JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

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Application +4 | -1

Let

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

and a vector

\vec{b}

be such that

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

and

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

. Then

|\vec{b}|

equals?

View answer and explanation

Correct option: A

\sqrt{\frac{11}{3}}

Option A (

\sqrt{\frac{11}{3}}

) is correct because it is the magnitude of vector

\vec{b}

calculated using the given cross product and dot product conditions.

More Questions from CROSS PRODUCT

Practice All Chapter MCQs →

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

\vec{a}, \vec{b}

\lambda

a = 3\hat i + 2\hat j + x\hat k\mathrm{ }\,\mathrm{\,\,and}\,\,\mathrm{ }b = \hat i - \hat j + \hat k

x.

\left| {a \times b} \right| = r

a = 2\widehat i + \widehat j - 2\widehat k{\text{ }}\,\,{\text{and }}\,\,b = \widehat i + \widehat j.

a \cdot c = |c|,\,\,|c - a\mid = 2\sqrt 2

a \times b

c

30^\circ ,

|(a \times b) \times c|

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

(\vec{A} \times \vec{B}) \times \vec{C}

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}

O

S

O A

O C

\frac{\text { area of the quadrilateral } O A B C}{\text { area of } S}