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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 61 questions
1 Hard
1 of 61
Analyzing +4 | -1
If (a+3b)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\mathbf{b} is
View answer and explanation
Correct option: A
60∘60^{\circ}
Option A (60∘60^{\circ}) is correct because the given conditions lead to the dot product of a and b being half the magnitude squared of b, implying a 60-degree angle.
2 Hard
2 of 61
Application +4 | -1
The unit vector perpendicular to the vectors 6i^+2j^+3k^6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} and 3i^−6j^−2k^3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, is
View answer and explanation
Correct option: C
2i^+3j^−6k^7\frac{2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}}{7}
Option C (2i^+3j^−6k^7\frac{2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}}{7}) is correct because it is the unit vector obtained by normalizing the cross product of the given vectors.
3 Hard
3 of 61
Analyzing +4 | -1
Let the cosine of angle between the vectors p\mathbf{p} and q\mathbf{q} be λ\lambda such that 2p+q=i^+j^2 \mathbf{p}+\mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}} and p+2q=i^−j^\mathbf{p}+2 \mathbf{q}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, then λ\lambda is equal to
View answer and explanation
Correct option: B
−45-\frac{4}{5}
Option B (-4/5) is correct because it is the cosine of the angle between the vectors p and q, calculated using the dot product and magnitudes of the vectors.
4 Hard
4 of 61
Application +4 | -1
A unit vector perpendicular to the vector −i^+2j^+2k^-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} and making equal angles with XX and YY-axes can be
View answer and explanation
Correct option: A
13(2i^+2j^−k^)\frac{1}{3}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})
Option A (13(2i^+2j^−k^)\frac{1}{3}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})) is correct because it is a unit vector perpendicular to the given vector and makes equal angles with the X and Y axes.
5 Hard
5 of 61
Application +4 | -1
In cartesian coordinates the point AA is (x1,y1)\left(x_1, y_1\right), where x1=1x_1=1 on the curve y=x2+x+10y=x^2+x+10. Then tangent at AA cuts the XX-axis at BB. The value of the dot product OA⋅AB\mathbf{O A} \cdot \mathbf{A B} is
View answer and explanation
Correct option: B
−148-148
Option B (-148) is correct because it is the result of the dot product of vectors OA and AB, calculated using the equation of the tangent to the curve at point A.
6 Hard
6 of 61
Application +4 | -1
Angle between diagonals of a parallelogram whose sides are represented by a=2i^+j^+k^\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} and b=i^−j^−k^\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}
View answer and explanation
Correct option: A
cos⁡−1(13)\cos ^{-1}\left(\frac{1}{3}\right)
Option A (cos⁡−1(13)\cos^{-1}(\frac{1}{3})) is correct because it is the angle between the diagonals calculated using the dot product formula.
7 Hard
7 of 61
Understanding +4 | -1
If A^\hat{\mathbf{A}} is a unit vector in a given direction, then the value of A^⋅dA^dt\hat{\mathbf{A}} \cdot \frac{d \hat{\mathbf{A}}}{d t} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the dot product of a unit vector and its derivative with respect to time is always zero, as the magnitude of the unit vector is constant.
8 Hard
8 of 61
Application +4 | -1
If ∣a∣=3,∣b∣=4|\mathbf{a}|=3,|\mathbf{b}|=4, then a value of λ\lambda for which a+λb\mathbf{a}+\lambda \mathbf{b} is perpendicular to a−λb\mathbf{a}-\lambda \mathbf{b}.
View answer and explanation
Correct option: B
3/43 / 4
Option B (3/43 / 4) is correct because it is the value of λ\lambda that makes the dot product of the two vectors zero, indicating perpendicularity.
9 Hard
9 of 61
Analyzing +4 | -1
If a=3i^−j^+5k^\mathbf{a}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}} and b=i^+2j^−3k^\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} are given vector. A vector c\mathbf{c} which is perpendicular to ZZ-axis satisfying c⋅a=9\mathbf{c} \cdot \mathbf{a}=9 and c⋅b=−4\mathbf{c} \cdot \mathbf{b}=-4. If inclination of c\mathbf{c} with XX-axis and γ\gamma-axis is α\alpha and β\beta respectively, then which of the following is not true?
View answer and explanation
Correct option: C
α>π2\alpha>\frac{\pi}{2}
Option C (α>π2\alpha > \frac{\pi}{2}) is correct because by finding the vector c and its components, we determine the angle it makes with the x-axis, which is less than π2\frac{\pi}{2}.
10 Hard
10 of 61
Analyzing +4 | -1
Two vectors A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} are inclined at an angle θ\theta. Now if the vectors are interchanged then the resultant turns through an angle β\beta. Which of the following relation is true
View answer and explanation
Correct option: B
tan⁡α2=(A−BA+B)tan⁡θ2\tan \frac{\alpha }{2} = {\left( {\frac{{A - B}}{{A + B}}} \right)}\tan \frac{\mathrm{\theta }}{2}
Option B (tan⁡α2=(A−BA+B)tan⁡θ2\tan \frac{\alpha}{2} = {(\frac{{A - B}}{{A + B}})}\tan \frac{{\theta}}{2}) is correct because it represents the correct relationship between the angle of rotation and the magnitudes of the vectors.
11 Hard
11 of 61
Application +4 | -1
In cartesian coordinates the point AA is (x1,y1)\left(x_1, y_1\right), where x1=1x_1=1 on the curve y=x2+x+10y=x^2+x+10. Then tangent at AA cuts the XX-axis at BB. The value of the dot product OA⋅AB\mathbf{O A} \cdot \mathbf{A B} is
View answer and explanation
Correct option: B
−148-148
Option B (-148) is correct because it is the dot product of vectors OA and AB, calculated using the tangent to the curve at point A.
12 Hard
12 of 61
Analyzing +4 | -1
If (a+3b)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\mathbf{b} is
View answer and explanation
Correct option: A
60∘60^{\circ}
Option A (60∘60^{\circ}) is correct because solving the dot product equations leads to cos⁡θ=12\cos \theta = \frac{1}{2}, which corresponds to an angle of 60∘60^{\circ}.
13 Hard
13 of 61
Application +4 | -1
Consider three vectors A=i^+j^−2k^,B=i^−j^+k^\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} and C=2i^−3j^+4k^\mathbf{C}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}. A vector X\mathbf{X} of the form αA+βB\alpha \mathbf{A}+\beta \mathbf{B} ( α\alpha and β\beta are numbers) is perpendicular to C\mathbf{C}. The ratio of α\alpha and β\beta is
View answer and explanation
Correct option: A
1:11: 1
Option A (1:11: 1) is correct because the ratio of α\alpha and β\beta is 1:1 when vector X\mathbf{X} is perpendicular to vector C\mathbf{C}.
14 Hard
14 of 61
Understanding +4 | -1
(a⋅i^)i^+(a⋅j^)j^+(a⋅k^)k^(\mathbf{a} \cdot \hat{\mathbf{i}}) \hat{\mathbf{i}}+(\mathbf{a} \cdot \hat{\mathbf{j}}) \hat{\mathbf{j}}+(\mathbf{a} \cdot \hat{\mathbf{k}}) \hat{\mathbf{k}} is equal to
View answer and explanation
Correct option: A
a\mathbf{a}
Option A (a\mathbf{a}) is correct because the expression represents the vector a\mathbf{a} resolved along the i, j, and k axes, which sums back to the original vector.
15 Hard
15 of 61
Analyzing +4 | -1
Statement I Two forces (P+Q) and (P-Q) where P⊥QP \bot Q, when act at an angle θ1{\theta _1} to each other, the magnitude of their resultant is 3(P2+Q2),\sqrt {3({P^2} + {Q^2})} , when they act at an angle θ2,{\theta _2}, 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. θ1=60∘  and  θ2=90∘{\theta _1} = {60^ \circ }\,\,\mathrm{and}\,\,{\theta _2} = {90^ \circ }
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: B
Both Statement I and Statement II are true
Option B (Both Statement I and Statement II are true) is correct because solving the equations for the resultant forces with the given angles confirms that both statements are true.

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16 Hard
16 of 61
Application +4 | -1
If ∣a∣=5,∣a−b∣=8|\mathbf{a}|=5,|\mathbf{a}-\mathbf{b}|=8 and ∣a+b∣=10|\mathbf{a}+\mathbf{b}|=10, then ∣b∣|\mathbf{b}| is equal to
View answer and explanation
Correct option: B
57\sqrt{57}
Option B (57\sqrt{57}) is correct because it is derived from the vector identity relating the magnitudes of the sum and difference of two vectors to their individual magnitudes.
17 Hard
17 of 61
Analyzing +4 | -1
The vectors a=2λ2i^+4λj^+k^\mathbf{a}=2 \lambda^2 \hat{\mathbf{i}}+4 \lambda \hat{\mathbf{j}}+\hat{\mathbf{k}} and b=7i^−2j^+λk^\mathbf{b}=7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} make an obtuse angle whereas the angle between b\mathbf{b} and k^\hat{\mathbf{k}} is acute and less than π/6\pi / 6,
View answer and explanation
Correct option: D
null set
Option D (null set) is correct because the conditions imposed on λ lead to contradictory requirements, resulting in no possible value for λ.
18 Hard
18 of 61
Application +4 | -1
If the unit vectors e1\mathbf{e}_1 and e2\mathbf{e}_2 are inclined at an angle 2θ2 \theta and ∣e1−e2∣<1\left|\mathbf{e}_1-\mathbf{e}_2\right|<1, then for θ∈[0,π],θ\theta \in[0, \pi], \theta may lie in the interval
View answer and explanation
Correct option: A
[0,π6)\left[0, \frac{\pi}{6}\right)
Option A ([0, π6\frac{\pi}{6})) is correct because it satisfies the inequality derived from the given condition |e1−e2\mathbf{e}_1 - \mathbf{e}_2| < 1.
19 Hard
19 of 61
Analyzing +4 | -1
If a,b,ca, b, c and A,B,C∈R−{0}A, B, C \in R-\{0\} such that aA+bB+cC+(a2+b2+c2)(A2+B2+C2)=0a A+b B+c C+\sqrt{\left(a^2+b^2+c^2\right)\left(A^2+B^2+C^2\right)}=0, then value of aBbA=bCcB+cAaC\frac{a B}{b A}=\frac{b C}{c B}+\frac{c A}{a C} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the given condition implies that the vectors are anti-parallel, leading to the value 3.
20 Hard
20 of 61
Analyzing +4 | -1
If a,b and c are unit vectors, then ∣a−b∣2+∣b−c∣2+∣c−a∣2|\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2 does not exceed
View answer and explanation
Correct option: B
9
Option B (9) is correct because by expanding the expression and using the fact that the vectors are unit vectors, we can find an upper bound for the given expression.
21 Hard
21 of 61
Application +4 | -1
The projection of a=2i^+3j^−2k^\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} on b=i^+2j^+3k^\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} is
View answer and explanation
Correct option: B
214\frac{2}{\sqrt{14}}
Option B (214\frac{2}{\sqrt{14}}) is correct because it is the result of applying the projection formula of vector a\mathbf{a} on b\mathbf{b}.
22 Hard
22 of 61
Analyzing +4 | -1
If x,yx, y and zz are distinct scalars such that [xa+yb+zc,xb+yc+za,xc+ya+zb]=0][x \mathbf{a}+y \mathbf{b}+z \mathbf{c}, x \mathbf{b}+y \mathbf{c}+z \mathbf{a}, x \mathbf{c}+y \mathbf{a}+z \mathbf{b}]=0] where a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} are non-coplanar vectors, then
View answer and explanation
Correct option: A
x+y+z=0x+y+z=0
The correct answer is A (x+y+z=0x+y+z=0) because the given condition implies that the determinant of the coefficients is zero, leading to the factor (x+y+z) being zero since x, y, and z are distinct.
23 Hard
23 of 61
Analyzing +4 | -1
If a,b,ca, b, c and A,B,C∈R−{0}A, B, C \in R-\{0\} such that aA+bB+cC+(a2+b2+c2)(A2+B2+C2)=0a A+b B+c C+\sqrt{\left(a^2+b^2+c^2\right)\left(A^2+B^2+C^2\right)}=0, then value of aBbA=bCcB+cAaC\frac{a B}{b A}=\frac{b C}{c B}+\frac{c A}{a C} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the given condition implies that the vectors are anti-parallel, leading to the ratio of components being equal, and thus the expression simplifies to 3.
24 Hard
24 of 61
Analyzing +4 | -1
If x,yx, y and zz are distinct scalars such that [xa+yb+zc,xb+yc+za,xc+ya+zb]=0][x \mathbf{a}+y \mathbf{b}+z \mathbf{c}, x \mathbf{b}+y \mathbf{c}+z \mathbf{a}, x \mathbf{c}+y \mathbf{a}+z \mathbf{b}]=0] where a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} are non-coplanar vectors, then
View answer and explanation
Correct option: A
x+y+z=0x+y+z=0
Option A (x+y+z=0x+y+z=0) is correct because the scalar triple product being zero implies that x+y+z=0x+y+z=0 given that x, y, and z are distinct.
25 Hard
25 of 61
Application +4 | -1
The unit vector in ZOX plane making angles 45∘45^{\circ} and 60∘60^{\circ} respectively, with a=2i^+2j^−k^a=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} and b=j^−k^\mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}, is
View answer and explanation
Correct option: B
12(i^−k^)\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})
Option B (12(i^−k^)\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})) is correct because it satisfies the conditions of being a unit vector in the ZOX plane and making the specified angles with vectors a and b.
26 Hard
26 of 61
Application +4 | -1
If ∣a∣=3,∣b∣=4|\mathbf{a}|=3,|\mathbf{b}|=4, then a value of λ\lambda for which a+λb\mathbf{a}+\lambda \mathbf{b} is perpendicular to a−λb\mathbf{a}-\lambda \mathbf{b}.
View answer and explanation
Correct option: B
3/43 / 4
Option B (3/43 / 4) is correct because it satisfies the condition for perpendicularity between the vectors a+λb\mathbf{a}+\lambda \mathbf{b} and a−λb\mathbf{a}-\lambda \mathbf{b}.
27 Hard
27 of 61
Analyzing +4 | -1
Let a, b,c, be vectors of length 3, 4, 5 respectively and a\mathbf{a} be perpendicular to (b+c),b(\mathbf{b}+\mathbf{c}), \mathbf{b} to (c+a)(\mathbf{c}+\mathbf{a}) and c\mathbf{c} to (a+b)(\mathbf{a}+\mathbf{b}), then the value of (a+b+c)(\mathbf{a}+\mathbf{b}+\mathbf{c}) is
View answer and explanation
Correct option: D
525 \sqrt{2}
The correct answer is D (525 \sqrt{2}) because by using the given perpendicularity conditions and vector algebra, we find the magnitude of the sum of the vectors.
28 Hard
28 of 61
Analyzing +4 | -1
If OABCO A B C is a tetrahedron such that OA2+BC2\mathrm{OA}^2+\mathrm{BC}^2 =OB2+CA2=OC2+AB2=\mathrm{OB}^2+\mathrm{CA}^2=\mathrm{OC}^2+\mathrm{AB}^2, then which of the following is not true?
View answer and explanation
Correct option: D
AB⊥ACA B \perp A C
Option D (AB⊥ACA B \perp A C) is correct because the given conditions imply that OA⊥BCOA \perp BC, OB⊥ACOB \perp AC, and OC⊥ABOC \perp AB, but not necessarily that AB⊥ACAB \perp AC.
29 Hard
29 of 61
Analyzing +4 | -1
Let A, B and C\mathbf{C} be unit vectors. Suppose A⋅B=A⋅C=0\mathbf{A} \cdot \mathbf{B}=\mathbf{A} \cdot \mathbf{C}=0 and the angle between B\mathbf{B} and C\mathbf{C} is π4\frac{\pi}{4}. Then,
View answer and explanation
Correct option: B
A=±2(B×C)\mathbf{A}=\pm \sqrt{2}(\mathbf{B} \times \mathbf{C})
The correct answer is B (A=±2(B×C)\mathbf{A}=\pm \sqrt{2}(\mathbf{B} \times \mathbf{C})) because it is derived from the fact that A is perpendicular to both B and C, and A is a unit vector.
30 Hard
30 of 61
Understanding +4 | -1
(a⋅i^)i^+(a⋅j^)j^+(a⋅k^)k^(\mathbf{a} \cdot \hat{\mathbf{i}}) \hat{\mathbf{i}}+(\mathbf{a} \cdot \hat{\mathbf{j}}) \hat{\mathbf{j}}+(\mathbf{a} \cdot \hat{\mathbf{k}}) \hat{\mathbf{k}} is equal to
View answer and explanation
Correct option: A
a\mathbf{a}
Option A (a\mathbf{a}) is correct because the expression simplifies to the original vector a\mathbf{a} using the properties of dot products with unit vectors.
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