The point of intersection of ${\bf{\vec r}} \times {\bf{\vec a}} = {\bf{\vec b}} \times {\bf{\vec a}}$ and ${\bf{\vec r}} \times {\bf{\vec b}} = {\bf{\vec a}} \times {\bf{\vec b}},$ where ${\bf{\vec a}} = \widehat {\bf{i}} + \widehat {\bf{j}}$ and ${\bf{\vec b}} = \widehat {\bf{i}} - {\bf{\hat k}}$ is

$3\widehat {\bf{i}} + \widehat {\bf{j}} - {\bf{\hat k}}$

$3\widehat {\bf{i}} - {\bf{\hat k}}$

$3\widehat {\bf{i}} + 2\widehat {\bf{j}} + {\bf{\hat k}}$

If ${\bf{\vec a}},{\bf{\vec b}}$ are any two vwctors, then $\left( {2\vec a + 3{\bf{\vec b}}} \right) \times \left( {5{\bf{\vec a}} + 7{\bf{\vec b}}} \right) + {\bf{\vec a}} \times {\bf{\vec b}}$ is equal to

${\bf{\vec a}} \times {\bf{\vec b}}$

${\bf{\vec b}} \times {\bf{\vec a}}$

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JEE • MATHEMATICS • CROSS PRODUCT

Cross product practice · Mixed difficulty

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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 151-180 of 195 questions

151 Medium

151 of 195

Application +4 | -1

The area of triangle whose vertices are

(1,2,3),(2,5,-1)

and

(-1,1,2)

View answer and explanation

Correct option: C

\frac{\sqrt{155}}{2}

sq. unit

Option C (

\frac{{\sqrt{{155}}}}{{2}}

sq. unit) is correct because it is half the magnitude of the cross product of vectors AB and AC.

152 Easy

152 of 195

Application +4 | -1

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

and

-5 \mathbf{i}+7 \mathbf{j}

are the vector sides of any triangle, then its area is given by

View answer and explanation

Correct option: C

\frac{41}{2}

Option C (

\frac{41}{2}

) is correct because the area of the triangle is half the magnitude of the cross product of the two side vectors.

153 Easy

153 of 195

Application +4 | -1

Area of rhombus is ……., where diagonals are

\mathbf{\vec a} = 2\widehat \mathbf{i} - 3\widehat \mathbf{j} + 5\mathbf{\hat k}

and

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

View answer and explanation

Correct option: C

\sqrt {28.5}

Option C (

\sqrt {28.5}

) is correct because it represents half the magnitude of the cross product of the diagonals of the rhombus.

154 Easy

154 of 195

Application +4 | -1

For any vector

{\mathbf{\vec a}}

, the value of

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

is equal to

View answer and explanation

Correct option: B

2\mathbf{\vec a}{\;^2}

Option B (

2\mathbf{\vec a}{\;^2}

) is correct because it follows from expanding the cross products and simplifying the expression.

155 Easy

155 of 195

Application +4 | -1

|\vec{a}|=2

and

|\vec{b}|=3

and

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

, then

((\vec{a} \times(\vec{a} \times(\vec{a} \times(\vec{a} \times \vec{b})))

is equal to

View answer and explanation

Correct option: A

48 \overrightarrow{{b}}

Option A (

48 \overrightarrow{{b}}

) is correct because applying the vector triple product repeatedly and using the orthogonality condition simplifies the expression to

48 \overrightarrow{{b}}

156 Easy

156 of 195

Application +4 | -1

\overrightarrow{O A}=3 \mathbf{i}+2 \mathbf{j}-\mathbf{k}

and

\overrightarrow{O B}=i+3 \mathbf{j}+\mathbf{k}

, then the area of the triangle

O A B

View answer and explanation

Correct option: C

\frac{3}{2} \sqrt{10}

Option C (

\frac{3}{2} \sqrt{10}

) is correct because the area of the triangle OAB is half the magnitude of the cross product of the position vectors OA and OB.

157 Easy

157 of 195

Analyzing +4 | -1

Let

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

and

b = \hat i + \hat j

, if c is a vector such that

a.\,c = \left| c \right|,\left| {c - a} \right| = 2\sqrt 2

and the angle between

\left( {a \times b} \right)

and c is

\frac{\pi }{6},

then the value of

\left| {\left( {a \times b} \right) \times c} \right|

View answer and explanation

Correct option: D

\frac{3}{2}

Option D (3/2) is correct because, after calculating |c| and |a x b|, the magnitude of the cross product (a x b) x c is found using the given angle.

158 Easy

158 of 195

Application +4 | -1

\mathbf{a}=2 \mathbf{i}-3 \mathbf{j}-\mathbf{k}

and

\mathbf{b}=\mathbf{i}+4 \mathbf{j}-2 \mathbf{k}

, then

\mathbf{a} \times \mathbf{b}

View answer and explanation

Correct option: B

10 \mathbf{i}+3 \mathbf{j}+11 \mathbf{k}

Option B (

10 \mathbf{i}+3 \mathbf{j}+11 \mathbf{k}

) is correct because it is the result of the cross product calculation of the two given vectors.

159 Easy

159 of 195

Analyzing +4 | -1

Let

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

and

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

. If

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

, then

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

is equal to

View answer and explanation

Correct option: B

Option B (5) is correct because after finding the value of alpha using the given dot product condition, the magnitude of the cross product of vector b and 2j is calculated to be 5.

160 Easy

160 of 195

Application +4 | -1

A unit vector

{n}

perpendicular to the plane determined by the points

A(0,-2,1), B(1,-1,-2)

and

C

(-1,1,0)

View answer and explanation

Correct option: B

1 / 4 \sqrt{6}(8{i}+4 {j}+4 {k})

Option B (

\cos^{-1} \frac{8}{5 \sqrt{29}}

) is correct because the angle between the two vectors is found using the dot product formula and the magnitudes of the vectors.

161 Easy

161 of 195

Application +4 | -1

\alpha=2 i+3 j-k, \beta=-i+2 j-4 k

and

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

, then

(\alpha \times \beta) .(\alpha \times \gamma)

is equal to

View answer and explanation

Correct option: D

-74

Option D (-74) is correct because it is the result of the scalar triple product calculation

(\alpha \times \beta) . (\alpha \times \gamma)

162 Easy

162 of 195

Remembering +4 | -1

The moment of the force

\mathbf{F}

acting at a point

P

, about the point

C

View answer and explanation

Correct option: D

\mathbf{C P} \times \mathrm{F}

Option D (

\mathbf{C P} \times \mathrm{F}

) is correct because the moment of a force is given by the cross product of the position vector and the force vector.

163 Easy

163 of 195

Application +4 | -1

Let,

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

and

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

, where

\alpha \in \mathbf{R}

. If the area of the parallelogram whose adjacent sides are represented by the vectors

\vec{a}

and

\vec{b}

\sqrt{15\left(\alpha^2+4\right)}

, then the value of

2|\vec{a}|^2+(\vec{a} \cdot \vec{b})|\vec{b}|^2

is equal to :

View answer and explanation

Correct option: D

Option D (14) is correct because it is the final result after calculating the cross product, solving for alpha, and substituting into the expression.

164 Medium

164 of 195

Application +4 | -1

\vec{a}=9 \hat{i}+2 \hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}

and

\vec{c}=7 \hat{i}-3 \hat{j}+2 \hat{k}

are three given vectors. Let there be a

\vec{r}

such that

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

and

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

then

\bar{r} \cdot \bar{c}

View answer and explanation

Correct option: A

\frac{280}{11}

Option A (280/11) is correct because it is the value of the dot product of vectors r and c, derived from the given conditions and vector algebra.

165 Easy

165 of 195

Analyzing +4 | -1

Le the vectors

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

and

{\mathbf{\vec d}}

be such that

\left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right) \times \left( {\mathbf{\vec c} \times \mathbf{\vec d}} \right) = 0

. Let

{P_1}

and

{P_2}

be planes determined by pair of vectors

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

and

\mathbf{\vec c},\mathbf{\vec d}

respectively. Then, the angle between

{P_1}

and

{P_2}

View answer and explanation

Correct option: A

The correct answer is A (0) because the condition

\left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right) \times \left( {\mathbf{\vec c} \times \mathbf{\vec d}} \right) = 0

implies that the normal vectors to the planes are parallel, hence the angle between the planes is 0.

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166 Easy

166 of 195

Application +4 | -1

The vector

[(\mathbf{i}-\mathbf{j}+\mathbf{k}) \times(2 \mathbf{i}-3 \mathbf{j}-\mathbf{k})] \times[(-3 \mathbf{i}+\mathbf{j}+\mathbf{k}) \times(2 \mathbf{j}+\mathbf{k})

] is given by

View answer and explanation

Correct option: B

-5(3 \mathbf{i}-5 \mathbf{j}-3 \mathbf{k})

Option B (

-5(3 \mathbf{i}-5 \mathbf{j}-3 \mathbf{k})

) is correct because it is the result of calculating the cross product of two cross products.

167 Easy

167 of 195

Application +4 | -1

\left| {\mathbf{\vec a}} \right| = 5,\;\left| {\mathbf{\vec b}} \right| = 6

and

\mathbf{\vec a}\cdot\mathbf{\vec b} = - 25,

then

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

is equal to

View answer and explanation

Correct option: D

5\sqrt {11}

Option D (

5\sqrt {11}

) is correct because it accurately calculates the magnitude of the cross product using the given dot product and magnitudes.

168 Easy

168 of 195

Application +4 | -1

The area of a parallelogram having diagonals

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

and

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

View answer and explanation

Correct option: A

5 \sqrt{3}

Option A (

5 \sqrt{3}

) is correct because it is the magnitude of the cross product of the adjacent sides of the parallelogram.

169 Medium

169 of 195

Understanding +4 | -1

The components of a vector a along and perpendicular to the non-zero vector

\mathbf{b}

are respectively

View answer and explanation

Correct option: B

\frac{\mathbf{a} . \mathbf{b}}{|\mathbf{b}|}, \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{b}|}

Option B is correct because it correctly represents the components of vector a along and perpendicular to vector b using dot and cross products respectively.

170 Easy

170 of 195

Application +4 | -1

The point of intersection of

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

and

\mathbf{\vec r} \times \mathbf{\vec b} = \mathbf{\vec a} \times \mathbf{\vec b},

where

\mathbf{\vec a} = \widehat \mathbf{i} + \widehat \mathbf{j}

and

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

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the vector sum of vectors a and b, which is the point of intersection.

171 Medium

171 of 195

Analyzing +4 | -1

\vec{a}, \vec{b}, \vec{c}

are unit vectors equally inclined to each other at an angle

\alpha(\neq 0)

, then the angle between

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

and the plane containing

\vec{b}

and

\vec{c}

View answer and explanation

Correct option: A

\sin ^{-1}\left(\tan \frac{\alpha}{2} \cdot|\cot \alpha|\right)

Option A (

\sin^{-1}(\tan \frac{\alpha}{2} \cdot |\cot \alpha|)

) is correct because it represents the sine of the angle between the vector product and the plane, derived using vector algebra and trigonometric identities.

172 Easy

172 of 195

Application +4 | -1

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

and

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

, then

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

is equal to

View answer and explanation

Correct option: A

24 \mathbf{i}+7 \mathbf{j}-5 \mathbf{k}

Option A (

24 \mathbf{i}+7 \mathbf{j}-5 \mathbf{k}

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

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

173 Medium

173 of 195

Application +4 | -1

For any two vectors

{u}

and

{v},(1-{u} \cdot {v})^2+\mid {u}+{v}+({u}

\times {v})\left.\right|^2

is equal to

View answer and explanation

Correct option: B

\left(1+|{u}|^2\right)\left(1+|{v}|^2\right)

Option B (

\left(1+|{u}|^2\right)\left(1+|{v}|^2\right)

) is correct because the given expression simplifies to this form after expanding and using vector identities.

174 Easy

174 of 195

Application +4 | -1

The vector

(({i}-{j}) \times({j}-{k})) \times({i}+5 {k})

is equal to

View answer and explanation

Correct option: A

5 {i}-4 {j}-{k}

Option D (

\frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k})

) is correct because it represents a unit vector perpendicular to a and coplanar with a and b, satisfying the given conditions.

175 Easy

175 of 195

Understanding +4 | -1

\vec{a}

and

\vec{b}

are the two vectors such that

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

and

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

, then

View answer and explanation

Correct option: C

either

\vec{a}

\vec{b}

is a null vector.

Option C (either

\vec{{a}}

\vec{{b}}

is a null vector) is correct because if both the dot product and cross product of two vectors are zero, at least one of the vectors must be the zero vector.

176 Easy

176 of 195

Application +4 | -1

|\vec{c}|^2=60

and

\overrightarrow{\mathrm{c}} \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=\overrightarrow{0}

, then a value of

\overrightarrow{\mathrm{c}} \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})

is:

View answer and explanation

Correct option: D

12 \sqrt{2}

Option A (0) is correct because the vectors are coplanar, meaning their scalar triple product is zero, which allows us to solve for

\lambda

177 Medium

177 of 195

Analyzing +4 | -1

A tetrahedron has vertices

O(0,0,0), A(1,2,1), B(2,1,3)

and

C(-1,1,2)

. Then the angle between the faces

O A B

and

A B C

will be

View answer and explanation

Correct option: A

\cos ^{-1}\left(\frac{19}{35}\right)

The angle between two faces of a tetrahedron is equal to the angle between their normal vectors, which is found using the dot product formula, resulting in

\cos ^{-1}(\frac{19}{35})

178 Medium

178 of 195

Application +4 | -1

If the non-zero vectors

{\vec a}

and

{\vec b}

are perpendicular to each other, then the solution of the equation,

\vec r \times \vec a = \vec b

is given by

View answer and explanation

Correct option: A

\vec r = x\vec a + \frac{{\vec a \times \vec b}}{{{{\left| {\vec a} \right|}^2}}}

Option A (

\vec r = x\vec a + \frac{{\vec a \times \vec b}}{{{{\left| {\vec a} \right|}^2}}})

is correct because it satisfies the given equation

\vec r \times \vec a = \vec b

when vectors a and b are perpendicular.

179 Easy

179 of 195

Application +4 | -1

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

are any two vwctors, then

\left( {2\vec a + 3\mathbf{\vec b}} \right) \times \left( {5\mathbf{\vec a} + 7\mathbf{\vec b}} \right) + \mathbf{\vec a} \times \mathbf{\vec b}

is equal to

View answer and explanation

Correct option: A

{\vec 0}

Option A (

{\vec 0}

) is correct because the cross product simplifies to zero after expanding and canceling terms.

180 Easy

180 of 195

Application +4 | -1

A vector perpendicular to both of the vectors

\mathbf{i}+\mathbf{j}+\mathbf{k}

and

\mathbf{i}+\mathbf{j}

View answer and explanation

Correct option: C

c(\mathbf{i}-\mathbf{j}), c

is a scalar

Option C (

c(\mathbf{i}-\mathbf{j}), c

is a scalar) is correct because the cross product of the two vectors results in a vector of the form

c(\mathbf{i}-\mathbf{j})

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