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Vector Algebra

JEE Physics · 18 questions · Page 1 of 2 · Click an option or "Show Solution" to reveal answer

Q1
A vector in x−yx-y plane makes an angle of 30∘30^{\circ} with yy-axis. The magnitude of y\mathrm{y}-component of vector is 232 \sqrt{3}. The magnitude of xx-component of the vector will be :
A 3\sqrt{3}
B 2
C 6
D 13\dfrac{1}{\sqrt{3}}
Correct Answer
Option B
Solution
Ay=Acos⁡30∘=23⇒A32=23⇒A=4\begin{aligned} & \mathrm{A}_{\mathrm{y}}=\mathrm{A} \cos 30^{\circ}=2 \sqrt{3} \\\\ & \Rightarrow \mathrm{A} \frac{\sqrt{3}}{2}=2 \sqrt{3} \\\\ & \Rightarrow \mathrm{A}=4 \end{aligned}

Now Ax=Asin⁡30∘=4×12=2A_x=A \sin 30^{\circ}=4 \times \dfrac{1}{2}=2

Q2
If two vectors A⃗\vec{A} and B⃗\vec{B} having equal magnitude RR are inclined at angle θ\theta, then
A ∣A⃗+B⃗∣=2Rcos⁡(θ2)|\vec{A}+\vec{B}|=2 R \cos \left(\dfrac{\theta}{2}\right)
B ∣A⃗−B⃗∣=2Rcos⁡(θ2)|\vec{A}-\vec{B}|=2 R \cos \left(\dfrac{\theta}{2}\right)
C ∣A⃗−B⃗∣=2Rsin⁡(θ2)|\vec{A}-\vec{B}|=\sqrt{2} R \sin \left(\dfrac{\theta}{2}\right)
D ∣A⃗+B⃗∣=2Rsin⁡(θ2)|\vec{A}+\vec{B}|=2 R \sin \left(\dfrac{\theta}{2}\right)
Correct Answer
Option A
Solution

The magnitude of resultant vector

R′=a2+b2+2abcos⁡θR^{\prime}=\sqrt{a^2+b^2+2 a b \cos \theta}

Here

a=b=Ra=b=R

Then

R′=R2+R2+2R2cos⁡θR^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos \theta}
=R21+cos⁡θ=2R2cos⁡2θ2=2Rcos⁡θ2\begin{aligned} & =R \sqrt{2} \sqrt{1+\cos \theta} \\ & =\sqrt{2} R \sqrt{2 \cos ^2 \frac{\theta}{2}} \\ & =2 R \cos \frac{\theta}{2} \end{aligned}
Q3
Two vectors A→\overrightarrow A and B→\overrightarrow B have equal magnitudes. If magnitude of A→\overrightarrow A + B→\overrightarrow B is equal to two times the magnitude of A→\overrightarrow A −- B→\overrightarrow B , then the angle between A→\overrightarrow A and B→\overrightarrow B will be :
A sin⁡−1(35){\sin ^{ - 1}}\left( {{3 \over 5}} \right)
B sin⁡−1(13){\sin ^{ - 1}}\left( {{1 \over 3}} \right)
C cos⁡−1(35){\cos ^{ - 1}}\left( {{3 \over 5}} \right)
D cos⁡−1(13){\cos ^{ - 1}}\left( {{1 \over 3}} \right)
Correct Answer
Option C
Solution
A2+A2+2A2cos⁡θ=2A2+A2+2A2(−cos⁡θ)\sqrt {{A^2} + {A^2} + 2{A^2}\cos \theta } = 2\sqrt {{A^2} + {A^2} + 2{A^2}( - \cos \theta )}
⇒2A2+2A2cos⁡θ=8A2+8A2−(−cos⁡θ)\Rightarrow 2{A^2} + 2{A^2}\cos \theta = 8{A^2} + 8{A^2} - ( - \cos \theta )
⇒5cos⁡θ=3\Rightarrow 5\cos \theta = 3
⇒θ=cos⁡−1(35)\Rightarrow \theta = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)
Q4
If A→×B→=B→×A→\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A , then the angle beetween A and B is
A π2{\pi \over 2}
B π3{\pi \over 3}
C π\pi
D π4{\pi \over 4}
Correct Answer
Option C
Solution
A→×B→=B→×A→\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A
A→×B→−B→×A→=0\overrightarrow A \times \overrightarrow B - \overrightarrow B \times \overrightarrow A = 0
⇒A→×B→+A→×B→=0\Rightarrow \overrightarrow A \times \overrightarrow B + \overrightarrow A \times \overrightarrow B = 0

∴\therefore

A→×B→=0\overrightarrow A \times \overrightarrow B = 0
⇒ABsin⁡θ=0\Rightarrow AB\sin \theta = 0
θ=\theta =
0,π,  0,\pi ,\,\,
2π2\pi

........ from the given options,

θ=π\theta = \pi
Q5
When vector A⃗=2i^+3j^+2k^\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k} is subtracted from vector B→\overrightarrow{\mathrm{B}}, it gives a vector equal to 2j^2 \hat{j}. Then the magnitude of vector B→\overrightarrow{\mathrm{B}} will be :
A 3
B 33\sqrt{33}
C 6\sqrt6
D 5\sqrt5
Correct Answer
Option B
Solution

Given that when vector

A⃗=2i^+3j^+2k^\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}

is subtracted from vector

B→\overrightarrow{\mathrm{B}}

, it gives a vector equal to

2j^2 \hat{j}

. We can write this as:

B⃗−A⃗=2j^\vec{B} - \vec{A} = 2 \hat{j}

Now, let's express the vector

B→\overrightarrow{\mathrm{B}}

in terms of its components:

B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}

Subtract vector

A⃗\vec{A}

from vector

B⃗\vec{B}

:

(Bxi^+Byj^+Bzk^)−(2i^+3j^+2k^)=2j^(B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) - (2 \hat{i}+3 \hat{j}+2 \hat{k}) = 2 \hat{j}

Comparing the components, we get:

Bx−2=0By−3=2Bz−2=0B_x - 2 = 0 \\ B_y - 3 = 2 \\ B_z - 2 = 0

Solving these equations, we find the components of vector

B⃗\vec{B}

:

Bx=2By=5Bz=2B_x = 2 \\ B_y = 5 \\ B_z = 2

Now, we can find the magnitude of vector

B⃗\vec{B}

:

∣B⃗∣=Bx2+By2+Bz2=22+52+22=4+25+4=33|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{2^2 + 5^2 + 2^2} = \sqrt{4 + 25 + 4} = \sqrt{33}

Therefore, the magnitude of vector

B→\overrightarrow{\mathrm{B}}

is

33\sqrt{33}
Q6
What will be the projection of vector A→=i^+j^+k^\overrightarrow A = \widehat i + \widehat j + \widehat k on vector B→=i^+j^\overrightarrow B = \widehat i + \widehat j ?
A 2(i^+j^+k^)\sqrt 2 (\widehat i + \widehat j + \widehat k)
B (i^+j^)(\widehat i + \widehat j)
C 2(i^+j^)\sqrt 2 (\widehat i + \widehat j)
D 2(i^+j^+k^)2(\widehat i + \widehat j + \widehat k)
Correct Answer
Option B
Solution

Projection =

A→.B→∣B→∣(B^){{\overrightarrow A .\overrightarrow B } \over {\left| {\overrightarrow B } \right|}}(\widehat B)
=(i^+j^+k^).(i^+j^)2(i^+j^)2= {{(\widehat i + \widehat j + \widehat k).(\widehat i + \widehat j)} \over {\sqrt 2 }}{{(\widehat i + \widehat j)} \over {\sqrt 2 }}
=22= {2 \over {\sqrt 2 }}

×\times

(i^+j^)2{{(\widehat i + \widehat j)} \over {\sqrt 2 }}
=(i^+j^)= (\widehat i + \widehat j)
Q7
Which of the following relations is true for two unit vector A^\widehat A and B^\widehat B making an angle θ\theta to each other?
A ∣A^+B^∣=∣A^−B^∣tan⁡θ2|\widehat A + \widehat B| = |\widehat A - \widehat B|\tan {\theta \over 2}
B ∣A^−B^∣=∣A^+B^∣tan⁡θ2|\widehat A - \widehat B| = |\widehat A + \widehat B|\tan {\theta \over 2}
C ∣A^+B^∣=∣A^−B^∣cosθ2|\widehat A + \widehat B| = |\widehat A - \widehat B|cos{\theta \over 2}
D ∣A^−B^∣=∣A^+B^∣cos⁡θ2|\widehat A - \widehat B| = |\widehat A + \widehat B|\cos {\theta \over 2}
Correct Answer
Option B
Solution

∵\because

∣A^−B^∣=2sin⁡(θ2)\left| {\widehat A - \widehat B} \right| = 2\sin \left( {{\theta \over 2}} \right)

and,

∣A^+B^∣=2cos⁡(θ2)\left| {\widehat A + \widehat B} \right| = 2\cos \left( {{\theta \over 2}} \right)
⇒∣A^−B^∣∣A^+B^∣=tan⁡(θ2)\Rightarrow {{\left| {\widehat A - \widehat B} \right|} \over {\left| {\widehat A + \widehat B} \right|}} = \tan \left( {{\theta \over 2}} \right)
Q8
If two vectors P→=i^+2mj^+mk^\overrightarrow P = \widehat i + 2m\widehat j + m\widehat k and Q→=4i^−2j^+mk^\overrightarrow Q = 4\widehat i - 2\widehat j + m\widehat k are perpendicular to each other. Then, the value of m will be :
A −1-1
B 3
C 1
D 2
Correct Answer
Option D
Solution

P⃗ & Q⃗\vec{P} \,\&\, \vec{Q} are perpendicular

P→⋅Q→=0(i^+2 mj^+mk^)⋅(4i^−2j^+mk^)=0⇒4−4 m+m2=0⇒(m−2)2=0⇒m=2\begin{aligned} & \overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}=0 \\\\ & (\hat{\mathrm{i}}+2 \mathrm{~m} \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}})=0 \\\\ & \Rightarrow 4-4 \mathrm{~m}+\mathrm{m}^2=0 \\\\ & \Rightarrow(\mathrm{m}-2)^2=0 \Rightarrow \mathrm{m}=2 \end{aligned}
Q9
Two vectors A→\overrightarrow A and B→\overrightarrow B have equal magnitudes. The magnitude of (A→+B→)\left( {\overrightarrow A + \overrightarrow B } \right) is 'n' times the magnitude of (A→−B→)\left( {\overrightarrow A - \overrightarrow B } \right) . The angle between A→{\overrightarrow A } and B→{\overrightarrow B } is -
A sin⁡−1[n−1n+1]{\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]
B sin⁡−1[n2−1n2+1]{\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]
C cos⁡−1[n2−1n2+1]{\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]
D cos⁡−1[n−1n+1]{\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]
Correct Answer
Option C
Solution
∣A→+B→∣=2acos⁡θ/2\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2

. . . (1)

∣A→−B→∣=2acos⁡(π−θ)2=2asin⁡θ/2\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2

. . . (2)

⇒   n(2acos⁡θ2)=2asin⁡θ2\Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}
⇒   tan⁡θ2=n\Rightarrow \,\,\,\tan {\theta \over 2} = n
Q10
Two vectors X→\overrightarrow X and Y→\overrightarrow Y have equal magnitude. The magnitude of (X→\overrightarrow X −- Y→\overrightarrow Y ) is n times the magnitude of (X→\overrightarrow X + Y→\overrightarrow Y ). The angle between X→\overrightarrow X and Y→\overrightarrow Y is :
A cos⁡−1(−n2−1n2−1){\cos ^{ - 1}}\left( {{{ - {n^2} - 1} \over {{n^2} - 1}}} \right)
B cos⁡−1(n2−1−n2−1){\cos ^{ - 1}}\left( {{{{n^2} - 1} \over { - {n^2} - 1}}} \right)
C cos⁡−1(n2+1−n2−1){\cos ^{ - 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 B
Solution

Given X = Y

X2+Y2−2×Ycos⁡θ\sqrt {{X^2} + {Y^2} - 2 \times Y\cos \theta }
=nX2+Y2+2×Ycos⁡θ= n\sqrt {{X^2} + {Y^2} + 2 \times Y\cos \theta }

Square both sides

2X2(1−cos⁡θ)=n2.2X2(1+cos⁡θ)2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )
1−cos⁡θ=n2+n2cos⁡θ1 - \cos \theta = {n^2} + {n^2}\cos \theta
cos⁡θ=1−n21+n2\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}
θ=cos⁡−1[n2−1−n2−1]\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]
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