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 · Mixed difficulty

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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 61-90 of 195 questions

61 Easy

61 of 195

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.

62 Easy

62 of 195

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

63 Medium

63 of 195

Analyzing +4 | -1

A non-zero vector

{\mathbf{\vec a}}

is parallel to the line of intersection of the plane determined by the vectors

\widehat \mathbf{i},\;\widehat \mathbf{i} + \widehat \mathbf{j}

and the plane determined by the vectors

\widehat \mathbf{i} - \widehat \mathbf{j},\;\widehat \mathbf{i} + \mathbf{\hat k}.

The angle between

{\mathbf{\vec a}}

and

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

View answer and explanation

Correct option: D

\frac{\pi }{4}

Option D is correct because the vector

\vec a

is parallel to the intersection of two planes, and the angle between

\vec a

and the given vector can be found using the dot product formula.

64 Medium

64 of 195

Application +4 | -1

A unit vector perpendicular to the vector

4 i-j+3 k

and

-2 i+j-2 k

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the unit vector obtained by taking the cross product of the two given vectors and normalizing it.

65 Medium

65 of 195

Application +4 | -1

\vec{a}=\hat{i}+\hat{j}+\hat{k} \& \vec{a} \cdot \vec{b}=1 \& \vec{a} \times \vec{b}=\hat{j}-\hat{k}

then

\vec{b}

is equal to

View answer and explanation

Correct option: C

\hat{\mathrm{i}}

Option C (

\hat{\mathrm{i}}

) is correct. By setting

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

and using the given dot and cross product conditions, we can solve for the components of

\vec{b}

66 Easy

66 of 195

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.

67 Easy

67 of 195

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})

68 Easy

68 of 195

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.

69 Medium

69 of 195

Application +4 | -1

The scalars

l

and

m

such that

l \mathbf{a}+m \mathbf{b}=\mathbf{c}

, where

\mathbf{a}, \mathbf{b}

and c are given vectors, are equal to

View answer and explanation

Correct option: A

I=\frac{(\mathbf{c} \times \mathbf{b}) .(\mathbf{a} \times \mathbf{b})}{(\mathbf{a} \times \mathbf{b})^2}, m=\frac{(\mathbf{c} \times \mathbf{a}) .(\mathbf{b} \times \mathbf{a})}{(\mathbf{b} \times \mathbf{a})^2}

Option A is correct because it provides the correct formulas for calculating scalars l and m using cross and dot products of the given vectors.

70 Easy

70 of 195

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.

71 Easy

71 of 195

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.

72 Easy

72 of 195

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.

73 Easy

73 of 195

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.

74 Easy

74 of 195

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.

75 Easy

75 of 195

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.

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76 Easy

76 of 195

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.

77 Medium

77 of 195

Application +4 | -1

\mathrm{A}(3,1,-1), \mathrm{B}\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), \mathrm{C}(2,2,1)

and

\mathrm{D}\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)

are the vertices of a quadrilateral

\mathrm{ABCD}

, then its area is

View answer and explanation

Correct option: A

\frac{4 \sqrt{2}}{3}

Option A (

\frac{{4 \sqrt{{2}}}}{{3}}

) is correct because the area of the quadrilateral is calculated using the cross product of the vectors AC and BD.

78 Easy

78 of 195

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.

79 Easy

79 of 195

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.

80 Easy

80 of 195

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.

81 Easy

81 of 195

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.

82 Medium

82 of 195

Application +4 | -1

Let

|\vec{a}|=2,|\vec{b}|=3

and the angle between the vectors

\vec{a}

and

\vec{b}

\frac{\pi}{4}

. Then

|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}

is equal to

View answer and explanation

Correct option: D

882

Option D (882) is correct because after simplifying the cross product and using the given magnitudes and angle, the final result is 882.

83 Medium

83 of 195

Application +4 | -1

Let

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

and

\mathrm{\;}\mathbf{\vec c} = 4\widehat \mathbf{i} - 3\widehat \mathbf{j} + 7\mathbf{\hat k}.\;\mathrm{If}\;\mathrm{\;}\overrightarrow {\mathbf{r}\;}

is a vector such that

\mathbf{\vec r} \times \mathbf{\vec b} = \mathbf{\vec c} \times \mathbf{\vec b}\;\mathrm{and\;}\mathbf{\vec r}\cdot\mathbf{\vec a} = 0,

then value of

\mathbf{\vec r}\cdot\mathbf{\vec b}

View answer and explanation

Correct option: B

- 7

Option B (-7) is correct because solving the vector equations

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

and

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

yields

\mathbf{\vec r}\cdot\mathbf{\vec b} = -7

84 Easy

84 of 195

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.

85 Easy

85 of 195

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.

86 Easy

86 of 195

Understanding +4 | -1

\quad(a-b) \times(a+b)=

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the cross product

(a-b) \times (a+b)

simplifies to

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

87 Easy

87 of 195

Application +4 | -1

|\mathrm{a}| = 2,|\mathrm{b}| = 5

and

\mathrm{a} \times \mathrm{b| = 8},

then

|\mathrm{a}.\mathrm{b}|

is equal to

View answer and explanation

Correct option: A

Option A (6) is correct because it's the absolute value of the dot product calculated using the magnitudes of the vectors and the angle between them, derived from the cross product.

88 Easy

88 of 195

Analyzing +4 | -1

\vec a \times \vec b = \vec c \times \vec d

and

\vec a \times \vec c = \vec b \times \vec d

, then

View answer and explanation

Correct option: A

\left( {\vec a \pm \vec d} \right) = \lambda \left( {\vec b \pm \vec c} \right)

Option A is correct because the given conditions imply that the vectors (

\vec a - \vec d

) and (

\vec b - \vec c

) are parallel, as are (

\vec a + \vec d

) and (

\vec b + \vec c

89 Medium

89 of 195

Application +4 | -1

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

and

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

, then

\lambda+\mu+v

is equal to

View answer and explanation

Correct option: A

8 d \cdot(a+b+c)

Option A (

8 d \cdot(a+b+c)

) is correct because it is derived by taking the dot product of d with c, a, and b, and then summing the results.

90 Easy

90 of 195

Application +4 | -1

\mathbf{a}

and

\mathbf{b}

are two vectors, then

(\mathbf{a} \times \mathbf{b})^2

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the correct expression for

(\mathbf{a} \times \mathbf{b})^2

|\mathbf{a}|^2 |\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2

, which can be represented by a determinant but not any of the given options.

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