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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 181-195 of 195 questions

181 Easy

181 of 195

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.

182 Easy

182 of 195

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.

183 Easy

183 of 195

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.

184 Easy

184 of 195

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

185 Medium

185 of 195

Application +4 | -1

\vec{a}=\hat{i}+\hat{j}+\hat{k}

and

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

, then the vector

\vec{c}

such that

\vec{a} \cdot \vec{c}=2

and

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

View answer and explanation

Correct option: B

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

Option B (

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

) is correct. Using the vector triple product identity and the given conditions, we can solve for

\vec{c}

186 Easy

186 of 195

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.

187 Easy

187 of 195

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.

188 Easy

188 of 195

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.

189 Easy

189 of 195

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.

190 Easy

190 of 195

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.

191 Easy

191 of 195

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.

192 Easy

192 of 195

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.

193 Medium

193 of 195

Application +4 | -1

|\vec{a}|=1,|\vec{b}|=2, \vec{a} \cdot \vec{b}=4, \vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}

then

\vec{b} \cdot \vec{c}

equals

View answer and explanation

Correct option: B

-12

Option B (-12) is correct because it is the value of the dot product of vectors b and c, calculated using the given conditions and vector algebra properties.

194 Easy

194 of 195

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.

195 Easy

195 of 195

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.

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