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 · 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 121-149 of 149 questions

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121 Easy

121 of 149

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.

122 Easy

122 of 149

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.

123 Easy

123 of 149

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)

124 Easy

124 of 149

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.

125 Easy

125 of 149

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.

126 Easy

126 of 149

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.

127 Easy

127 of 149

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.

128 Easy

128 of 149

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.

129 Easy

129 of 149

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.

130 Easy

130 of 149

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.

131 Easy

131 of 149

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}

132 Easy

132 of 149

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.

133 Easy

133 of 149

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.

134 Easy

134 of 149

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

135 Easy

135 of 149

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.

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136 Easy

136 of 149

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})

137 Easy

137 of 149

Analyzing +4 | -1

\vec r \times \vec a = \vec b \times \vec a\;;\vec r \times \vec b = \vec a \times \vec b\;;\vec a \ne 0\;;\vec b \ne 0\;;\vec a \ne \lambda \;\vec b,\;\vec a

is not perpendicular to

{\vec b}

, then

\vec r =

View answer and explanation

Correct option: B

\vec a + \vec b

Option B (

\vec a + \vec b

) is correct because it satisfies both given vector equations.

138 Easy

138 of 149

Application +4 | -1

The area of a parallelogram whose two adjacent sides are represented by the vector

3 \mathbf{i}-\mathbf{k}

and

\mathbf{i}+2 \mathbf{j}

View answer and explanation

Correct option: C

\sqrt{41}

Option C (

\sqrt{41}

) is correct because the area of the parallelogram is found by taking the magnitude of the cross product of the two adjacent side vectors.

139 Easy

139 of 149

Application +4 | -1

\mathbf{u}=\mathbf{i} \times(\mathbf{a} \times \mathbf{i})+\mathbf{j} \times(\mathbf{a} \times \mathbf{j})+\mathbf{k} \times(\mathbf{a} \times \mathbf{k})

, then

View answer and explanation

Correct option: C

u=2 a

Option C (

u=2 a

) is correct because the expression simplifies to 2a using vector triple product identities.

140 Easy

140 of 149

Application +4 | -1

{\vec a}

is a vector perpendicular to the vectors

\vec b = \hat i + 2\hat j + 3\hat k

and

c = - 2\hat i + 4\hat j + \hat k

and satisfies the condition

\vec a.\left( {\hat i - 2\hat j + \hat k} \right) = - 6

, then

\vec a =

View answer and explanation

Correct option: A

5\hat i + \frac{7}{2}\hat j - 4\hat k

Option A (

5\hat i + \frac{7}{2}\hat j - 4\hat k

) is correct because it is the vector that satisfies the given conditions of being perpendicular to both

\vec b

and

\vec c

and having a specific dot product with

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

141 Easy

141 of 149

Application +4 | -1

The area of the parallelogram whose diagonals are

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

and

2 i-6 j+8 k

View answer and explanation

Correct option: A

5 \sqrt{3}

Option A (

5 \sqrt{{3}}

) is correct because it is the result of calculating half the magnitude of the cross product of the diagonals.

142 Easy

142 of 149

Application +4 | -1

If the vectors

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

represents the diagonals of a parallelogram, then its area will be

View answer and explanation

Correct option: B

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

Option B (

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

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

143 Easy

143 of 149

Application +4 | -1

The area of triangle having verities as

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

View answer and explanation

Correct option: B

0 sq units

Option B (0 sq units) is correct because the area of the triangle is 0, indicating that the points are collinear.

144 Easy

144 of 149

Application +4 | -1

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

and

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

then

|{a} \cdot {b}|

is equal to

View answer and explanation

Correct option: B

Option B (6) is correct because the magnitude of the dot product is calculated using the relationship between the dot product, cross product, and magnitudes of the vectors.

145 Easy

145 of 149

Application +4 | -1

If the distance from the point

P(1,1,1)

to the line passing through the points

Q(0,6,8)

and

R(-1,4,7)

is expressed in the form

\sqrt{p / q}

, where

p

and

q

are co-prime, then the value of

\frac{(p+q)(p+q-1)}{2}

is equal to

View answer and explanation

Correct option: A

4950

Option A (4950) is correct because the distance from the point to the line is calculated correctly, and then the value of

\frac{(p+q)(p+q-1)}{2}

is found.

146 Easy

146 of 149

Application +4 | -1

Given

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

and

{c}

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

. A unit vector perpendicular to both

{a}+{b}

and

{b}+{c}

View answer and explanation

Correct option: C

{k}

Option C (k) is correct because the cross product of (a+b) and (b+c) results in a vector parallel to k, and normalizing gives k.

147 Easy

147 of 149

Application +4 | -1

A force of magnitude 6 acts along the vector

(9,6,-2)

and passes through a point

A(4,-1,-7)

. Then moment of force about the point

O(1,-3,2)

View answer and explanation

Correct option: A

\frac{150}{11}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})

Option A (

\frac{150}{11}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})

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

148 Easy

148 of 149

Application +4 | -1

\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k} \quad|\vec{b}|^2=6

and angle between

\vec{a}

and

\vec{b}

\frac{\pi}{4}

. If

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

then

\left(\alpha^2+\beta^2\right)|\vec{a} \times \vec{b}|^2

View answer and explanation

Correct option: D

Option D (18) is correct because it is calculated by first finding

\alpha^2 + \beta^2

using the dot product and then using the cross product magnitude formula.

149 Easy

149 of 149

Application +4 | -1

The area of the parallelogram whose diagonals are

\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: B

5 \sqrt{3}

Option B (

5 \sqrt{3}

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

Revision notes and key takeaways

Review CROSS PRODUCT in your textbook or class notes, then use this set to check your understanding.

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