x

x = 7

And, 1 + 9 = 3 + y

y = 7

And, 1 + 11 = 5 + z

z = 7

∴ Position vector of fourth vertex is 7(i + j + k)

Concept:

• \(\vec a\) and \(\vec b\) are two vectors perpendicular to each other
• ∴ \(\vec a.{\rm{\;}}\vec b = 0\)
• \(\vec a\) and \(\vec b\) are two vectors parallel to each other
• ∴ \(\vec a \times \vec b = 0\)

Calculation:

Given:

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

\( \Rightarrow \;{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + 2\vec a.{\rm{\;}}\vec b = \;{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2}\;\)

\(\Rightarrow 2\vec a.{\rm{\;}}\vec b = 0\)

\(\Rightarrow \vec a.{\rm{\;}}\vec b = 0\)

∴ \(\vec a\) and \(\vec b\) are perpendicular.

Concept:

One algebraic property of real numbers is the distributive law. The distributive law for the real numbers says: "For all real numbers x, y, and z, \(x.( y+ z)=x. y+ x.z\)

The vector dot product is distributive over addition. In general: \(\vec a.( \vec b+ \vec c)=\vec a. \vec b+ \vec a.\vec c\)  

The vector cross product is distributive over addition. In general: \(\vec a × (\vec b + \vec c) = \vec a × \vec b + \vec a × \vec c\)

Associative property: (p × q) × r = p × (q × r) (where p, q, and r are any three natural/whole numbers)

Calculation:

Let

\(\displaystyle \vec a = a_x \overline i+a_y \overline j+a_z \overline k\\\vec b = b_x \overline i+b_y \overline j+b_z \overline k\\\vec c = c_x \overline i+c_y \overline j+c_z \overline k\)

Statement I: Dot product over vector addition is distributive

We have to prove \(\vec a.( \vec b+ \vec c)=\vec a. \vec b+ \vec a.\vec c\)

• \(\displaystyle \vec a.( \vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k).[(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(c_x ̂ i+c_y ̂ j+c_z ̂ k)] \)
• \(\displaystyle \vec a.( \vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k).[(b_x+c_x)̂ i+(b_y+c_y)̂ j+(b_z +c_z )̂ k] \)
• \(\vec a.( \vec b+ \vec c)=\) a_x (b_x + c_x) + a_y (b_y + c_y) + a_z (b_z + c_z)
• \(\vec a.( \vec b+ \vec c)=\) ax bx + ax cx + ay by + ay cy + az bz + az cz................................... (1)
• \(\displaystyle \vec a. \vec b+ \vec a.\vec c=(a_x ̂ i+a_y ̂ j+a_z ̂ k).(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(a_x ̂ i+a_y ̂ j+a_z ̂ k).(c_x ̂ i+c_y ̂ j+c_z ̂ k)\)
• \(\displaystyle \vec a. \vec b+ \vec a.\vec c=(a_x.b_x+a_y.b_y +a_z.b_z)+(a_x.c_x +a_y.c_y+a_z.c_z)\)
• \(\vec a. \vec b+ \vec a.\vec c=\) ax bx + ax cx + ay by + ay cy + az bz + az cz................................... (2) 
• From equation (1) and (2)
• ∴ \(\vec a.( \vec b+ \vec c)=\vec a. \vec b+ \vec a.\vec c\)  

Statement II: Cross product over vector addition is distributive

We have to prove \(\vec a × (\vec b + \vec c) = \vec a × \vec b + \vec a × \vec c\)

• \(\displaystyle \vec a\times(\vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times[(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(c_x ̂ i+c_y ̂ j+c_z ̂ k)] \)
• ​\(\displaystyle \vec a\times( \vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times[(b_x+c_x)̂ i+(b_y+c_y)̂ j+(b_z +c_z )̂ k] \)​
• \(\vec a\times( \vec b+ \vec c)=\begin{bmatrix} \overline i & \overline j & \overline k \\[0.3em] a_x & a_y & a_z \\[0.3em] b_x+c_x &b_y+c_y & b_z+c_z \end{bmatrix}\)
• \(\vec a\times( \vec b+ \vec c)=\) ̂̂î [ay (bz + cz) - az (by + cy)] - ĵ [ax (bz + cz) - az (bx + cx)] + k̂ [ax (by + cy) - ay (bx + cx)] ............(3) 
• \(\displaystyle \vec a\times \vec b+ \vec a\times\vec c=(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times(c_x ̂ i+c_y ̂ j+c_z ̂ k)\)
• \(\displaystyle \vec a\times \vec b+ \vec a\times\vec c=\begin{bmatrix} ̂ i & ̂ j & ̂ k \\[0.3em] a_x & a_y & a_z \\[0.3em] b_x &b_y& b_z \end{bmatrix}+\begin{bmatrix} ̂ i & ̂ j & ̂ k \\[0.3em] a_x & a_y & a_z \\[0.3em] c_x &c_y& c_z \end{bmatrix}\)
• \((\vec a\times \vec b+\vec a\times \vec c)=\) ̂̂î [aybz - azby)] - ĵ (axbz - a_zbx)] + k̂ [ax by - aybx] + î [aycz - azcy)] - ĵ (axcz - azcx)] + k̂ [ax cy - aycx]
• \((\vec a\times \vec b+\vec a\times \vec c)=\) î [ay (bz + cz) - az (by + cy)] - ĵ [ax (bz + cz) - az (bx + cx)] + k̂ [ax (by + cy) - ay (bx + cx)] ..............(4) 
• ∴ \(\vec a × (\vec b + \vec c) = \vec a × \vec b + \vec a × \vec c\)

Statement III: Cross product of vectors is associative

• Consider two non-zero perpendicular vectors, a and b.
• We have (a × a) × b = 0 × b = 0
• However, a × b is perpendicular to a and is not the zero vector, so
• a × (a × b) ≠ 0
• (a × a) × b ≠ a × (a × b)
• Cross product of vectors is not associative

∴ Only Statements I and II are correct.

Concept:

Properties of vectors:

• If \(\rm \vec {a} \;and\; \vec{b}\) are two vectors parallel to each other then \({\rm{\vec a}} \times {\rm{\vec b}} = 0\)
• If \(\rm \vec {a} \;and\; \vec{b}\) are two vectors perpendicular to each other then \({\rm{\vec a}} \cdot {\rm{\vec b}} = 0\)
• A cross product is not commutative \({\rm{\vec a}} \times {\rm{\vec b}} = - {\rm{\;\vec b}} \times {\rm{\vec a}}\)

Hence option 1 is the correct answer.