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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Medium

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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 55 questions

1 Medium

1 of 55

Application +4 | -1

The distance between the planes given by

\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

and

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

View answer and explanation

Correct option: B

\frac{13}{3}

unit

Option B is correct because the distance between two parallel planes is calculated using the formula involving the constants and the magnitude of the normal vector.

2 Medium

2 of 55

Application +4 | -1

Find the equation of the perpendicular drown from the origin to the plane

2x + 4y - 5z = 10

View answer and explanation

Correct option: B

\mathbf{\vec r} = \left( {2k,4k, - 5k} \right)k \in R

Option B (

\mathbf{\vec r} = \left( {2k,4k, - 5k} \right)k \in R

) is correct because it represents a line passing through the origin and perpendicular to the given plane, with direction ratios matching the coefficients of x, y, and z in the plane's equation.

3 Medium

3 of 55

Application +4 | -1

Let

a, b>0

and

\vec{\alpha}=\frac{\hat{i}}{a}+\frac{4 \hat{j}}{b}+b k

and

\vec{\beta}=b \hat{i}+a \hat{j}+\frac{1}{b} \hat{k}

then the maximum value of

\frac{10}{5+\vec{\alpha} \cdot \vec{\beta}}

View answer and explanation

Correct option: A

Option A (1) is correct because the maximum value of the expression is found by minimizing the denominator using AM-GM inequality.

4 Medium

4 of 55

Application +4 | -1

The projection of the vector

\hat i - 2\hat j + \hat k

on the vector

4\hat i - 4\hat j + 7\hat k

, is

View answer and explanation

Correct option: B

\frac{{19}}{9}

Option B (

\frac{{19}}{9}

) is correct because it is the result of projecting the vector

\hat i - 2\hat j + \hat k

onto the vector

4\hat i - 4\hat j + 7\hat k

using the projection formula.

5 Medium

5 of 55

Analyzing +4 | -1

The vector

\vec{a}=\hat{i}+2 \hat{j}+\hat{k}

lying in the plane of the vectors

\vec{b}=\hat{i}+\hat{j}

and

\vec{c}

bisects the angle between

\vec{b}

and

\vec{c}

. If

|\vec{c}|=\sqrt{2}

, then

\vec{c} \cdot \vec{a}

View answer and explanation

Correct option: C

Option C (3) is correct because by expressing

\vec{a}

as a linear combination of

\frac{\vec{b}}{|\vec{b}|}

and

\vec{c}

, and using the given conditions, we find

\vec{c} = \hat{j} + \hat{k}

, so

\vec{c} \cdot \vec{a} = 3

6 Medium

6 of 55

Application +4 | -1

If the plane

\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1

, cuts the coordinate axes in

A, B, C

, then the area of

\triangle A B C

View answer and explanation

Correct option: C

\sqrt{61} \mathrm{sq}

. Units

Option C (

\sqrt{{61}}

sq. Units) is correct because it is the area of the triangle formed by the intercepts of the plane with the coordinate axes.

7 Medium

7 of 55

Application +4 | -1

\overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+\mu(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})+v(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})

and

[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=\frac{1}{8}

, then

\lambda+\mu+v

is equal to -

View answer and explanation

Correct option: D

8 \overrightarrow{\mathrm{d}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})

Option D (

8 \overrightarrow{{\mathrm{{d}}}} \cdot (\overrightarrow{{\mathrm{{a}}}} + \overrightarrow{{\mathrm{{b}}}} + \overrightarrow{{\mathrm{{c}}}})

) is correct because by taking the dot product of

\overrightarrow{{\mathrm{{d}}}}

with

(\overrightarrow{{\mathrm{{a}}}} + \overrightarrow{{\mathrm{{b}}}} + \overrightarrow{{\mathrm{{c}}}})

and using the given scalar triple product, we can find the value of

\lambda + \mu + \nu

8 Medium

8 of 55

Application +4 | -1

The vector equation of the plane through the point

(2,1,-1)

and passing through the line of intersection of the plane

\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0

and

\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=0

is -

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the correct equation of the plane is

\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+11 \hat{\mathrm{j}}+15 \hat{\mathrm{k}})=6

, which is not listed.

9 Medium

9 of 55

Analyzing +4 | -1

Let two non-collinear vectors

\mathbf{a}

and

\mathbf{b}

inclined at an angle

\frac{2 \pi}{3}

be such that

|\mathbf{a}|=3

and

|\mathbf{b}|=2

. If a point

P

moves so that at any time

t

its position vector

\mathrm{OP}

(where

O

is the origin) is given as

\mathbf{O P}=\left(t+\frac{1}{t}\right) \mathbf{a}+\left(t-\frac{1}{t}\right) \mathbf{b}

, then least distance of

P

from the origin is

View answer and explanation

Correct option: B

\sqrt{2 \sqrt{133}+10}

Option B (

\sqrt{2 \sqrt{133}+10}

) is correct because it represents the minimum distance of point P from the origin, obtained by minimizing the square of the distance using AM-GM inequality.

10 Medium

10 of 55

Application +4 | -1

The point of intersection of the plane

\vec{r} \cdot(3 \hat{i}-5 \hat{j}+2 \hat{k})=6

with the straight line passing through the origin and perpendicular to the plane

2 x-y-z=4

, is

\left(x_3, y_0, z_0\right)

. The value of

\left(2 x_0-3 y_0+z_0\right)

, is:

View answer and explanation

Correct option: D

Option D (4) is correct because by finding the intersection point of the line and the plane, and then substituting the coordinates into the given expression, the result is 4.

11 Medium

11 of 55

Application +4 | -1

Let

\mathbf{a}

and

\mathbf{b}

be two unit vectors. If the vectors

\mathbf{c}=\mathbf{a}+2 \mathbf{b}

and

\mathbf{d}=5 \mathrm{a}-4 \mathrm{a}

are perpendicular to each other, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: C

\frac{\pi}{3}

Option C (

\frac{\pi}{3}

) is correct because it is the angle between the vectors a and b, derived from the perpendicularity condition of c and d.

12 Medium

12 of 55

Analyzing +4 | -1

Let

\overrightarrow a

and

\overrightarrow b

be the vectors along the diagonals of a parallelogram having area

2\sqrt 2 .

Let the angle between

\overrightarrow a

and

\overrightarrow b

be acute,

\left| {\overrightarrow a } \right| = 1,

and

\left| {\overrightarrow a \cdot \overrightarrow b } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|.

\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b ,

then an angle between

\overrightarrow b

and

\overrightarrow c

View answer and explanation

Correct option: D

\frac{{3\pi }}{4}

Option D (

\frac{{3\pi }}{{4}}

) is correct because it is the angle between vectors

\overrightarrow b

and

\overrightarrow c

calculated using the given conditions and vector properties.

13 Medium

13 of 55

Analyzing +4 | -1

Let

\hat{a}

and

\hat{b}

be two unit vectors such that the angle between them is

\frac{\pi}{4}

. If

\theta

is the angle between the vectors

(\hat{a}+\hat{b})

and

(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))

, then the value of

164 \cos ^2 \theta

is equal to :

View answer and explanation

Correct option: A

90+27 \sqrt{2}

Option A (

90+27 \sqrt{2}

) is correct because it is the value of

164 \cos^2 \theta

calculated using the dot product and magnitudes of the given vectors.

14 Medium

14 of 55

Analyzing +4 | -1

\mathbf{a} \cdot \mathbf{b}

and

\mathbf{c}

are any three non-zero vectors, then the component of

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})

perpendicular to

\mathrm{b}

View answer and explanation

Correct option: D

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{c})}{|b|^2} \mathbf{b}

Option A (1) is correct because the minimum value of

\alpha \cdot \beta

is 5, which maximizes the denominator and minimizes the overall expression.

15 Medium

15 of 55

Application +4 | -1

Let

\vec{a}=\alpha \hat{i}+\hat{j}-k

and

\vec{b}=2 \hat{i}+\hat{j}-\alpha k, \alpha>0

. If the projection of

\vec{a} \times \vec{b}

on the vector

-\hat{i}+2 \hat{j}-2 k

is 30 , then

\alpha

is equal to

View answer and explanation

Correct option: D

7

Option D (7) is correct because solving the projection equation yields α = 7 after rejecting the negative solution.

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16 Medium

16 of 55

Application +4 | -1

The vector equation of the plane containing the lines

\mathbf{r}=(\mathbf{i}+\mathbf{j})+\lambda(\mathbf{i}+2 \mathbf{j}-\mathbf{k})

and

\mathbf{r}=(\mathbf{i}+\mathbf{j})+\mu(-\mathbf{i}+\mathbf{j}-2 \mathbf{k})

View answer and explanation

Correct option: B

\mathbf{r} \cdot(\mathbf{i}-\mathbf{j}-\mathbf{k})=0

Option B,

\mathbf{r} \cdot(\mathbf{i}-\mathbf{j}-\mathbf{k})=0

, is correct because it represents the equation of a plane containing the given lines, found by taking the cross product of the direction vectors.

17 Medium

17 of 55

Application +4 | -1

Constant forces

{{\mathbf{\vec P}}_1} = \widehat \mathbf{i} - \widehat \mathbf{j} + \mathbf{\hat k}\;

{{\mathbf{\vec P}}_2} = - \widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}

and

{{\mathbf{\vec P}}_3} = \widehat \mathbf{j} - \mathbf{\hat k}

act on a particle at point

A

. The work done when the particle is displaced from the point

A

B

where

{{\mathbf{\vec A}}_\;} = 4\widehat \mathbf{i} - 3\widehat \mathbf{j} - 2\mathbf{\hat k}

and

{{\mathbf{\vec B}}_\;} = 6\widehat \mathbf{i} + \widehat \mathbf{j} - 3\mathbf{\hat k}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the work done calculated using the dot product of the total force and displacement is 8, which is not among the given options.

18 Medium

18 of 55

Analyzing +4 | -1

Let

\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}

and

\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}

be three vectors if

\vec{r}

is a vector such that,

\vec{r} \times \vec{b}=\vec{c} \times \vec{b}

and

\vec{r} \cdot \vec{a}=0

, then

25|\vec{r}|^{2}

is equal to

View answer and explanation

Correct option: D

339

Option D (339) is correct because by solving the vector equation and using the dot product condition, we find that

25|\vec{r}|^{2} = 339

19 Medium

19 of 55

Analyzing +4 | -1

\vec{a}=\hat{i}+2 \hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=7 \hat{i}-3 \hat{j}+4 \hat{k}

\vec{r} \times \vec{b}+\vec{b} \times \vec{c}=\overrightarrow{0}

and

\vec{r} \cdot \vec{a}=0

. Then

\vec{r} \cdot \vec{c}

is equal to

View answer and explanation

Correct option: C

Option C (34) is correct because it is the value obtained by correctly solving for vector r using the given conditions and then calculating the dot product.

20 Medium

20 of 55

Analyzing +4 | -1

The value of '

a

' for which the straight line

4 x-3 y+4 z-2=0=3 x-2 y+z-5

is parallel to the plane

2 x-y+a z-7=0

, is:

View answer and explanation

Correct option: B

-2

Option B (-2) is correct because the dot product of the parallel vector of the line and the normal vector of the plane must be zero for the line to be parallel to the plane.

21 Medium

21 of 55

Application +4 | -1

The unit vector which is orthogonal to the vector

3\widehat i + 2\widehat j + 6\widehat k

and is coplanar with the vectors

2\widehat i + \widehat j + \widehat k

and

\widehat i - \widehat j + \widehat k

View answer and explanation

Correct option: C

\frac{{3\widehat j - \widehat k}}{{\sqrt {10} }}

Option C (

\frac{{3\widehat j - \widehat k}}{{\sqrt {10} }}

) is correct because it represents a unit vector orthogonal to the given vector and coplanar with the other two vectors, derived using vector triple product.

22 Medium

22 of 55

Application +4 | -1

\vec a,\;\,\vec b,\;\,\vec c

are three mutually perpendicular vectors of equal magnitude, then the angle

\theta

which

\vec a + \vec b + \vec c

makes with any one of three given vectors is given by

View answer and explanation

Correct option: A

{\cos ^{ - 1}}\,\,\frac{1}{{\sqrt 3 }}

Option A (

{\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}

) is correct because it is derived by finding the cosine of the angle between the vector sum and one of the mutually perpendicular vectors.

23 Medium

23 of 55

Application +4 | -1

\vec{a}=-\hat{\imath}+2 \hat{\jmath}+\hat{k}

is rotated by

90^{\circ}

about origin passing through

y

-axis. If new vector is

\vec{b}

then projection of

\vec{b}

\vec{c}=5 \hat{\imath}+4 \hat{\jmath}+3 \hat{k}

is equal to:

View answer and explanation

Correct option: A

\frac{6}{5}

Option A (6/5) is correct because the projection of the rotated vector b onto vector c is calculated to be 6/5.

24 Medium

24 of 55

Application +4 | -1

If a unit vector lies in

y z

-plane and makes angles of

30^{\circ}

and

60^{\circ}

with the positive

y

-axis and

z

-axis respectively, then its components along the co-ordinate axes will be

View answer and explanation

Correct option: B

0, \frac{\sqrt{3}}{2}, \frac{1}{2}

Option B (0,

\frac{\sqrt{3}}{2}

\frac{1}{2}

) is correct because the components of the unit vector are found using the given angles with the y and z axes and the fact that it lies in the yz-plane.

25 Medium

25 of 55

Analyzing +4 | -1

a, b, c

and

A, B, C \in R-\{0\}

such that

a 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

\frac{a B}{b A}=\frac{b C}{c B}+\frac{c A}{a C}

View answer and explanation

Correct option: A

Option A (3) is correct because the given condition implies that the vectors are anti-parallel, leading to a constant ratio between corresponding components, and thus the expression evaluates to 3.

26 Medium

26 of 55

Application +4 | -1

The position vectors of

\mathrm{P}, \mathrm{Q}, \mathrm{R}

with respect to

\mathrm{O}

are

3 \vec{i}-2 \vec{j}+\vec{k}, \vec{i}+3 \vec{j}+4 \vec{k}

and

2 \vec{i}+\vec{j}-2 \vec{k}

respectively. Then the distance of

\mathrm{P}

from the plane

\mathrm{OQR}

View answer and explanation

Correct option: B

\frac{11}{3}

Option B (

\frac{{11}}{{3}}

) is correct because it represents the perpendicular distance from point P to the plane OQR, calculated using the projection of OP onto the normal vector of the plane.

27 Medium

27 of 55

Application +4 | -1

Let

\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}, \vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}

be three given vectors. If

\vec{r}

is a vector such that

\vec{r} \times \vec{a}=\vec{c} \times \vec{a}

and

\vec{r} \cdot \vec{b}=0

, then

|\vec{r}|

is equal to

View answer and explanation

Correct option: D

\frac{11}{7} \sqrt{2}

Option D (

\frac{11}{7} \sqrt{2}

) is correct because it is the magnitude of vector

\vec{r}

found by solving the given vector equations.

28 Medium

28 of 55

Application +4 | -1

The position vectors of the vertices

A, B

and

C

of triangle are

2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k}

-\hat{i}+\hat{j}+3 \hat{k}

respectively. Let

I

denotes the length of the angle bisector

A D

\angle B A C

where

D

is on the line segment

B C,

then

2 I^2

equals:

View answer and explanation

Correct option: C

Option C (45) is correct because it is the result of calculating 2 times the square of the length of the angle bisector AD using vector methods.

29 Medium

29 of 55

Analyzing +4 | -1

Let

\vec{a}=\hat{i}-\hat{j}+2 \hat{k}

and let

\vec{b}

be a vector such that

\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}

and

\vec{a} \cdot \vec{b}=3

. Then the projection of

\vec{b}

on the vector

\vec{a}-\vec{b}

is:

View answer and explanation

Correct option: A

\frac{2}{\sqrt{21}}

Option A (

\frac{2}{\sqrt{21}}

) is correct because it follows from calculating

|\vec{b}|^2

using the given dot and cross products, then finding

|\vec{a}-\vec{b}|

and finally calculating the projection of

\vec{b}

\vec{a}-\vec{b}

30 Medium

30 of 55

Analyzing +4 | -1

The value of

x

for which the angle between the vectors

\mathbf{a}=-3 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\mathbf{b}=x \hat{\mathbf{i}}+2 x \hat{\mathbf{j}}+\hat{\mathbf{k}}

is acute and the angle between

\mathbf{b}

and

X

-axis lies between

\pi / 2

and

\pi

satisfy

View answer and explanation

Correct option: B

x<0

Option B (

x<0

) is correct because it satisfies both conditions: the angle between vectors

\mathbf{a}

and

\mathbf{b}

is acute, and the angle between

\mathbf{b}

and the X-axis lies between

\pi/2

and

\pi

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 55 single-correct-answer MCQs on DOT PRODUCT OF VECTORS.
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