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Vector Algebra

JEE Physics · 18 questions · Page 2 of 2 · Click an option or "Show Solution" to reveal answer

Q11
If A→\overrightarrow A and B→\overrightarrow B are two vectors satisfying the relation A→\overrightarrow A . B→\overrightarrow B = ∣A→×B→∣\left| {\overrightarrow A \times \overrightarrow B } \right|. Then the value of ∣A→−B→∣\left| {\overrightarrow A - \overrightarrow B } \right| will be :
A A2+B2+2AB\sqrt {{A^2} + {B^2} + \sqrt 2 AB}
B A2+B2\sqrt {{A^2} + {B^2}}
C A2+B2−2AB\sqrt {{A^2} + {B^2} - \sqrt 2 AB}
D A2+B2+2AB\sqrt {{A^2} + {B^2} + 2AB}
Correct Answer
Option C
Solution

Given,

A→\overrightarrow A

.

B→\overrightarrow B

=

∣A→×B→∣\left| {\overrightarrow A \times \overrightarrow B } \right|

..... (i) Also, we know that

A→\overrightarrow A

.

B→\overrightarrow B

=

∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|

cosθ\theta .... (ii) and

A→×B→\overrightarrow A \times \overrightarrow B

=

∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|

sinθ\theta ..... (iii) From Eqs. (i), (ii) and (iii), we get

∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|

cosθ\theta =

∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|

sinθ\theta

⇒cos⁡θ=sin⁡θ⇒sin⁡θcos⁡θ=1\Rightarrow \cos \theta = \sin \theta \Rightarrow {{\sin \theta } \over {\cos \theta }} = 1
⇒tan⁡θ=1\Rightarrow \tan \theta = 1
⇒tan⁡θ=tan⁡45∘⇒θ=45∘\Rightarrow \tan \theta = \tan 45^\circ \Rightarrow \theta = 45^\circ

∴\therefore

∣A→−B→∣=A2+B2−2∣A→∣∣B→∣cos⁡θ\left| {\overrightarrow A - \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta }
=A2+B2−2∣A→∣∣B→∣cos⁡(45∘)= \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos (45^\circ )}
=A2+B2−2AB= \sqrt {{A^2} + {B^2} - \sqrt 2 AB}
Q12
A→\overrightarrow A is a vector quantity such that ∣A→∣|\overrightarrow A | = non-zero constant. Which of the following expression is true for A→\overrightarrow A ?
A A→ . A→=0\overrightarrow A \,.\,\overrightarrow A = 0
B A→×A→<0\overrightarrow A \times \overrightarrow A < 0
C A→×A→=0\overrightarrow A \times \overrightarrow A = 0
D A→×A→>0\overrightarrow A \times \overrightarrow A > 0
Correct Answer
Option C
Solution
A→×A→=A×A×sin⁡0∘\overrightarrow A \times \overrightarrow A = A \times A \times \sin 0^\circ
=0= 0
Q13
Two vectors P→{\overrightarrow P } and Q→{\overrightarrow Q } have equal magnitudes. If the magnitude of P→+Q→{\overrightarrow P + \overrightarrow Q } is n times the magnitude of P→−Q→{\overrightarrow P - \overrightarrow Q }, then angle between P→{\overrightarrow P } and Q→{\overrightarrow Q } is :
A sin⁡−1(n−1n+1){\sin ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)
B cos⁡−1(n−1n+1){\cos ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)
C sin⁡−1(n2−1n2+1){\sin ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)
D cos⁡−1(n2−1n2+1){\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)
Correct Answer
Option D
Solution
∣P→∣=∣Q→∣=x\left| {\overrightarrow P } \right| = \left| {\overrightarrow Q } \right| = x

..... (i)

∣P→+Q→∣=n∣P→−Q→∣\left| {\overrightarrow P + \overrightarrow Q } \right| = n\left| {\overrightarrow P - \overrightarrow Q } \right|
P2+Q2+2PQcos⁡θ=n2(P2+Q2−2PQcos⁡θ){P^2} + {Q^2} + 2PQ\cos \theta = {n^2}({P^2} + {Q^2} - 2PQ\cos \theta )

Using (i) in above equation

cos⁡θ=n2−11+n2\cos \theta = {{{n^2} - 1} \over {1 + {n^2}}}
θ=cos⁡−1(n2−1n2+1)\theta = {\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)
Q14
Two forces having magnitude AA and A2\dfrac{A}{2} are perpendicular to each other. The magnitude of their resultant is:
A 5A2\dfrac{5 A}{2}
B 5A4\dfrac{\sqrt{5} A}{4}
C 5A2\dfrac{\sqrt{5} A}{2}
D 5A22\dfrac{\sqrt{5} A^{2}}{2}
Correct Answer
Option C
Solution

The resultant of two perpendicular vectors is given by the Pythagorean theorem.

If we have two vectors of magnitudes AA and A2\dfrac{A}{2}, the resultant RR is: R=A2+(A2)2=A2+A24=5A24=5A2R = \sqrt{A^{2} + \left(\dfrac{A}{2}\right)^{2}} = \sqrt{A^{2} + \dfrac{A^{2}}{4}} = \sqrt{\dfrac{5A^{2}}{4}} = \dfrac{\sqrt{5} A}{2}.

Q15
Statement I : Two forces (P→+Q→)\left( {\overrightarrow P + \overrightarrow Q } \right) and (P→−Q→)\left( {\overrightarrow P - \overrightarrow Q } \right) where P→⊥Q→\overrightarrow P \bot \overrightarrow Q , when act at an angle θ\theta1 to each other, the magnitude of their resultant is 3(P2+Q2)\sqrt {3({P^2} + {Q^2})} , when they act at an angle θ\theta2, the magnitude of their resultant becomes 2(P2+Q2)\sqrt {2({P^2} + {Q^2})} . This is possible only when θ1<θ2{\theta _1} < {\theta _2}. Statement II : In the situation given above. θ\theta1 = 60∘^\circ and θ\theta2 = 90∘^\circ In the light of the above statements, choose the most appropriate answer from the options given below :-
A Statement I is false but Statement II is true
B Both Statement I and Statement II are true
C Statement I is true but Statement II is false
D Both Statement I and Statement II are false.
Correct Answer
Option B
Solution
A→=P→+Q→\overrightarrow A = \overrightarrow P + \overrightarrow Q
B→=P→−Q→\overrightarrow B = \overrightarrow P - \overrightarrow Q
P→⊥Q→\overrightarrow P \bot \overrightarrow Q
∣A→∣=∣B→∣=2(P2+Q2)(1+cos⁡θ)\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )}

For

∣A→+B→∣=3(P2+Q2)\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})}
θ1=60∘{\theta _1} = 60^\circ

For

∣A→+B→∣=2(P2+Q2)\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})}
θ2=90∘{\theta _2} = 90^\circ
Q16
Let A→\overrightarrow A = (i^+j^)\left( {\widehat i + \widehat j} \right) and, B→=(2i^−j^).\overrightarrow B = \left( {2\widehat i - \widehat j} \right). The magnitude of a coplanar vector C→\overrightarrow C such that A→.C→=B→.C→=A→.B→,\overrightarrow A .\overrightarrow C = \overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B , is given by :
A 109\sqrt {{{10} \over 9}}
B 59\sqrt {{{5} \over 9}}
C 209\sqrt {{{20} \over 9}}
D 912\sqrt {{{9} \over 12}}
Correct Answer
Option B
Solution

Let

C→\overrightarrow C

= a

i^\widehat i

+ b

j^\widehat j

Given,

A→.C→=A→.B→\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B

⇒\Rightarrow

    \,\,\,\,

a + b = 2 −- 1 ⇒\Rightarrow

    \,\,\,\,

a + b = 1 . . . . .(1) also given

B→.C→=A→.B→\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B

⇒\Rightarrow

    \,\,\,\,

2a −- b = 1 . . . . (2) Solving (1) and (2), we get, a =

13{1 \over 3}

and b =

23{2 \over 3}
∴    \therefore\,\,\,\,
C→=13i^+23j^\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j
∣C→∣=(13)2+(23)2\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}}

=

59\sqrt {{5 \over 9}}
Q17
Let ∣A1→∣=3\left| {\mathop {{A_1}}\limits^ \to } \right| = 3, ∣A2→∣=5\left| {\mathop {{A_2}}\limits^ \to } \right| = 5 and ∣A1→+A2→∣=5\left| {\mathop {{A_1}}\limits^ \to + \mathop {{A_2}}\limits^ \to } \right| = 5. The value of (2A1→+3A2→)(3A1→−2A2→)\left( {2\mathop {{A_1}}\limits^ \to + 3\mathop {{A_2}}\limits^ \to } \right)\left( {3\mathop {{A_1}}\limits^ \to - \mathop {2{A_2}}\limits^ \to } \right) is :-
A –118.5
B –112.5
C –99.5
D –106.5
Correct Answer
Option A
Solution
∣A1→∣=3,∣A2→∣=5 and ∣A1→+A2→∣=5\left| {\overrightarrow {{A_1}} } \right| = 3,\left| {\overrightarrow {{A_2}} } \right| = 5\,and\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = 5
 ∣A1→+A2→∣=∣A1→∣2+∣A2→∣2+2∣A1→∣∣A2→∣cos⁡θ\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = {\left| {\overrightarrow {{A_1}} } \right|^2} + {\left| {\overrightarrow {{A_2}} } \right|^2} + 2\left| {\overrightarrow {{A_1}} } \right|\left| {\overrightarrow {{A_2}} } \right|\cos \theta
cos⁡θ=−310\cos \theta = - {3 \over {10}}
(2A1→+3A2→).(3A1→−2A2→)\left( {2\overrightarrow {{A_1}} + 3\overrightarrow {{A_2}} } \right).\left( {3\overrightarrow {{A_1}} - 2\overrightarrow {{A_2}} } \right)

=

6∣A1→∣2+9A1→.A2→−4A1→.A2→−6∣A2→∣26{\left| {\overrightarrow {{A_1}} } \right|^2} + 9\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 4\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 6{\left| {\overrightarrow {{A_2}} } \right|^2}

= - 118.5

Q18
The angle between vector Q⃗\vec{Q} and the resultant of (2Q⃗+2P⃗)(2 \vec{Q}+2 \vec{P}) and (2Q⃗−2P⃗)(2 \vec{Q}-2 \vec{P}) is :
A tan⁡−1(P/Q) \tan ^{-1}(\mathrm{P} / \mathrm{Q})
B 0∘^\circ
C tan⁡−1(2Q⃗−2P⃗)2Q⃗+2P⃗ \tan ^{-1} \dfrac{(2 \vec{Q}-2 \vec{P})}{2 \vec{Q}+2 \vec{P}}
D tan⁡−1(2Q/P) \tan ^{-1}(2 Q / \mathrm{P})
Correct Answer
Option B
Solution

To find the angle between the vector

Q⃗\vec{Q}

and the resultant of

(2Q⃗+2P⃗)(2 \vec{Q}+2 \vec{P})

and

(2Q⃗−2P⃗)(2 \vec{Q}-2 \vec{P})

, we first find the resultant vector of

(2Q⃗+2P⃗)(2 \vec{Q}+2 \vec{P})

and

(2Q⃗−2P⃗)(2 \vec{Q}-2 \vec{P})

. The resultant vector of

(2Q⃗+2P⃗)(2 \vec{Q}+2 \vec{P})

and

(2Q⃗−2P⃗)(2 \vec{Q}-2 \vec{P})

can be simply found by adding these two vectors:

Resultant=(2Q⃗+2P⃗)+(2Q⃗−2P⃗)=4Q⃗\text{Resultant} = (2 \vec{Q}+2 \vec{P}) + (2 \vec{Q}-2 \vec{P}) = 4 \vec{Q}

Now, we need to find the angle between the vector

Q⃗\vec{Q}

and this resultant vector

4Q⃗4\vec{Q}

. Since the resultant vector is just a scaled version of

Q⃗\vec{Q}

, they are in the same direction.

The angle between any vector and another vector that is a scaled version of the first vector is always

0∘0^\circ

, because they are parallel to each other. Therefore, the angle between

Q⃗\vec{Q}

and the resultant of

(2Q⃗+2P⃗)(2 \vec{Q}+2 \vec{P})

and

(2Q⃗−2P⃗)(2 \vec{Q}-2 \vec{P})

is

0∘0^\circ

. So, the correct option is: Option B:

0∘0^\circ
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