If ${\bf{\vec a}} = 2\widehat {{\bf{\dot i}}} + 3\widehat {{\bf{\dot j}}} - {\bf{\hat k}}$̂, $\;{\bf{\vec b}} = \widehat {{\bf{\dot i}}} + 2\widehat {\,{\bf{\dot j}}} - 5\,{\bf{\hat k}}$, $\;{\bf{\vec c}} = 3\,\widehat {{\bf{\dot i}}} + 5\widehat {\,{\bf{\dot j}}} - {\bf{\hat k}}$̂, then a vector perpendicular to ${{\bf{\vec a}}}$ and in the plane containing ${\bf{\vec b}}\,\;{\rm{and\;}}\,\,{\bf{\vec c}}$ is

$ - 17\,\widehat {{\bf{\dot i}}} - 21\,\widehat {{\bf{\dot j}}} - 97\,{\bf{\hat k}}$

$ - 17\widehat {\,{\bf{\dot i}}} + 21\widehat {\,{\bf{\dot j}}} - 97\,{\bf{\hat k}}$

$17\,\widehat {{\bf{\dot i}}} + 21\widehat {\,{\bf{\dot j}}} - 123\,{\bf{\hat k}}$

$ - 17\,\widehat {{\bf{\dot i}}} - 21\,\widehat {{\bf{\dot j}}} + 97\,{\bf{\hat k}}$

If ${\bf{\vec a}} \times {\bf{\vec b}} = {\bf{\vec c}} \times {\bf{\vec d}}\;{\rm{and\;}}{\bf{\vec a}} \times {\bf{\vec c}} = {\bf{\vec b}} \times {\bf{\vec d}},$ then

$\left( {{\bf{\vec a}} - {\bf{\vec d}}} \right) = {\rm{\lambda }}\left( {{\bf{\vec b}} - {\bf{\vec c}}} \right)$

$\left( {{\bf{\vec a}} + {\bf{\vec d}}} \right) = {\rm{\lambda }}\left( {{\bf{\vec b}} + {\bf{\vec c}}} \right)$

$\left( {{\bf{\vec a}} - {\bf{\vec b}}} \right) = {\rm{\lambda }}\left( {{\bf{\vec c}} + {\bf{\vec d}}} \right)$

$\left( {{\bf{\vec a}} + {\bf{\vec b}}} \right) = {\rm{\lambda }}\left( {{\bf{\vec c}} - {\bf{\vec d}}} \right)$

If ${\bf{\vec a}} = \widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}},\;{\bf{\vec b}} = \widehat {\bf{i}} + 3\widehat {\bf{j}} + 5{\bf{\hat k}}$ and ${\bf{\vec c}} = 7\widehat {\bf{i}} + 9\widehat {\bf{j}} + 11{\bf{\hat k}}$ then the area of parallelogram having diagonals ${\bf{\vec a}} + {\bf{\vec b}}$ and ${\bf{\vec b}} + {\bf{\vec c}}$ is

$\frac{1}{2}\sqrt {21} \;{\rm{sq\;units}}$

$\frac{{\sqrt 6 }}{2}{\rm{sq\;units}}$

If the vectors ${\bf{\vec a}},{\bf{\vec b}}$ and ${\bf{\vec c}}$ form the sides $BC,CA\;{\rm{and\;}}AB$ respectively of a triangle $ABC$ then

${\bf{\vec a}} \times {\bf{\vec b}} = {\bf{\vec b}} \times {\bf{\vec c}} = {\bf{\vec c}} \times {\bf{\vec a}} = 0$

${\bf{\vec a}} \cdot {\bf{\vec b}} = {\bf{\vec b}} \cdot {\bf{\vec c}} = {\bf{\vec c}} \cdot {\bf{\vec b}} = 0$

${\bf{\vec a}} \cdot {\bf{\vec b}} = {\bf{\vec b}} \cdot {\bf{\vec c}} = {\bf{\vec c}} \cdot {\bf{\vec a}} = 0$

${\bf{\vec a}} \times {\bf{\vec a}} + {\bf{\vec a}} \times {\bf{\vec c}} + {\bf{\vec c}} \times {\bf{\vec a}} = 0$

A force $\;{\bf{\vec F}} = 2\,{\bf{\hat i}} + {\bf{\hat j}} - {\bf{\hat k}}$̂ acts at a point $A$, whose position vectors is $2\,{\bf{\hat i}} - {\bf{\hat j}}$. The moment of ${\bf{\vec F}}$ about the origin is

${\bf{\hat i}} + 2\,{\bf{\hat j}} + 4\,{\bf{\hat k}}$

${\bf{\hat i}} + 2\,{\bf{\hat j}} - 4\,{\bf{\hat k}}$

${\bf{\hat i}} - 2\,{\bf{\hat j}} - 4\,{\bf{\hat k}}$

${\bf{\hat i}} - 2\,{\bf{\hat j}} + 4\,{\bf{\hat k}}$

If ${\bf{\vec a}} \times {\bf{\vec b}} = {\bf{\vec c}}{\rm{\;and\;}}{\bf{\vec b}} \times {\bf{\vec c}} = {\bf{\vec a}}$, then

$\left| {{\bf{\vec b}}\left| { = 1,{\rm{\;}}} \right|{\bf{\vec c}}\left| { = {\rm{\;}}} \right|{\bf{\vec a}}} \right|$

$\left| {{\bf{\vec a}}\left| { = 1,{\rm{\;}}} \right|{\bf{\vec b}}\left| { = {\rm{\;}}} \right|{\bf{\vec c}}} \right|$

$\left| {{\bf{\vec c}}\left| { = 1,} \right|{\bf{\vec a}}} \right| = 1$

$\left| {{\bf{\vec b}}\left| { = 2,{\rm{\;}}} \right|{\bf{\vec b}}\left| { = 2} \right|{\bf{\vec a}}} \right|$

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JEE • MATHEMATICS • CROSS PRODUCT

Cross product practice · Easy

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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 31-60 of 149 questions

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31 Easy

31 of 149

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}

32 Easy

32 of 149

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.

33 Easy

33 of 149

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.

34 Easy

34 of 149

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}

35 Easy

35 of 149

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.

36 Easy

36 of 149

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.

37 Easy

37 of 149

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.

38 Easy

38 of 149

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

39 Easy

39 of 149

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

40 Easy

40 of 149

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.

41 Easy

41 of 149

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.

42 Easy

42 of 149

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}

43 Easy

43 of 149

Analyzing +4 | -1

Let

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

and

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

.
Let

\vec{d}

be a vector which is perpendicular to both

\vec{a}

and

\vec{b}

, and

\vec{c} \cdot \vec{d}=12

. Then

(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})

is equal to

View answer and explanation

Correct option: B

Option B (44) is the correct answer. The vector

\vec{d}

is found using the cross product of

\vec{a}

and

\vec{b}

, then scaled to satisfy the condition

\vec{c} \cdot \vec{d} = 12

. Finally, the dot product

(-\hat{i}+\hat{j}-\hat{k}) \cdot (\vec{c} \times \vec{d})

is computed.

44 Easy

44 of 149

Application +4 | -1

\mathbf{\vec a} = 2\widehat {\mathbf{\dot i}} + 3\widehat {\mathbf{\dot j}} - \mathbf{\hat k}

̂,

\;\mathbf{\vec b} = \widehat {\mathbf{\dot i}} + 2\widehat {\,\mathbf{\dot j}} - 5\,\mathbf{\hat k}

\;\mathbf{\vec c} = 3\,\widehat {\mathbf{\dot i}} + 5\widehat {\,\mathbf{\dot j}} - \mathbf{\hat k}

̂, then a vector perpendicular to

{\mathbf{\vec a}}

and in the plane containing

\mathbf{\vec b}\,\;\mathrm{and\;}\,\,\mathbf{\vec c}

View answer and explanation

Correct option: D

- 17\,\widehat {\mathbf{\dot i}} - 21\,\widehat {\mathbf{\dot j}} - 97\,\mathbf{\hat k}

Option D (

- 17\,\widehat {\mathbf{\dot i}} - 21\,\widehat {\mathbf{\dot j}} - 97\,\mathbf{\hat k}

) is correct because it represents a vector perpendicular to

\vec a

and lies in the plane of

\vec b

and

\vec c

45 Easy

45 of 149

Understanding +4 | -1

For any vectors

\mathbf{a}, \mathbf{b} ;|\mathbf{a} \times \mathbf{b}|^2+(\mathbf{a} \cdot \mathbf{b})^2

is equal to

View answer and explanation

Correct option: A

|\mathbf{a}|^2|\mathbf{b}|^2

Option A (

|\mathbf{a}|^2|\mathbf{b}|^2

) is correct because using the definitions of dot and cross products, the given expression simplifies to the square of the magnitudes of the vectors.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

46 Easy

46 of 149

Application +4 | -1

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

and

\mathbf{c}=\mathbf{i}+3 \mathbf{j}-\mathbf{k}

, then

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

is equal to

View answer and explanation

Correct option: A

20 \mathrm{i}-3 \mathrm{j}+7 \mathrm{k}

Option A (

20 \mathbf{i}-3 \mathbf{j}+7 \mathbf{k}

) is correct because it is the result of calculating the vector triple product

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

47 Easy

47 of 149

Application +4 | -1

A force

\mathbf{F}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}

acts at a point

A

, whose position vector is

2 \hat{\mathbf{i}}-\hat{\mathbf{j}}

. The moment of

\mathbf{F}

about the origin is

View answer and explanation

Correct option: C

\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}

Option C (

\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}

) is correct because the moment of a force about the origin is calculated by taking the cross product of the position vector and the force vector.

48 Easy

48 of 149

Analyzing +4 | -1

\mathbf{\vec a} \times \mathbf{\vec b} = \mathbf{\vec c} \times \mathbf{\vec d}\;\mathrm{and\;}\mathbf{\vec a} \times \mathbf{\vec c} = \mathbf{\vec b} \times \mathbf{\vec d},

then

View answer and explanation

Correct option: A

\left( {\mathbf{\vec a} - \mathbf{\vec d}} \right) = \mathrm{\lambda }\left( {\mathbf{\vec b} - \mathbf{\vec c}} \right)

Option A (

\left( {\mathbf{\vec a} - \mathbf{\vec d}} \right) = \mathrm{\lambda }\left( {\mathbf{\vec b} - \mathbf{\vec c}} \right)

) is correct because it follows from the given conditions and properties of cross products.

49 Easy

49 of 149

Application +4 | -1

\mathbf{a}=\mathbf{i}+\mathbf{j}-\mathbf{k}, \mathbf{b}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{c}=\mathbf{i}-\mathbf{j}-\mathbf{k}

, then

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

View answer and explanation

Correct option: B

2 \mathrm{i}-2 \mathrm{j}

Option B (

2 \mathbf{i}-2 \mathbf{j}

) is correct because it is the result of calculating the vector triple product

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

using the given vectors.

50 Easy

50 of 149

Application +4 | -1

\mathbf{a}=2 \mathbf{i}+3 \mathbf{j}-5 \mathbf{k}, \mathbf{b}=m \mathbf{i}+n \mathbf{j}+12 \mathbf{k}

and

\mathbf{a} \times \mathbf{b}=0

, then

(m, n)=

View answer and explanation

Correct option: C

\left(-\frac{24}{5},-\frac{36}{5}\right)

Option C ((-24/5, -36/5)) is correct because when the cross product of two vectors is zero, the vectors are parallel, allowing us to find m and n.

51 Easy

51 of 149

Analyzing +4 | -1

Q(0, - 1, - 3)

is the image of the point

P

in the plane

3x - y + 4z = 2

and

R

is the point

(3, - 1, - 2),

then the area (in sq units) of

\Delta PQR

View answer and explanation

Correct option: A

\frac{{\sqrt {91} }}{2}

Option A (

\frac{{\sqrt {91} }}{{2}}

) is correct because it is the area of the triangle calculated using the cross product of vectors PQ and PR.

52 Easy

52 of 149

Application +4 | -1

{a}, {b}, {c}

be three units vectors such that

{a} \times({b} \times {c})=(1 / 2) {b} ; {b}

and

{c}

being non-parallel then

View answer and explanation

Correct option: A

The angle between

{a}

and

{c}

\pi / 3

Option A (The angle between a and c is

\pi / 3

) is correct because it is derived from the vector triple product and the condition that b and c are non-parallel.

53 Easy

53 of 149

Application +4 | -1

The area of the parallelogram whose adjacent sides are

2 \hat{i}-3 \hat{k}

and

4 \hat{j}+2 \hat{k}

View answer and explanation

Correct option: B

4 \sqrt{14} \mathrm{sq}

, units

Option B is correct because the area of the parallelogram is the magnitude of the cross product of the adjacent sides.

54 Easy

54 of 149

Application +4 | -1

\mathbf{\vec a} = \widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k},\;\mathbf{\vec b} = \widehat \mathbf{i} + 3\widehat \mathbf{j} + 5\mathbf{\hat k}

and

\mathbf{\vec c} = 7\widehat \mathbf{i} + 9\widehat \mathbf{j} + 11\mathbf{\hat k}

then the area of parallelogram having diagonals

\mathbf{\vec a} + \mathbf{\vec b}

and

\mathbf{\vec b} + \mathbf{\vec c}

View answer and explanation

Correct option: A

4\sqrt 6 \;\mathrm{sq\;units}

Option A (

4\sqrt 6 \;\mathrm{sq\;units}

) is correct because it is the result of calculating the area of the parallelogram using the cross product of its diagonals.

55 Easy

55 of 149

Understanding +4 | -1

If the vectors

\mathbf{\vec a},\mathbf{\vec b}

and

\mathbf{\vec c}

form the sides

BC,CA\;\mathrm{and\;}AB

respectively of a triangle

ABC

then

View answer and explanation

Correct option: B

\mathbf{\vec a} \times \mathbf{\vec b} = \mathbf{\vec b} \times \mathbf{\vec c} = \mathbf{\vec c} \times \mathbf{\vec a} = 0

Option B is correct because the cross product of vectors representing sides of a closed triangle, taken in order, sums to zero, implying pairwise equality of cross products.

56 Easy

56 of 149

Application +4 | -1

\mathbf{\vec a}\,\,\left| { = 10,\,} \right|\,\mathbf{\vec b}| = 2\;

and

\mathrm{\;}\mathbf{\vec a}\,\, \cdot \,\,\mathbf{\vec b} = 12

then

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

is equal to

View answer and explanation

Correct option: C

Option C (16) is correct because it uses the relationship between the dot product, cross product, and magnitudes of the vectors to find the magnitude of the cross product.

57 Easy

57 of 149

Application +4 | -1

Let

\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}

and

\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}

be two vectors such that

|\vec{a}|=1, \vec{a} \cdot \vec{b}=2

and

|\vec{b}|=4

. If

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

, then the angle between

\vec{b}

and

\vec{c}

is equal to :

View answer and explanation

Correct option: C

\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)

The correct answer is C (

\cos^{-1}(-\frac{\sqrt{3}}{2})

) because it is found by calculating the dot product of vectors b and c, and then using the formula for the angle between two vectors.

58 Easy

58 of 149

Application +4 | -1

A force

\;\mathbf{\vec F} = 2\,\mathbf{\hat i} + \mathbf{\hat j} - \mathbf{\hat k}

̂ acts at a point

A

, whose position vectors is

2\,\mathbf{\hat i} - \mathbf{\hat j}

. The moment of

\mathbf{\vec F}

about the origin is

View answer and explanation

Correct option: C

\mathbf{\hat i} + 2\,\mathbf{\hat j} + 4\,\mathbf{\hat k}

Option C (

\mathbf{\hat i} + 2\,\mathbf{\hat j} + 4\,\mathbf{\hat k}

) is correct because it is calculated by taking the cross product of the position vector and the force vector.

59 Easy

59 of 149

Application +4 | -1

The moment of a force represented by

\mathbf{F}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}

about the point

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

is equal to

View answer and explanation

Correct option: C

-5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}

Option C (

-5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}

) is correct because it is the moment of the force about the given point, calculated using the cross product of the position vector and the force vector.

60 Easy

60 of 149

Analyzing +4 | -1

\mathbf{\vec a} \times \mathbf{\vec b} = \mathbf{\vec c}\mathrm{\;and\;}\mathbf{\vec b} \times \mathbf{\vec c} = \mathbf{\vec a}

, then

View answer and explanation

Correct option: D

\left| {\mathbf{\vec b}\left| { = 1,\mathrm{\;}} \right|\mathbf{\vec c}\left| { = \mathrm{\;}} \right|\mathbf{\vec a}} \right|

Option D (

\left| {\mathbf{\vec b}\left| { = 1,\;}\right|\mathbf{\vec c}\left| { = \;}\right|\mathbf{\vec a}} \right|

) is correct because it satisfies the given cross product conditions and mutual perpendicularity of the vectors.

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