Which one of the following vectors is of magnitude 6 and perpendicular to both ${\bf{\vec a}} = 2\widehat {\bf{i}} + 2\widehat {\bf{j}} + {\bf{\hat k}}$ and ${\bf{\vec b}} = \widehat {\bf{i}} - 2\widehat {\bf{j}} + 2{\bf{\hat k}}$?

$2\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

$2\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}$

$2\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$3\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

A tetrahedron has vertices at $O\left( {0,0,0} \right),$ $A\left( {1,2,1} \right),\;B\left( {2,1,3} \right){\rm{and\;}}C\left( { - 1,1,2} \right).$ Then, the angle between the faces $OAB$ and $ABC$ will be

${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{19}}{{35}}} \right)$

${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{17}}{{31}}} \right)$

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JEE • MATHEMATICS • CROSS PRODUCT

Cross product practice · Mixed difficulty

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Showing 31-60 of 195 questions

31 Easy

31 of 195

Application +4 | -1

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

and

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

, then

\vec{b}

is equal to

View answer and explanation

Correct option: A

\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k})

Option A (

\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k})

) is correct because it is the vector \vec{b} that satisfies the given conditions.

32 Easy

32 of 195

Application +4 | -1

Let

\hat a

\hat b

and

\hat c

be three unit vectors such that

\hat a \times \left( {\hat b \times \hat c} \right) = \frac{{\sqrt 3 }}{2}\left( {\hat b + \hat c} \right)

. If

\hat b

is not parallel to

\hat c

, then the angle between

\hat a

and

\hat b

View answer and explanation

Correct option: D

\frac{{5\pi }}{6}

Option D (

\frac{{5\pi }}{6}

) is correct because it is the angle between

\hat a

and

\hat b

calculated from the given vector equation.

33 Easy

33 of 195

Analyzing +4 | -1

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

then

\vec{a}+\vec{b}+\vec{c}=

View answer and explanation

Correct option: C

The correct answer is C (0) because the given condition implies that the vectors are coplanar and their sum must be zero.

34 Easy

34 of 195

Analyzing +4 | -1

Let

\vec a = \hat i + \hat j - \hat k

and

\vec c = 2\hat i - 3\hat j + 2\hat k

Then the number of vectors

\overrightarrow b

such that

\overrightarrow b \times \overrightarrow c = \overrightarrow a

and

\left| {\overrightarrow b } \right| \in \{ 1,2,....,10\}

is:

View answer and explanation

Correct option: A

Option A (0) is correct because the dot product of the given vectors

\vec c

and

\vec a

is not zero, which contradicts the vector triple product property.

35 Medium

35 of 195

Analyzing +4 | -1

Let

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

and

\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}

. If

\vec{d}

is a vector perpendicular to both

\vec{b}

and

\vec{c}

, and

\vec{a} \cdot \vec{d}=18

, then

|\vec{a} \times \vec{d}|^{2}

is equal to

View answer and explanation

Correct option: C

720

Option C (720) is correct because it is the result of calculating the magnitude squared of the cross product of vector a and d, given the conditions.

36 Easy

36 of 195

Application +4 | -1

For any vector

\mathbf{a}

, the value of

(\mathbf{a} \times \hat{\mathbf{i}})^2+(\mathbf{a} \times \hat{\mathbf{j}})^2+(\mathbf{a} \times \hat{\mathbf{k}})^2

is equal to

View answer and explanation

Correct option: B

2 \mathrm{a}^2

Option B (

2 \mathrm{a}^2

) is correct because expanding the expression and simplifying using vector identities results in

2(a_1^2 + a_2^2 + a_3^2) = 2a^2

37 Easy

37 of 195

Application +4 | -1

|\mathbf{a} \cdot \mathbf{b}|=3

and

|\mathbf{a} \times \mathbf{b}|=4

, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: B

\cos ^{-1} \frac{3}{5}

Option B (

\cos^{{-1}} \frac{{3}}{{5}}

) is correct because it represents the angle between the vectors, derived from the ratio of the dot product to the magnitude of the cross product.

38 Easy

38 of 195

Application +4 | -1

A vector perpendicular to the plane containing the points

A\left( {1. - 1,2} \right),\;B\left( {2,0, - 1} \right),\;C\left( {0,2,1} \right)

View answer and explanation

Correct option: B

8\widehat \mathbf{i} + 4\widehat \mathbf{j} + 4\mathbf{\hat k}

Option B (

8\widehat{\mathbf{i}} + 4\widehat{\mathbf{j}} + 4\mathbf{\hat k}

) is correct because it represents a vector perpendicular to the plane containing the points A, B, and C, found by taking the cross product of vectors formed by these points.

39 Easy

39 of 195

Application +4 | -1

The area of the parallelogram whose diagonals are the vectors

2 a-b

and

4 a-5 b

, where

a

and

b

are the unit vectors forming an angle of

45^{\circ}

, is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the area of the parallelogram is half the magnitude of the cross product of its diagonals, considering the angle between the unit vectors.

40 Hard

40 of 195

Application +4 | -1

A rigid body rotates with constant angular velocity

\omega

about the line whose vector equation is,

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

. The speed of the particle at the instant it passes through the point with position vector

(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})

is equal to

View answer and explanation

Correct option: A

\omega \sqrt{2}

Option A (

\omega \sqrt{2}

) is correct because it represents the speed of the particle, calculated using the cross product of angular velocity and position vector.

41 Medium

41 of 195

Application +4 | -1

\vec{a}

and

\vec{b}

are two vectors such that

|\vec{a}|=1

|\vec{b}|=4, \vec{a} \cdot \vec{b}=2

. If

\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}

then angle between

\vec{b}

and

\vec{c}

is :

View answer and explanation

Correct option: D

\frac{5 \pi}{6}

Option D (

\frac{{5 \pi}}{{6}}

) is correct because the angle between the vectors

\vec{{b}}

and

\vec{{c}}

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

42 Easy

42 of 195

Application +4 | -1

\mathbf{u}=\mathbf{i}+\mathbf{j}-\mathbf{k}, \mathbf{v}=2 \mathbf{i}+\mathbf{j}+\mathbf{k}

and

\mathbf{w}=\mathbf{i}+\mathbf{j}+

2 \mathbf{k}

then the magnitude of projection of

\mathbf{u} \times \mathbf{v}

\mathbf{w}

is given by

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it represents the magnitude of the projection of

\mathbf{u} \times \mathbf{v}

\mathbf{w}

43 Easy

43 of 195

Application +4 | -1

If force represented by

5 {i}+{k}

is acting through the point

9 {i}-{j}+2 {k}

, then the moment about the point

3 {i}+2 {j}

+{k}

View answer and explanation

Correct option: C

-3 {i}-{j}+15 {k}

Option C (

-3 {i}-{j}+15 {k}

) is correct because it is the result of the cross product between the force vector and the displacement vector from the point of application to the point about which the moment is calculated.

44 Medium

44 of 195

Application +4 | -1

Let

\vec{a}

and

\vec{b}

be two vectors such that

|\vec{b}|=1

and

|\vec{b} \times \vec{a}|=2

. Then

|(\vec{b} \times \vec{a})-\vec{b}|^2

is equal to

View answer and explanation

Correct option: A

5

The correct answer is A (5) because by expanding the square and using the properties of the cross product, we can find the value of the expression.

45 Easy

45 of 195

Application +4 | -1

The area of the triangle having vertices as

\hat{i}-2 \hat{j}+3 \hat{k},-2 \hat{i}+3 \hat{j}-\hat{k}, 4 \hat{i}-7 \hat{j}+7 \hat{k}

is:

View answer and explanation

Correct option: D

0 sq unit

Option D (0 sq unit) is correct because the vectors AB and AC are scalar multiples of each other, indicating the points are collinear and the area of the triangle is zero.

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46 Easy

46 of 195

Application +4 | -1

{a}={i}+{j}-{k}, {b}=-{i}+2 {j}+2 {k}

and

{c}=-{i}+2 {j}-{k}

then a unit vector normal to the vectors

{a}+{b}

and

{b}

-{c}

View answer and explanation

Correct option: A

{i}

Option A (

{i}

) is correct because the cross product of the vectors

{a}+{b}

and

{b}-{c}

is parallel to the i-axis, resulting in a unit vector of

{i}

47 Hard

47 of 195

Application +4 | -1

Let

\vec{a}=\alpha_1 \hat{i}+\alpha_2 \hat{j}+\alpha_3 \hat{k}, \vec{b}=\beta_1 \hat{i}+\beta_2 \hat{j}+\beta_3 \hat{k}

and

\vec{c}=\gamma_1 \hat{i}+\gamma_2 \hat{j}+\gamma_3 \hat{k}

and

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

makes the angle

\pi / 3

with the plane of

\overrightarrow{\mathrm{b}}

and

\overrightarrow{\mathrm{c}}

and angle between

\overrightarrow{\mathrm{b}} \& \overrightarrow{\mathrm{c}}

\pi / 6

then the value of

\left|\begin{array}{lll}\alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3\end{array}\right|

equals

View answer and explanation

Correct option: C

\left(\frac{\sqrt{3}(|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid)}{\sqrt{2}}\right)^{\mathrm{n}}

Option C (

\left(\frac{{\sqrt{{3}}(|\overrightarrow{{\mathrm{{b}}}}||\overrightarrow{{\mathrm{{c}}}}|)}}{{\sqrt{{2}}}}\right)^{\mathrm{{n}}}

) is correct because it correctly represents the magnitude of the scalar triple product of the vectors in terms of their magnitudes and the angles between them.

48 Medium

48 of 195

Application +4 | -1

a \times(b \times c)+b \times(c \times a)+c \times(a \times b)=

View answer and explanation

Correct option: A

2[\mathbf{a} \mathbf{b} \mathbf{c}]

Option D (0) is correct because the expression simplifies to 0 due to the properties of the scalar triple product when the vectors are coplanar.

49 Easy

49 of 195

Application +4 | -1

If the non-zero vectors

{a}

and

{b}

are perpendicular to each other then the solution of the equation

{r} \times {a}={b}

View answer and explanation

Correct option: A

r=x {a}+\frac{1}{{a} \cdot {a}}({a} \times {b})

Option A (

r=x {a}+\frac{{1}}{{{a} \cdot {a}}}({a} \times {b})

) is correct because it satisfies the given equation

{r} \times {a}={b}

and incorporates the condition that

{a}

and

{b}

are perpendicular.

50 Easy

50 of 195

Application +4 | -1

If the diagonals of a parallelogram are represented by the vectors

3 \mathbf{i}+\mathbf{j}-2 \mathbf{k}

and

\mathbf{i}+3 \mathbf{j}-4 \mathbf{k}

, then its area in square unit is

View answer and explanation

Correct option: D

\sqrt{42}

Option D (

\sqrt{42}

) is correct because the area of the parallelogram is half the magnitude of the cross product of its diagonals.

51 Easy

51 of 195

Application +4 | -1

Unit vector perpendicular to the plane of

\Delta ABC

with position vectors a, b, c of the vertices A, B, C is

View answer and explanation

Correct option: B

\frac{{a \times b + b \times c + c \times a}}{{2\Delta }}

Option B (

\frac{{a \times b + b \times c + c \times a}}{{2\Delta }}

) is correct because it represents the unit vector perpendicular to the plane of the triangle, normalized by twice the area of the triangle.

52 Easy

52 of 195

Application +4 | -1

\vec{a} \perp \vec{b}

, then vector

\vec{v}

in terms of

\vec{a}

and

\vec{b}

satisfying the condition

\vec{v} \cdot \vec{a}

and

\vec{v} \cdot \vec{b}=1

and

\left[ {\begin{array}{cccccccccccccccccccc} {\vec v}&{\vec a}&{\vec b} \end{array}} \right] = 1

View answer and explanation

Correct option: A

\frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|^{2}}+\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}}

Option A (

\frac{{\overrightarrow{{\mathrm{{b}}}}}}{{\|\overrightarrow{{\mathrm{{b}}}}\|^2}} + \frac{{\overrightarrow{{\mathrm{{a}}}} \times \overrightarrow{{\mathrm{{b}}}}}}{{\|\overrightarrow{{\mathrm{{a}}}} \times \overrightarrow{{\mathrm{{b}}}}\|^2}}

) is correct because it satisfies the given conditions

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

\vec{v} \cdot \vec{b} = 1

, and

\left[ {\begin{{array}}{{*{20}{c}}} {\vec v}&{\vec a}&{\vec b} \end{{array}}} \right] = 1

53 Medium

53 of 195

Application +4 | -1

Which one of the following vectors is of magnitude 6 and perpendicular to both

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

and

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

View answer and explanation

Correct option: D

2\left( {2\widehat \mathbf{i} - \widehat \mathbf{j} - 2\mathbf{\hat k}} \right)

Option D (

2(2\widehat{\mathbf{i}} - \widehat{\mathbf{j}} - 2\mathbf{\hat k})

) is correct because it represents a vector with magnitude 6 that is perpendicular to both given vectors, found by taking the cross product and scaling.

54 Easy

54 of 195

Application +4 | -1

|\vec{a}|=10,|\vec{b}|=2

and

\vec{a} . \vec{b}=12

, then the value of

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

View answer and explanation

Correct option: D

Option D (16) is correct because it is calculated using the formula

|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta

, where

\sin \theta

is derived from

\vec{a} . \vec{b}=|\vec{a}||\vec{b}| \cos \theta

55 Hard

55 of 195

Analyzing +4 | -1

{\vec a}

and

{\vec b}

are not perpendicular to each other and

\vec r \times \vec a = \vec b \times \vec a,\;\vec r.\vec c = 0

, then

{\vec r}

is equal to

View answer and explanation

Correct option: C

\vec b - \frac{{\left( {\vec b.\vec c} \right)}}{{\left( {\vec a.\vec c} \right)}}\vec a

Option C (

\vec b - \frac{{\left( {\vec b.\vec c} \right)}}{{\left( {\vec a.\vec c} \right)}}\vec a

) is correct because it satisfies both the cross product and dot product conditions.

56 Easy

56 of 195

Application +4 | -1

\mathbf{r}

satisfies the equation

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

, then for any scalar

t, \mathbf{r}

is equal to

View answer and explanation

Correct option: B

\mathbf{j}+t(\mathbf{i}+2 \mathbf{j}+\mathbf{k})

Option B (

\mathbf{j}+t(\mathbf{i}+2 \mathbf{j}+\mathbf{k})

) is correct because it satisfies the given equation for any scalar t.

57 Hard

57 of 195

Application +4 | -1

A tetrahedron has vertices at

O\left( {0,0,0} \right),

A\left( {1,2,1} \right),\;B\left( {2,1,3} \right)\mathrm{and\;}C\left( { - 1,1,2} \right).

Then, the angle between the faces

OAB

and

ABC

will be

View answer and explanation

Correct option: A

\mathrm{co}{\mathrm{s}^{ - 1}}\left( {\frac{{19}}{{35}}} \right)

Option A (

\mathrm{co}{\mathrm{s}^{ - 1}}(\frac{{19}}{{35}}))

is correct because the angle between two faces is found by calculating the dot product of their normal vectors.

58 Medium

58 of 195

Analyzing +4 | -1

Let

\vec{a}=2 i-j+5 k

and

\vec{b}=\alpha i+\beta j+2 k

. If

((\vec{a} \times \vec{b}) \times i) \cdot k=\frac{23}{2}

, then

|\vec{b} \times 2 j|

is equal to

View answer and explanation

Correct option: B

Option B (5) is correct because by calculating the cross products and using the given dot product, we can find the values of

\alpha

and

\beta

, and then calculate

|\vec{b} \times 2 \vec{j}|

59 Easy

59 of 195

Understanding +4 | -1

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

is coplanar with

View answer and explanation

Correct option: A

\mathbf{b}

and

\mathbf{c}

Option A is correct because the vector triple product a × (b × c) lies in the plane defined by vectors b and c.

60 Easy

60 of 195

Application +4 | -1

\left| {\mathbf{\vec a} \times \mathbf{\vec b}} \right| = \left| {\mathbf{\vec a} \cdot \mathbf{\vec b}} \right|,

then the angle between

\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b}

View answer and explanation

Correct option: C

\frac{\mathrm{\pi }}{4}

The correct answer is C. When the magnitude of the cross product equals the magnitude of the dot product, the angle between the vectors is

\frac{\mathrm{\pi }}{4}

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