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Correct choice is (d) \(\VEC{a}\) and \(\vec{c}\)

The best I can explain: The vectors which start from the same initial point are CALLED coinitial vectors. In the given FIGURE, vectors \(\vec{a}\) and \(\vec{c}\) start from the same point but are of DIFFERENT magnitude. Therefore, vectors \(\vec{a}\) and \(\vec{c}\) AE coinitial but not equal.

The correct answer is (b) |\(\VEC{A}\)|

EASY explanation: The correct way of representing the MAGNITUDE of a given vector A is|\(\vec{A}\)|. The magnitude of a vector is the DISTANCE between the initial and the terminal points of a vector.

Correct choice is (a) TRUE

For explanation I would say: The GIVEN STATEMENT is true. The QUANTITY which has only magnitude and no direction is CALLED scalar quantity.

Example: Time. Time is a scalar quantity which has only magnitude and no direction.

Right OPTION is (a) \(\vec{a}\), \(\vec{c}\) and \(\vec{b}\)

Explanation: Two VECTORS are said to coinitial if they have the same initial point. In the GIVEN figure, vectors \(\vec{a}\), \(\vec{c}\) and \(\vec{b}\) are starting from the point, HENCE they are coinitial.

The CORRECT choice is (b) False

Explanation: The given STATEMENT is false. Collinear vectors are those vectors that are parallel same line irrespective of the direction and magnitude. The vectors which have the same INITIAL point are CALLED coinitial vectors.

Correct choice is (c) l^2+m^2+n^2=1

Explanation: Consider \(\vec{a}\) is the position VECTOR of a POINT M(x,y,z) andα, β, γ are the angles, MADE by the vector \(\vec{a}\) with the positive directions of x, y and z respectively. The cosines of the angles, cos⁡α, cos⁡β, cos⁡γ are the DIRECTION cosines of the vector \(\vec{a}\) denoted by l, m, n, then

cos^2⁡α+cos^2⁡β+cos^2⁡γ=1 i.e.l^2+m^2+n^2=1.

The correct choice is (b) DENSITY

For EXPLANATION: Density is a scalar quantity as it has only MAGNITUDE but no direction. SPEED, FORCE, velocity has both magnitude and direction. Therefore, they all are vectors.

The correct option is (d) \(cos^{-1}⁡\frac{\sqrt{58}}{3\sqrt{3}}\)

To explain: The angle between the two VECTORS is GIVEN by

\(cos⁡θ=\frac{|\vec{a}|.|\vec{B}|}{\vec{a}.\vec{b}}\)

\(|\vec{a}|=\sqrt{1^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt{6}\)

\(|\vec{b}|=\sqrt{3^2+2^2+(-4)^2}=\sqrt{9+4+16}=\sqrt{29}\)

\(\vec{a}.\vec{b}\)=1(3)-1(2)+2(4)=9

∴\(cos⁡θ=\frac{\sqrt{6}.\sqrt{29}}{9}=\frac{\sqrt{58}}{3\sqrt{3}}\)

∴\(θ=cos^{-1}\frac{⁡\sqrt{58}}{3\sqrt{3}}\)

Correct answer is (a) 11\(\hat{i}\)+\(\hat{j}\)

The EXPLANATION: Given that, \(\VEC{a}\)=6\(\hat{i}\)-3\(\hat{j}\) and \(\vec{b}\)=5\(\hat{i}\)+4\(\hat{j}\)

The sum of the VECTORS is given by \(\vec{a}+\vec{b}\).

∴\(\vec{a}+\vec{b}\)=(6\(\hat{i}\)-3\(\hat{i}\))+(5\(\hat{i}\)+4\(\hat{j}\))

=(6+5) \(\hat{i}\)+(-3+4)\(\hat{j}\)

=11\(\hat{i}\)+\(\hat{j}\)

The correct option is (c) positive

To explain: Direction of λ\(\vec{a}\) and \(\vec{a}\) is same if value λ is positive as it GIVES it a direction which is positive in NATURE. If the value of λ is NEGATIVE then the direction of the result after MULTIPLICATION becomes in opposite direction. WHEREAS the value of the product vector becomes zero if value of λ is 0.

The correct option is (d) zero VECTOR

Explanation: The vector whose initial and final POINTS COINCIDE is called zero vector or a null vector. It is denoted by \(\VEC{0}\). It has zero magnitude and can be considered to have any direction.

The CORRECT option is (c) \(\FRAC{π}{3}\)

To EXPLAIN: Given that, \(|\vec{a}|=2, \,|\vec{b}|=\frac{1}{2√3}\) and \(\vec{a}×\vec{b}=\frac{1}{2}\)

We know that, \(\vec{a}×\vec{b}=\vec{a}.\vec{b} \,sin⁡θ\)

∴ \(sin⁡θ=\frac{(\vec{a}×\vec{b})}{|\vec{a}|.|\vec{b}|}\)

sin⁡θ=\(\frac{\frac{1}{2}}{2×\frac{1}{2√3}}=\frac{\SQRT{3}}{2}\)

θ=\(sin^{-1}⁡\frac{\sqrt{3}}{2}=\frac{π}{3}\)

Right choice is (b) –\(\FRAC{7}{2}\)

Explanation: Given that: \(\VEC{a}=\hat{i}-\hat{j}+3\hat{k}, \,\vec{b}=5\hat{i}-2\hat{j}+\hat{k} \,and \,\vec{C}=\hat{i}-\hat{j}\)

Also given, \(\vec{a}+μ\vec{b}\) is perpendicular to \(\vec{c}\)

Therefore, \((\vec{a}+μ\vec{b}).\vec{c}=0\)

i.e. \((\hat{i}-\hat{j}+3\hat{k}+μ(5\hat{i}-2\hat{j}+\hat{k})).(\hat{i}-\hat{j})\)=0

\(((1+5μ) \,\hat{i}-(1+2μ) \,\hat{j}+(μ+3) \,\hat{k}).(\hat{i}-\hat{j})\)=0

1+5μ+1+2μ=0

μ=-\(\frac{7}{2}\).

The correct answer is (b) \(\SQRT{37}\)

The explanation is: \(|\VEC{a}+\vec{b}|^2=(\vec{a}+\vec{b}).(\vec{a}+\vec{b})\)

=\(\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}\)

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

=(3)^2+2(6)+(4)^2

=9+12+16=37

∴\(|\vec{a}+\vec{b}|=\sqrt{37}\)

The correct OPTION is (a) |λ\(\vec{a}\)|

The explanation is: |λ| TIMES the magnitude of VECTOR \(\vec{a}\) is denoted as |λ\(\vec{a}\)| = |λ||\(\vec{a}\)|

As we KNOW that the magnitude of vector \(\vec{a}\) is denoted by |\(\vec{a}\)|, if we multiply the magnitude of vector \(\vec{a}\) with magnitude of λ we get |λ\(\vec{a}\)|.

The correct answer is (a) 1

The EXPLANATION is: If \(\VEC{a} \,and \,\vec{b}\) are TWO vectors, where \(a_1, a_2, a_3\) are the components of VECTOR \(\vec{a} \,and \,b_1, b_2, b_3\) are the components of vector \(\vec{b}\), then the scalar product is given by

\(\vec{a}.\vec{b}=a_1 b_1+a_1 b_2+a_3 b_3\)

\((6\hat{i}-7\hat{j}+5\hat{k}).(6\hat{i}-7\hat{k})\)=6(6)-7(0)+5(-7)=36-35=1.

The correct option is (d) \(cos^{-1}⁡-\sqrt{\frac{3}{2}}\)

To explain: The angle between TWO VECTORS is given by

\(cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}\)

\(|\vec{a}|=\sqrt{(-1)^2+(1)^2+(-1)^2}=\sqrt{3}\)

\(|\vec{b}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}\)

\(\vec{a}.\vec{b}\)=(-1)(1)+1(-1)+0=-2

\(cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{-2}=-\sqrt{\frac{3}{2}}\)

∴\(θ=cos^{-1}⁡-\sqrt{\frac{3}{2}}\)

Correct choice is (a) True

Easiest explanation: The given STATEMENT is true. If the angle between TWO vectors is \(\frac{π}{2}\) i.e. they are perpendicular to each other, their SCALAR product will be zero.

Right CHOICE is (d) 4\(\hat{i}\)+\(\hat{j}\)

To elaborate: Given that, \(\vec{a}+\vec{B}\)+\(\vec{c}\)=8\(\hat{i}\)+2\(\hat{j}\) -(1)

Given: \(\vec{a}\)=\(\hat{i}\)-6\(\hat{j}\) and \(\vec{c}\)=3\(\hat{i}\)+7\(\hat{j}\)

Substituting the VALUES of \(\vec{a}\) and \(\vec{b}\) in equation (1), we get

\(\vec{a}+\vec{b}\)+\(\vec{c}\)=8\(\hat{i}\)+2\(\hat{j}\)

(\(\hat{i}\)-6\(\hat{j}\))+\(\vec{b}\)+(3\(\hat{i}\)+7\(\hat{j}\))=8\(\hat{i}\)+2\(\hat{j}\)

∴\(\vec{c}\)=(8\(\hat{i}\)+2\(\hat{j}\))-(\(\hat{i}\)-6\(\hat{j}\))-(3\(\hat{i}\)+7\(\hat{j}\))

=(8-1-3) \(\hat{i}\)+(2+6-7) \(\hat{j}\)

=4\(\hat{i}\)+\(\hat{j}\)

The correct option is (b) \(\HAT{i}\)+3\(\hat{j}\)-6\(\hat{k}\)

For explanation: GIVEN that, \(\vec{a}\)=4\(\hat{i}\)-4\(\hat{j}\), \(\vec{b}\)=-3\(\hat{i}\)+2K, \(\vec{c}\)=7\(\hat{j}\)-8\(\hat{k}\)

To FIND: \(\vec{a}+\vec{b}\)+\(\vec{c}\)

∴\(\vec{a}+\vec{b}\)+\(\vec{c}\)=(4\(\hat{i}\)-4\(\hat{j}\)) +(-3\(\hat{i}\)+2k) +(7\(\hat{j}\)-8\(\hat{k}\))

=(4-3) \(\hat{i}\)+(-4+7) \(\hat{j}\)+(2-8)\(\hat{k}\)

=\(\hat{i}\)+3\(\hat{j}\)-6\(\hat{k}\)

The CORRECT choice is (b) 14\(\hat{i}\)–\(\hat{j}\)+4\(\hat{k}\)

The best I can EXPLAIN: Given that, \(\vec{a}\)=9\(\hat{i}\)-2\(\hat{j}\)+7\(\hat{k}\), \(\vec{b}\)=5\(\hat{i}\)+\(\hat{j}\)-3\(\hat{k}\)

We have to find \(\vec{a}+\vec{b}\)

∴\(\vec{a}+\vec{b}\)=(9\(\hat{i}\)-2\(\hat{j}\)+7\(\hat{k}\))+(5\(\hat{i}\)+\(\hat{j}\)-3\(\hat{k}\))

=(9+5) \(\hat{i}\)+(-2+1) \(\hat{j}\)+(7-3)\(\hat{k}\)

=14\(\hat{i}\)–\(\hat{j}\)+4\(\hat{k}\)

The correct answer is (a) λ\(\vec{a}\)

The best explanation: Multiplication of VECTOR \(\vec{a}\) and SCALAR λ is denoted as λ\(\vec{a}\), as \(\vec{a}\) is the original vector. λ is the scalar which can have any integer value which is to be multiplied to the given vector\(\vec{a}\), whereas 0 can only be the answer if the scalar λ = 0.

The correct CHOICE is (b) They have the same MAGNITUDE and direction

The best EXPLANATION: Two vectors are said to be equal, if they have the same magnitude and direction. They will be coinitial if they have the same initial point and they will be collinear if both the vectors are parallel to the same LINE.

Correct option is (b) \(\vec{a}\), \(\vec{b}\) and \(\vec{d}\)

For EXPLANATION I would say: Collinear vectors are those vectors which are PARALLEL to the same LINE irrespective of the magnitude and direction. In the GIVEN figure, \(\vec{a}\), \(\vec{b}\) and \(\vec{d}\) are collinear vectors.

Correct answer is (a) \(cos^{-1}⁡\frac{1}{\SQRT{3}}\)

Explanation: Given that, \(|\vec{a}|=\sqrt{3} \,and \,|\vec{b}|=\sqrt{2}\)

Also, \(\vec{a.} \vec{b}=3\sqrt{2}\)

The ANGLE between two VECTORS is given by

\(cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}\)

∴\(cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{3\sqrt{2}}=\frac{1}{\sqrt{3}}\)

∴\(θ=cos^{-1}⁡\frac{1}{\sqrt{3}}\).

The correct answer is (d) \(\FRAC{3}{\sqrt{13}} \HAT{i}-\frac{2}{\sqrt{13}} \hat{j}\)

The explanation: Given that, \(\VEC{a}\)=2\(\hat{i}\)+7\(\hat{j}\) and \(\vec{b}\)=\(\hat{i}\)-9\(\hat{j}\)

The sum of the two vectors will be

\(\vec{a}+\vec{b}\)=(2\(\hat{i}\)+7\(\hat{j}\))+(\(\hat{i}\)-9\(\hat{j}\))

=(2+1) \(\hat{i}\)+(7-9)\(\hat{j}\)

=3\(\hat{i}\)-2\(\hat{j}\)

The unit vector in the direction of the sum of the vectors is

\(\frac{1}{|\vec{a}+\vec{b}|}(\vec{a}+\vec{b})=\frac{3\hat{i}-2\hat{j}}{\sqrt{3^2+(-2)^2}}=\frac{3\hat{i}-2\hat{j}}{\sqrt{13}}=\frac{3}{1\sqrt{3}} \hat{i}-\frac{2}{\sqrt{13}}\hat{j}\)

The CORRECT option is (C) \(\vec{A}\)

For explanation I would say: The correct way denoting a VECTOR GIVEN vector A is \(\vec{A}\).

Right CHOICE is (B) 5\(\hat{i}\)-6\(\hat{j}\)+3\(\hat{k}\)

The BEST I can explain: It is given that, \(\vec{a}\)=3\(\hat{i}\)+2\(\hat{j}\)+2\(\hat{k}\), \(\vec{b}\)=2\(\hat{i}\)-8\(\hat{j}\)+\(\hat{k}\)

To find: \(\vec{a}+\vec{b}\)

∴\(\vec{a}+\vec{b}\)=(3\(\hat{i}\)+2\(\hat{j}\)+2\(\hat{k}\))+(2\(\hat{i}\)-8\(\hat{j}\)+\(\hat{k}\))

=(3+2) \(\hat{i}\)+(2-8) \(\hat{j}\)+(2+1)\(\hat{k}\)

=5\(\hat{i}\)-6\(\hat{j}\)+3\(\hat{k}\)

Correct ANSWER is (c) coinitial vectors

Easiest EXPLANATION: The vectors which have the same initial POINTS are CALLED coinitial vectors. It is not necessary that they should have the same magnitude and direction.

The correct option is (a) \(2(\vec{a}×\vec{B})\)

To EXPLAIN: Consider \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})\)

=\((\vec{a}-\vec{b})×\vec{a}+(\vec{a}+\vec{b})×\vec{b}\)

=\(\vec{a}×\vec{a}-\vec{b}×\vec{a}+\vec{a}×\vec{b}-\vec{b}×\vec{b}\)

We KNOW that, \(\vec{a}×\vec{a}=0,\vec{b}×\vec{b}=0 \,and \,\vec{a}×\vec{b}=-\vec{b}×\vec{a}\)

∴ \(\vec{a}×\vec{a}-\vec{b}×\vec{a}+\vec{a}×\vec{b}-\vec{b}×\vec{b}=0+2(\vec{a}×\vec{b})+0\)

Hence, \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})=2(\vec{a}×\vec{b})\)

The CORRECT choice is (d) \(\sqrt{17}\)

Explanation: Given that, \(\vec{a}\)=\(\hat{i}\)+4\(\hat{j}\) and \(\vec{B}\)=3\(\hat{i}\)-3\(\hat{j}\)

∴\(\vec{a}+\vec{b}\)=(1+3) \(\hat{i}\)+(4-3) \(\hat{j}\)

=4\(\hat{i}\)+\(\hat{j}\)

|\(\vec{a}+\vec{b}\)|=\(\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}\)

Correct option is (B) -23

Best explanation: If \(\vec{a} \,and \,\vec{b}\) are two vectors, where a1, A2 are the components of vector \(\vec{a} \,and \,b_1, \,b_2\) are the components of vector \(\vec{b}\), then the scalar PRODUCT is given by

\(\vec{a}.\vec{b}=a_1 \,b_1+a_1 \,b_2\)

∴\((2\hat{i}+5\hat{j}).(6\hat{i}-7\hat{j})\)=2(6)+5(-7)=12-35=-23.

The CORRECT option is (b) 6\(\hat{i}\)+11\(\hat{j}\)

Explanation: GIVEN that, \(\vec{a}\)=8\(\hat{i}\)+5\(\hat{j}\) and \(\vec{b}\)=-2\(\hat{i}\)+6\(\hat{j}\)

∴The sum of the VECTORS will be

\(\vec{a}+\vec{b}\)=(8\(\hat{i}\)+5\(\hat{j}\))+(-2\(\hat{i}\)+6\(\hat{j}\))

=(8-2) \(\hat{i}\)+(5+6)\(\hat{j}\)

=6\(\hat{i}\)+11\(\hat{j}\)

Correct ANSWER is (a) \(\SQRT{181}\)

To explain I would SAY: Given that, \(\vec{a}\)=4\(\hat{i}\)+9\(\hat{j}\) and \(\vec{b}\)=6\(\hat{i}\)

∴\(\vec{a}+\vec{b}\)=(4+6) \(\hat{i}\)+9\(\hat{j}\)

=10\(\hat{i}\)+9\(\hat{j}\)

|\(\vec{a}+\vec{b}\)|=\(\sqrt{10^2+9^2}=\sqrt{100+81}=\sqrt{181}\)

The CORRECT option is (b) False

Explanation: The given STATEMENT is false.|\(\vec{A}\)| denotes the magnitude of the VECTOR and since LENGTH of the LINE of the line can never be negative,|\(\vec{A}\)|<0 has no meaning.

Correct CHOICE is (b) \(\sqrt{645}\)

To explain: GIVEN that, \(\vec{a}=2\hat{i}+3\hat{J}+4\hat{k}\) and \(\vec{b}=4\hat{i}-2\hat{j}+3\hat{k}\)

∴ \(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\4&-2&3\end{vmatrix}\)

=\(\hat{i}(9—8)-\hat{j}(6-16)+\hat{k}(-4-12)\)

=\(17\hat{i}+10\hat{j}-16\hat{k}\)

∴\(|\vec{a}×\vec{b}|=\sqrt{17^2+10^2+(-16)^2}\)

=\(\sqrt{289+100+256}\)

=\(\sqrt{645}\)

The correct choice is (d) \(2\hat{i}-\hat{j}-\hat{k}\)

To elaborate: Given that, \(\vec{a}=-\hat{j}+\hat{k}\) and \(\vec{B}=-\hat{i}-\hat{j}-\hat{k}\)

Calculating the vector PRODUCT, we get

\(\vec{a}×\vec{b}=\BEGIN{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&-1&1\\-1&-1&-1\end{vmatrix}\)

=\(\hat{i}(1-(-1))-\hat{j}(0-(-1))+\hat{k}(0-1)\)

=\(2\hat{i}-\hat{j}-\hat{k}\)

Right CHOICE is (d) Opposite

Explanation: If the vector is multiplied with –λ then its DIRECTION become opposite as the direction in which it was PREVIOUS may be POSITIVE or negative. After it is multiplied with a negative value then its direction BECOMES exactly opposite to the previous direction.

Right OPTION is (b) 5\(\hat{i}\) + 5\(\hat{j}\) + 5\(\hat{K}\)

To explain I would SAY: Multiplication of VECTOR \(\vec{a}\) =\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) by scalar value 5 RESULTS in 5\(\hat{i}\) + 5\(\hat{j}\) + 5\(\hat{k}\), as in these type of questions we multiply\(\hat{i}\), \(\hat{j,}\)

\(\hat{k}\) with the constant given and the answer comes out to be 5\(\hat{i}\) + 5\(\hat{j}\) + 5\(\hat{k}\).

Right choice is (d) \(\frac{29}{5}\)

For explanation: The projection of a VECTOR \(\vec{a}\) on vector \(\vec{B}\) is given by

\(\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})\)

\(|\vec{b}|=\sqrt{4^2+3^2}=\sqrt{16+9}\)=5

\(\vec{a}.\vec{b}\)=8(4)-1(3)+0=32-3=29

The projection of vector \(8\hat{i}-\hat{j}+6\hat{k}\) on vector \(4\hat{i}+3\hat{j}\) will be

\(\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})=\frac{1}{5} (29)=\frac{29}{5}\)

The correct answer is (a) x=2, y=2, z=1

For explanation I would say: As both the vectors are equal HENCE, we can EQUATE their constants and get the value of x, y and z. Now we equate the coefficients of \(\hat{i}\), \(\hat{J}\), \(\hat{k}\) of both the EQUATIONS and get the values x=2, y=2, z=1.