If ${\bf{\vec a}} = \left( {\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}} \right),\;{\bf{\vec a}}\cdot{\bf{\vec b}} = 1\;$ and ${\bf{\vec a}} \times {\bf{\vec b}} = \widehat {\bf{j}} - {\bf{\hat k}},$ then ${{\bf{\vec b}}}$ is

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

$2\widehat {\bf{j}} - {\bf{\hat k}}$

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JEE • MATHEMATICS • VECTOR TRIPLE PRODUCT

Vector triple product practice · Medium

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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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1 Medium

1 of 14

Analyzing +4 | -1

Let

\lambda \in \mathrm{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}

((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}

then

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

is equal to

View answer and explanation

Correct option: C

140

Option C (140) is correct because it is obtained by simplifying the vector expression, finding the value of lambda, and then calculating the magnitude squared.

2 Medium

2 of 14

Analyzing +4 | -1

Let

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

such that

\vec{a} \cdot \vec{b}=1

and

\vec{a} \times \vec{b}=\hat{i}-\hat{j}

then

\vec{a}-6 \vec{b}

equals

View answer and explanation

Correct option: A

3(\hat{i}+\hat{j}+\hat{k})

Option A (3(\hat{i}+\hat{j}+\hat{k})) is correct because it is the result of the vector operation and simplification as shown in the solution.

3 Medium

3 of 14

Application +4 | -1

\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}

are coplanar vectors, then

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

View answer and explanation

Correct option: D

Option D (0) is correct because if vectors

\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}

are coplanar, the scalar triple products

[\,\mathbf{a} \,\mathbf{b} \,\mathbf{d}

and

[\,\mathbf{a} \,\mathbf{b} \,\mathbf{c}

are zero, leading to

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

4 Medium

4 of 14

Analyzing +4 | -1

Let

a = \widehat i - \widehat j,\,\,\,b = \widehat j - \widehat k

and

c = \widehat k - \widehat i.

If d is a unit vectors such that

a \cdot d = 0 = (bcd),

then d is equal to

View answer and explanation

Correct option: A

\pm \frac{{(\widehat i + \widehat j - 2\widehat k)}}{{\sqrt 6 }}

The correct answer is A (

\pm \frac{{(\widehat i + \widehat j - 2\widehat k)}}{{{\sqrt 6 }}}

) because it is derived by finding a vector perpendicular to a and coplanar with b and c, then normalizing it to obtain the unit vector d.

5 Medium

5 of 14

Understanding +4 | -1

The value of

[\mathbf{a}-\mathbf{b} \mathbf{b}-\mathbf{c} \mathbf{c}-\mathbf{a}]

, where

|\mathbf{a}|=1,|\mathbf{b}|=5

and

|c|=3

View answer and explanation

Correct option: A

\mathrm{O}

Option A is correct because the scalar triple product of (a-b), (b-c), and (c-a) is always zero.

6 Medium

6 of 14

Application +4 | -1

\vec{a}

is perpendicular to both

\vec{b}

and

\vec{c}

such that

|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=4

and the angle between

\vec{b}

and

\vec{c}

\frac{2 \pi}{3}

, then

|[\vec{a} \vec{b} \vec{c}]|

is equal to

View answer and explanation

Correct option: C

12 \sqrt{3}

Option C (

12 \sqrt{3}

) is correct because the scalar triple product

|[\vec{a} \vec{b} \vec{c}]|

is equal to

|\vec{a}| |\vec{b}| |\vec{c}| \sin \theta

, where

\theta

is the angle between

\vec{b}

and

\vec{c}

, given that

\vec{a}

is perpendicular to both

\vec{b}

and

\vec{c}

7 Medium

7 of 14

Analyzing +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be three non zero vectors such that

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

and

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

. If

\vec{d}

be a vector such that

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

, then

(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})

is equal to

View answer and explanation

Correct option: A

\frac{1}{4}

Option A (

\frac{1}{4}

) is correct because by using vector triple product and given conditions, we can find the value of the expression.

8 Medium

8 of 14

Application +4 | -1

\mathbf{\vec a} = \left( {\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}} \right),\;\mathbf{\vec a}\cdot\mathbf{\vec b} = 1\;

and

\mathbf{\vec a} \times \mathbf{\vec b} = \widehat \mathbf{j} - \mathbf{\hat k},

then

{\mathbf{\vec b}}

View answer and explanation

Correct option: C

\widehat \mathbf{i}

Option C (

\widehat \mathbf{i}

) is the correct answer because it is derived by using the vector triple product identity and solving for

\mathbf{\vec b}

9 Medium

9 of 14

Application +4 | -1

A vector

c

perpendicular to the vectors

2 {i}+3 {j}-{k}

and

{i}-2 {j}+3 {k}

satisfying

{c}(2 {i}-{j}+{k})=-6

View answer and explanation

Correct option: C

-3 {i}+3 {j}+3 {k}

Option C (-3i + 3j + 3k) is correct because it is perpendicular to both given vectors and satisfies the dot product condition.

10 Medium

10 of 14

Application +4 | -1

\vec{a}=4 \hat{i}+6 \hat{j}

and

\vec{b}=3 \hat{j}+4 \hat{k}

, the component of

\vec{a}

along

\vec{b}

\mathrm{K}

then the value of

[\mathrm{K}]

(where [.] represents G.I.F.)

View answer and explanation

Correct option: A

Option A (3) is correct because the component of vector a along vector b is calculated using the formula (a.b)/|b| and then taking the greatest integer function.

11 Medium

11 of 14

Application +4 | -1

Given

\vec{A}=2 \hat{i}+3 \hat{j}+6 \hat{k}, \vec{B}=\hat{i}+\hat{j}-2 \hat{k}

and

\vec{C}=\hat{i}+2 \hat{j}+\hat{k}

. Compute the value of

|\vec{A} \times[\vec{A} \times(\vec{A} \times \vec{B})] \cdot \vec{C}|

View answer and explanation

Correct option: A

343

Option A (343) is correct because the scalar triple product simplifies to

|-|\vec{{A}}|^2[\vec{{A}} \vec{{B}} \vec{{C}}] |

and the calculation yields 343.

12 Medium

12 of 14

Analyzing +4 | -1

\vec{n}

is a vector,

\vec{a} \neq 0, \vec{b} \neq 0

. If

\vec{n} \perp \vec{c}, \vec{a}=\alpha \vec{b}-\hat{n}

and

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

, then the value of

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

equals: (where

\hat n

represents unit vector in direction of

\vec{n}

)

View answer and explanation

Correct option: C

Option C (12) is correct because by simplifying the vector triple product and using the given conditions, the magnitude of the resulting vector is 12.

13 Medium

13 of 14

Application +4 | -1

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

and

\left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right) \times \mathbf{\vec c} = \mathrm{\lambda \;}\mathbf{\vec a} + \mathrm{\mu }\;\mathbf{\vec b},

then

\lambda + \mu

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the value of λ + μ is 0, derived from calculating the vector triple product and equating coefficients.

14 Medium

14 of 14

Analyzing +4 | -1

\bar{u}, \bar{m}, \bar{r}

be three mutually perpendicular vectors with same magnitude. If

\bar{e}

satisfies the relation

\bar{u} \times\{(\bar{e}-\bar{m}) \times \bar{u}\}+\bar{m} \times\{(\bar{e}-\bar{r}) \times \bar{m}\}+\bar{r} \times\{(\bar{e}-\bar{u}) \times \bar{r}\}=\overline{0}

then

\bar{e}=

View answer and explanation

Correct option: B

\frac{1}{2}(\bar{u}+\bar{m}+\bar{r})

Option B (

\frac{1}{2}(\bar{u}+\bar{m}+\bar{r})

) is correct because it satisfies the given vector equation after simplification using vector triple product.

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Key Takeaways

✓ This set contains 14 single-correct-answer MCQs on VECTOR TRIPLE PRODUCT.
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