Skip to main content
Back to DOT PRODUCT OF VECTORS

Your Progress

0 / 241 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 181-210 of 241 questions

Preparing practice and saved progress…

181 Easy
181 of 241
Application +4 | -1
If ∣  a⃗×b⃗∣=4  and  ∣  a⃗⋅b⃗∣=2,  then  ∣a⃗∣2∣b⃗∣2\left| {\mathrm{\;}\mathbf{\vec a} \times \mathbf{\vec b}} \right| = 4\mathrm{\;and\;}\left| {\mathrm{\;}\mathbf{\vec a}\cdot\mathbf{\vec b}} \right| = 2,\mathrm{\;then\;}{\left| {\mathbf{\vec a}} \right|^2}{\left| {\mathbf{\vec b}} \right|^2} is equal to
View answer and explanation
Correct option: D
20
Option D (20) is correct because it is the result of using the identity relating the magnitudes of the cross product and dot product of two vectors.
182 Easy
182 of 241
Application +4 | -1
If sdaa a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are vectors such that a⃗+b⃗+c⃗=0\vec{a}+\vec{b}+\vec{c}=0 and ∣a⃗∣=7,∣b⃗∣=5,∣c⃗∣=3|\vec{a}|=7,|\vec{b}|=5,|\vec{c}|=3 then angle between vector b⃗\vec{b} and c⃗\vec{c} is
View answer and explanation
Correct option: A
60∘60^{\circ}
Option A (60∘60^{\circ}) is correct because the angle between vectors b⃗\vec{b} and c⃗\vec{c} is found using the dot product and the given magnitudes.
183 Easy
183 of 241
Understanding +4 | -1
If a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} are three vectors such that a=b+c\mathbf{a}=\mathbf{b}+\mathbf{c} and the angle between b\mathbf{b} and c\mathbf{c} is π2\frac{\pi}{2}, then
View answer and explanation
Correct option: A
a2=b2+c2a^2=b^2+c^2
Option A (a2=b2+c2a^2=b^2+c^2) is correct because, with a=b+c\mathbf{a}=\mathbf{b}+\mathbf{c} and the angle between b\mathbf{b} and c\mathbf{c} being π2\frac{\pi}{2}, the Pythagorean theorem applies.
184 Easy
184 of 241
Application +4 | -1
Forces acting on a particle have magnitude 5, 3 and 1 unit and act in the direction of the vectors 6i^+2j^+3k^\mathbf{\hat i} + 2\mathbf{\hat j} + 3\mathbf{\hat k}, 3i^−2j^+6k^3\mathbf{\hat i} - 2\mathbf{\hat j} + 6\mathbf{\hat k} and 2i^−3j^−6k^2\mathbf{\hat i} - 3\mathbf{\hat j} - 6\mathbf{\hat k} respectively. They remain constant while the particle is displaced from the points A(2,−1,−3)A\left( {2, - 1, - 3} \right) to B(5,−1,1)B\left( {5, - 1,1} \right). The work done is
View answer and explanation
Correct option: B
33 unit
The correct answer is B (33 unit) because it is the result of calculating the work done by the forces during the displacement, using the dot product of the resultant force and the displacement vector.
185 Easy
185 of 241
Application +4 | -1
If the scalar product of the vector \widehat \mathbf{i} + \widehat \mathbf{j} + 2\mathbf{\hat k}\;\widehat \mathbf{i} + \widehat \mathbf{j} + 2\mathbf{\hat k}\; with the unit vector along m\widehat \mathbf{i} + 2\widehat \mathbf{j} + 3\mathbf{\hat k}m\widehat \mathbf{i} + 2\widehat \mathbf{j} + 3\mathbf{\hat k} is equal to 2, then one of the value of mm is
View answer and explanation
Correct option: D
6
Option D (6) is correct because it is one of the solutions for m obtained by setting up the scalar product equation and solving the resulting quadratic equation.
186 Easy
186 of 241
Analyzing +4 | -1
Let three vectors aa, bb and cc be such that a×b=c,  b×c=aa \times b = c,\,\,b \times c = a and ∣a∣=2\left| a \right| = 2. Then, which one of the following is not true?
View answer and explanation
Correct option: D
∣3a+b−2c∣2=51{\left| {3a + b - 2c} \right|^2} = 51
Option D (∣3a+b−2c∣2=51{\left| {3a + b - 2c} \right|^2} = 51) is not true because by using the given conditions and vector properties, the correct value of the expression is 53, not 51.
187 Easy
187 of 241
Application +4 | -1
If a=i+j+k{a}={i}+{j}+{k} and b=i−j+2k{b}={i}-{j}+2 {k} then the projection of a{a} on b{b} is given by
View answer and explanation
Correct option: D
13(i−j+2k)\frac{1}{3}({i}-{j}+2 {k})
Option D (13(i−j+2k)\frac{1}{3}({i}-{j}+2 {k})) is correct because the projection of {a} on {b} is calculated using the formula and results in 13(i−j+2k)\frac{1}{3}({i}-{j}+2 {k}).
188 Easy
188 of 241
Application +4 | -1
If forces of magnitudes 6 and 7 units acting in the directions i^−2j^+2k^\hat i - 2\hat j + 2\hat k and 2i^−3j^−6k^2\hat i - 3\hat j - 6\hat k respectively act on a particle which is displaced from the point P(2,  −1,  −3)P\left( {2,\; - 1,\; - 3} \right) to Q(5,  −1,  1)Q\left( {5,\; - 1,\;1} \right), then the work done by the forces is
View answer and explanation
Correct option: A
4 units
Option A (4 units) is correct because the work done is calculated by finding the resultant force, displacement vector, and then taking their dot product.
189 Easy
189 of 241
Application +4 | -1
Let uˉ,vˉ,wˉ\bar{u}, \bar{v}, \bar{w} be such that ∣uˉ∣=1,∣vˉ∣=2,∣wˉ∣=3|\bar{u}|=1,|\bar{v}|=2,|\bar{w}|=3. If the projection vˉ\bar{v} along uˉ\bar{u} is equal to that of wˉ\bar{w} along uˉ\bar{u} and vˉ,wˉ\bar{v}, \bar{w} are perpendicular to each other then ∣uˉ−vˉ+wˉ∣|\bar{u}-\bar{v}+\bar{w}| equals
View answer and explanation
Correct option: C
14\sqrt{14}
Option C (14\sqrt{{14}}) is correct because it is the magnitude of the vector uˉ−vˉ+wˉ\bar{{u}}-\bar{{v}}+\bar{{w}} calculated using the given conditions.
190 Easy
190 of 241
Analyzing +4 | -1
If aa is a real constant and A,B,C\mathrm{A}, \mathrm{B}, \mathrm{C} are variable angles and a2−4Tan⁡A+aTan⁡B+a2+4Tan⁡C=6a\sqrt{a^2-4} \operatorname{Tan} A+a \operatorname{Tan} B+\sqrt{a^2+4} \operatorname{Tan} C=6 a Then the least value of Tan⁡2A+Tan⁡2B+Tan⁡2C=\operatorname{Tan}^2 A+\operatorname{Tan}^2 B+\operatorname{Tan}^2 C=
View answer and explanation
Correct option: C
12
Option A (10,11) is correct because the number of solutions depends on the range of the greatest integer function, and the given conditions are satisfied when n is 10 or 11.
191 Easy
191 of 241
Application +4 | -1
Suppose \mathbf{\vec a} = \lambda \;\widehat \mathbf{i} - 7\widehat \mathbf{j} + 3\mathbf{\hat k}\mathbf{\vec a} = \lambda \;\widehat \mathbf{i} - 7\widehat \mathbf{j} + 3\mathbf{\hat k}, \mathbf{\vec b} = \lambda \;\widehat \mathbf{i} + \widehat \mathbf{j} + 2\lambda \;\mathbf{\hat k}.\mathbf{\vec b} = \lambda \;\widehat \mathbf{i} + \widehat \mathbf{j} + 2\lambda \;\mathbf{\hat k}. If the angle between a⃗{\mathbf{\vec a}} and b⃗{\mathbf{\vec b}} is grater than 90∘90^\circ, then λ satisfies the inequality
View answer and explanation
Correct option: A
−7<λ<1- 7 < \lambda < 1
The correct answer is A. The angle between two vectors is greater than 90 degrees when their dot product is negative, leading to the inequality −7<λ<1- 7 < \lambda < 1.
192 Easy
192 of 241
Application +4 | -1
\mathbf{\vec a}\cdot\widehat \mathbf{i} = \mathbf{\vec a}\cdot\left( {2\widehat \mathbf{i} + \widehat \mathbf{j}} \right) = \mathbf{\vec a}\cdot\left( {\widehat \mathbf{i} + \widehat \mathbf{j} + 3\mathbf{\hat k}} \right) = 1,\;\mathbf{\vec a}\cdot\widehat \mathbf{i} = \mathbf{\vec a}\cdot\left( {2\widehat \mathbf{i} + \widehat \mathbf{j}} \right) = \mathbf{\vec a}\cdot\left( {\widehat \mathbf{i} + \widehat \mathbf{j} + 3\mathbf{\hat k}} \right) = 1,\; then a⃗\mathbf{\vec a} is equal to
View answer and explanation
Correct option: D
\left( {3\widehat \mathbf{i} - 3\widehat \mathbf{j} + \mathbf{\hat k}} \right)/3\left( {3\widehat \mathbf{i} - 3\widehat \mathbf{j} + \mathbf{\hat k}} \right)/3
Option D ((3i^−3j^+k^)/3\left( {3\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} + \mathbf{\hat k}} \right)/3) is correct because solving the system of equations derived from the dot product conditions yields a⃗=13(3i^−3j^+k^)\mathbf{\vec a} = \frac{1}{3}\left( {3\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} + \mathbf{\hat k}} \right).
193 Easy
193 of 241
Application +4 | -1
If a⃗{\vec a} and b⃗{\vec b} are unit vectors and θ\theta is the angle between them then ∣a⃗−b⃗2∣\left| {\frac{{\vec a - \vec b}}{2}} \right|, is
View answer and explanation
Correct option: A
sin⁡θ2\sin \frac{\theta }{2}
Option A (sin⁡θ2\sin \frac{\theta }{{2}}) is correct because it is derived from the properties of unit vectors and the angle between them, using the half-angle formula.
194 Easy
194 of 241
Analyzing +4 | -1
Given four vectors a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c} and d\mathbf{d}. The vectors a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c} are coplanar but not collinear pair by pair and the vector d\mathbf{d} is not coplanar with the vecotrs a,b\mathbf{a}, \mathbf{b} and c\mathbf{c}. If it is known that the angle between a\mathbf{a} and b\mathbf{b} is equal to that between bb and c, each being equal to 60∘60^{\circ}. The angle between dd and a\mathbf{a} is α\alpha and between dd and bb is β\beta. Then, the angle between the vectors dd and cc
View answer and explanation
Correct option: A
cos⁡−1(cos⁡β−cos⁡α)\cos ^{-1}(\cos \beta-\cos \alpha)
Option A is correct because the angle between d and c can be found using the given angle conditions and coplanarity of a, b, and c.
195 Easy
195 of 241
Application +4 | -1
If a⃗=(i^+j^+k^),   a⃗⋅b⃗=1    and   a⃗×b⃗=j^−k^,     then    b⃗\mathbf{\vec a} = \left( {\mathbf{\hat i} + \mathbf{\hat j} + \mathbf{\hat k}} \right),\;\,\mathbf{\vec a} \cdot \mathbf{\vec b} = 1\mathrm{\;}\,\,\mathrm{and}\,\mathrm{\;}\mathbf{\vec a} \times \mathbf{\vec b} = \mathbf{\hat j} - \mathbf{\hat k},\mathrm{\;\;}\,\mathrm{then\;}\,\,\mathbf{\vec b} is

View answer and explanation
Correct option: C
i^{\mathbf{\hat i}}
Option C (i^{\mathbf{\hat i}}) is correct because by using the properties of dot and cross products, we can solve for the vector b⃗\mathbf{\vec b}.
196 Easy
196 of 241
Application +4 | -1
If ∣a⃗∣=5,∣b⃗∣=4,∣c⃗∣=3|\vec{a}|=5,|\vec{b}|=4,|\vec{c}|=3 thus what will be the value of ∣a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗∣|\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}|, given that a⃗+b⃗+c⃗=0\vec{a}+\vec{b}+\vec{c}=0
View answer and explanation
Correct option: A
25
Option A (25) is correct because it is the absolute value of the dot product sum, calculated using the condition a⃗+b⃗+c⃗=0\vec{{a}}+\vec{{b}}+\vec{{c}}=0.
197 Easy
197 of 241
Application +4 | -1
A tetrahedron has vertices at O  (0,  0),  A(1,  2,  1),  B(2,  1,  3)O\;\left( {0,\;0} \right),\;A\left( {1,\;2,\;1} \right),\;B\left( {2,\;1,\;3} \right) and C(−1,  1,  2)C\left( { - 1,\;1,\;2} \right). Then, the angle between the faces OAB  and  ABCOAB\;\mathrm{and}\;ABC will be
View answer and explanation
Correct option: A
cos⁡−1(1935){\cos ^{ - 1}}\left( {\frac{{19}}{{35}}} \right)
Option A (cos⁡−1(1935){\cos ^{ - 1}}\left( {\frac{{19}}{{35}}} \right)) is correct because it represents the angle between the normal vectors of the two faces, which is the angle between the faces.
198 Easy
198 of 241
Analyzing +4 | -1
Let the unit vectors a⃗  and  b⃗\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b} be perpendicular to each other and the unit vector c⃗{\mathbf{\vec c}} be inclined at an angle θ\mathrm{\theta } to both a⃗{\mathbf{\vec a}} and b⃗\mathbf{\vec b}. If c⃗=α,a⃗+β,  b⃗+γ(a⃗⋅b⃗),\mathbf{\vec c} = \alpha ,\mathbf{\vec a} + \mathrm{\beta },\mathrm{\;}\mathbf{\vec b} + \gamma \left( {\mathbf{\vec a}\cdot\mathbf{\vec b}} \right), where α,β,γ\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma } are scalars, then
View answer and explanation
Correct option: B
α=cos⁡θ,  β=cos⁡θ,  γ2=cos⁡2θ\mathrm{\alpha } = \cos \mathrm{\theta },\mathrm{\;\beta } = \cos \mathrm{\theta },\mathrm{\;}{\mathrm{\gamma }^2} = \cos 2\mathrm{\theta }
Option B is correct because by taking dot products of c⃗\vec c with a⃗\vec a and b⃗\vec b, we can find α\alpha and β\beta, and then use the magnitude of c⃗\vec c to find the relation for γ2\gamma^2.
199 Easy
199 of 241
Application +4 | -1
If a⃗,   b⃗,   c⃗\vec a,\;\,\vec b,\;\,\vec c are three mutually perpendicular vectors each of magnitude unity, then ∣ a⃗+b⃗+c⃗ ∣\left| {\,\vec a + \vec b + \vec c\,} \right| is equal to
View answer and explanation
Correct option: C
3\sqrt 3
Option C (3\sqrt 3) is correct because the magnitude of the sum of three mutually perpendicular unit vectors is 3\sqrt 3.
200 Easy
200 of 241
Application +4 | -1
If a⃗,  b⃗\mathbf{\vec a},\,\,\mathbf{\vec b} and b⃗+c⃗,  c⃗+a⃗\mathbf{\vec b} + \mathbf{\vec c},\,\,\mathbf{\vec c} + \mathbf{\vec a} are perpendicular to a⃗+b⃗\mathbf{\vec a} + \mathbf{\vec b} and ∣ a⃗+b⃗ ∣=6\left| {\,\mathbf{\vec a} + \mathbf{\vec b}\,} \right| = 6 respectively and if ∣ a⃗+b⃗ ∣=6\left| {\,\mathbf{\vec a} + \mathbf{\vec b}\,} \right| = 6, ∣ b⃗+c⃗ ∣=8\left| {\,\mathbf{\vec b} + \mathbf{\vec c}\,} \right| = 8 and ∣ c⃗+a⃗ ∣=10\left| {\,\mathbf{\vec c} + \mathbf{\vec a}\,} \right| = 10, then ∣ a⃗+b⃗+c⃗ ∣\left| {\,\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}\,} \right| is equal to
View answer and explanation
Correct option: D
10
The correct answer is D (10) because by using the given perpendicularity conditions and magnitudes, we can find the magnitude of the sum of the vectors.
201 Easy
201 of 241
Application +4 | -1
A parallelogram is constructed on the vectors a⃗=3α⃗−β⃗,  b⃗=α⃗+3β⃗\mathbf{\vec a} = 3\vec \alpha - \mathrm{\vec \beta },\;\mathbf{\vec b} = \mathrm{\vec \alpha } + 3\mathrm{\vec \beta }, if ∣α⃗∣=∣β⃗∣=2\left| {\mathrm{\vec \alpha }} \right| = \left| {\mathrm{\vec \beta }} \right| = 2 and angle between α⃗{\mathrm{\vec \alpha }} and β⃗  isπ3,\mathrm{\vec \beta }\;\mathrm{is}\frac{\mathrm{\pi }}{3}, then length of diagonal of the parallelogram is
View answer and explanation
Correct option: B
434\sqrt 3
Option A (474\sqrt{7}) is correct because it is the length of one of the diagonals of the parallelogram, calculated using vector addition and the given magnitudes and angle.
202 Easy
202 of 241
Analyzing +4 | -1
A vector a⃗\vec{a} is parallel to the line of intersection of the plane determined by the vectors i^,i^+j^\hat{i}, \hat{i}+\hat{j} and the plane determined by the vectors i^−j^,i^+k^\hat{i}-\hat{j}, \hat{i}+\hat{k}. The obtuse angle between a⃗\vec{a} and the vector b⃗=i^−2j^+2k^\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k} is
View answer and explanation
Correct option: A
3π4\frac{3 \pi}{4}
Option A (3π4\frac{{3 \pi}}{{4}}) is correct because the vector a⃗\vec{{a}} is parallel to the intersection of two planes, and the obtuse angle between a⃗\vec{{a}} and b⃗\vec{{b}} is found using the dot product formula.
203 Easy
203 of 241
Application +4 | -1
If n^1,  n^2{{\hat n}_1},\;{{\hat n}_2} are two unit vectors and θ\theta is the angle between them, then cos⁡θ/2=\cos \theta /2 =
View answer and explanation
Correct option: A
12∣n^1+n^2∣\frac{1}{2}\left| {{{\hat n}_1} + {{\hat n}_2}} \right|
The correct answer is A (12∣n^1+n^2∣\frac{1}{2}\left| {{{\hat n}_1} + {{\hat n}_2}} \right|) because it is the correct expression for cos⁡θ/2\cos \theta /2 in terms of the unit vectors n^1{\hat n}_1 and n^2{\hat n}_2 and the angle between them.
204 Easy
204 of 241
Application +4 | -1
If ∣  a⃗∣=3,  ∣b⃗∣=4\left| {\;\mathbf{\vec a}} \right| = 3,\mathrm{\;}\left| {\mathbf{\vec b}} \right| = 4, then a value of λ\lambda for which a⃗+λb⃗\mathbf{\vec a} + \mathrm{\lambda }\mathbf{\vec b} is perpendicular to a⃗−λb⃗\mathbf{\vec a} - \mathrm{\lambda }\mathbf{\vec b}, is
View answer and explanation
Correct option: B
34\frac{3}{4}
Option B (34\frac{3}{4}) is correct because the value of λ\lambda is found by using the perpendicularity condition (dot product equals zero) and the given magnitudes.
205 Easy
205 of 241
Analyzing +4 | -1
If lines x=ay+b,z=cy+dx = ay + b,z = cy + d\mathrm{ } and x=a′z+b′,y=c′z+d′x = a'z + b',y = c'z + d' are perpendicular, then
View answer and explanation
Correct option: C
aa′+c+c′=0aa' + c + c' = 0
Option C (aa′+c+c′=0aa' + c + c' = 0) is correct because it represents the condition for perpendicularity between two lines in 3D space, derived from the dot product of their direction vectors.
206 Easy
206 of 241
Analyzing +4 | -1
If ∣a+b∣>∣a−b∣|\mathbf{a}+\mathbf{b}|>|\mathbf{a}-\mathbf{b}|, then the angle between a\mathbf{a} and b\mathbf{b} is
View answer and explanation
Correct option: A
acute
Option A (acute) is correct because the given inequality implies that the dot product of a\mathbf{a} and b\mathbf{b} is positive, which means the angle between them is acute.
207 Easy
207 of 241
Application +4 | -1
If a\mathbf{a} and b\mathbf{b} are two unit vectors such that a+2b\mathbf{a}+2 \mathbf{b} and 5a−4b5 a-4 b are perpendicular to each other, then the angle between a\mathbf{a} and b\mathbf{b} is
View answer and explanation
Correct option: B
60∘60^{\circ}
Option B (60∘60^{\circ}) is correct because by using the perpendicularity condition and the fact that a and b are unit vectors, we can find the angle between them.
208 Easy
208 of 241
Application +4 | -1
If a=(1,−1,2),b=(−2,3,5),c=(2,−2,4)\mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4) and i\mathbf{i} is the unit vector in the xx-direction, then (a−2b+3c).i=(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) . \mathbf{i}=
View answer and explanation
Correct option: A
11
Option A (11) is correct because it is the x-component of the resulting vector (a-2b+3c) after performing the vector operations.
209 Easy
209 of 241
Application +4 | -1
If ai+6j−ka \mathbf{i}+6 \mathbf{j}-\mathbf{k} and 7i−3j+17k7 \mathbf{i}-3 \mathbf{j}+17 \mathbf{k} are perpendicular vectors, then the value of aa is
View answer and explanation
Correct option: A
5
Option A (5) is correct because the dot product of perpendicular vectors is zero, allowing us to solve for 'a'.
210 Easy
210 of 241
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 the correct answer because the given condition implies that OA⊥BCOA \perp BC, OB⊥ACOB \perp AC, and OC⊥ABOC \perp AB, but not necessarily AB⊥ACAB \perp AC.
Revision notes and key takeaways

Review DOT PRODUCT OF VECTORS in your textbook or class notes, then use this set to check your understanding.

Showing 181-210 of 241 questions

Your practice

0 / 241 answered · 241 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.